More About Chemical Equilibria

Size: px
Start display at page:

Download "More About Chemical Equilibria"

Transcription

1 1 More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 15 & 16 1 Objectives Chapter 15 Define the Common Ion Effect (15.1) Define buffer and show how a buffer controls ph of a solution ( ) Identify and Evaluate titration curves (15.4) Use of Indicators (15.5) 2 Stomach Acidity & Acid-Base Reactions 3 Acid-Base Reactions Strong acid + strong base HCl + NaOH Strong acid + weak base HCl + NH 3 Weak acid + strong base HOAc + NaOH Weak acid + weak base HOAc + NH 3 What is relative ph before, during, & after reaction? Need to study: a) Common ion effect and buffers b) Titrations 4 The Common Ion Effect Section ph of Aqueous NH 3 6 QUESTION: What is the effect on the ph of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) QUESTION: What is the effect on the ph of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) Here we are adding NH 4+, an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left (1), right (2), no change (3). The ph will go up (1), down (2), no change (3). NH 4+ is an acid! Let us first calculate the ph of a 0.25 M NH 3 solution. [NH 3 ] [NH 4+ ] [OH - ] initial change -x +x +x equilib x x x

2 2 ph of Aqueous NH 3 7 ph of NH 3 /NH 4+ Mixture 8 QUESTION: What is the effect on the ph of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) K b = 1.8 x 10-5 = [NH + 4 ][OH - ] x 2 = [NH 3 ] x Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = M This gives poh = 2.67 and so ph = Problem: What is the ph of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) We expect that the ph will decline on adding NH 4 Cl. Let s test that! [NH 3 ] [NH 4+ ] [OH - ] initial change equilib = for 0.25 M NH 3 ph of NH 3 /NH 4+ Mixture 9 ph of NH 3 /NH 4+ Mixture 10 Problem: What is the ph of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) We expect that the ph will decline on adding NH 4 Cl. Let s test that! [NH 3 ] [NH 4+ ] [OH - ] initial change -x +x +x equilib x x x Problem: What is the ph of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O NH 4+ (aq) + OH - (aq) K b = 1.8 x 10-5 = [NH + 4 ][OH - ] x( x) = [NH 3 ] x Assuming x is very small, [OH - ] = x = (0.25 / 0.10)(K b ) = 4.5 x 10-5 M This gives poh = 4.35 and ph = 9.65 ph drops from to 9.65 on adding a common ion Common Ion Effect The addition of a common ion in a base such as ammonia, limits the ionization of the base and the production of [OH - ] (Le Châtelier s Principle). B(aq) + H 2 O(l) HB + (aq) + OH - (aq) For acid, addition of conjugate base from salt limit ionization of acid, which limits [H 3 O + ] HA(aq) + H 2 O(l) A - (aq) + H 3 O + (aq) Doesn t work if the conjugate base is from a strong acid because the conjugate base is too weak. In other words, adding Cl - to above does not change the ph because the Cl- comes from strong acid HCl, which has a very large Ka. HCl + H 2 O H 3 O + + Cl - 11 Common Ion Effect Practice Calculate the ph of 0.30 M formic acid (HCO 2 H) in water and with enough sodium formate, NaHCO 2, to make the solution 0.10 M in the salt. The K a of formic acid is 1.8x10-4. HCO 2 H(aq) + H 2 O(l) HCO (aq) + H 3 O + 12

3 3 Common Ion Effect Practice Calculate the ph of 0.30 M formic acid (HCO 2 H) in water and with enough sodium formate, NaHCO 2, to make the solution 0.10 M in the salt. The K a of formic acid is 1.8x10-4. HCO 2 H(aq) + H 2 O(l) HCO (aq) + H 3 O + With salt: 13 Buffered Solutions Section 15.2 HCl is added to pure water. 14 HCl is added to a solution of a weak acid H 2 PO 4- and its conjugate base HPO 4. Buffer Solutions A buffer solution is a special case of the common ion effect. The function of a buffer is to resist changes in the ph of a solution, either by H + or by OH -. Buffer Composition Weak Acid + Conj. Base HOAc + OAc - H 2 PO HPO 4 NH NH 3 15 Buffer Solutions Consider HOAc/OAc - to see how buffers work ACID USES UP ADDED OH - We know that OAc - + H 2 O HOAc + OH - has K b = 5.6 x Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9 K reverse is VERY LARGE, so HOAc completely consumes OH -!!!! 16 Buffer Solutions 17 Buffer Solutions 18 Consider HOAc/OAc - to see how buffers work CONJ. BASE USES UP ADDED H + HOAc + H 2 O OAc - + H 3 O + has K a = 1.8 x 10-5 Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4 K reverse is VERY LARGE, so OAc - completely consumes H 3 O +! Problem: What is the ph of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x M HOAc has ph = 2.45 The ph of the buffer will have 1. ph < ph > ph = 2.45

4 4 Buffer Solutions 19 Buffer Solutions 20 Problem: What is the ph of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10-5 initial change equilib [HOAc] [OAc - ] [H 3 O + ] x +x +x x x x Problem: What is the ph of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10-5 [HOAc] [OAc - ] [H 3 O + ] equilib x x x Assuming that x << and 0.600, we have K a = 1.8 x 10-5 = [H 3 O+ ](0.600) [H 3 O + ] = 2.1 x 10-5 and ph = 4.68 Buffer Solutions Notice that the expression for calculating the H + conc. of the buffer is 21 Henderson-Hasselbalch Equation or 22 [H 3 O + Orig. conc. of HOAc ] = x K Orig. conc. of OAc - a Take the negative log of both sides of this equation [H 3 O + [Acid] ] = [Conj. base] x K a Notice that the [H 3 O + ] depends on: 1. the K a of the acid 2. the ratio of the [acid] and [conjugate base]. Note: HA and A - are generic terms for acid and conjugate base, respectively. The ph is determined largely by the pk a of the acid and then adjusted by the ratio of conjugate base and acid. Henderson-Hasselbalch Equation Works the same for base! B(aq) + H 2 O(l) HB + (aq) + OH - (aq) Solve for [OH - ] = Take the negative log of both sides of this equation to get: b The poh (and ph) is determined largely by the pk b of the base and then adjusted by the ratio of the conjugate acid and base. 23 Buffer Calculation Example (Acid/Conj. Base) What is the ph of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate (NaC 3 H 5 O 3 )? K a for lactic acid is

5 5 Buffer Calculation Example HC 3 H 5 O 3 + H 2 O C 3 H 5 O 3- + H 3 O + I 0.12 M E 0.1x x x x( x) = 1.4x x We ll make usual assumption that x is small. x = 0.12(1.40 x 10-4 ) = 1.68 x 10-4 = [H 3 O + ] 0.10 So the ph = Buffer Calculation using the Henderson Hasselbalch Equation HC 3 H 5 O 3 + H 2 O C 3 H 5 O 3- + H 3 O + acid conj base ph = log ( )+ log (0.10) (0.12) ph = ( 0.08) 26 ph = 3.77 (same as before!!) Buffer Calculation using the Henderson Hasselbalch Equation Calculate the ph of a buffer that has is M ammonia and M ammonium chloride. The K b for ammonia is 1.8 x NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) [base] [conj. acid] Buffer Practice 1 Calculate the ph of a solution that is 0.50 M HC 2 H 3 O 2 (acetic acid) and 0.25 M NaC 2 H 3 O 2 (K a = 1.8 x 10-5 ). Use H-H Eqn. 28 poh = log (1.8 x 10-5 ) + log (0.045) (0.080) poh = ( 0.25) poh = 4.49 ph = = 9.51 Buffer Practice 2 Calculate the concentration of sodium benzoate (NaC 7 H 5 O 2 ) that must be present in a 0.20 M solution of benzoic acid (HC 7 H 5 O 2 ) to produce a ph of (Benzoic acid K a = 6.3 x 10-5 ). Use Henderson-Hasselbalch equation. HC 7 H 5 O 2 + H 2 O H 3 O + + C 7 H 5 O 2-29 Adding an Acid to a Buffer (How a Buffer Maintains ph) Problem: What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (before HCl, ph = 7.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) (See slide 20) Solution to Part (a) Calc. [HCl] after adding 1.00 ml of HCl to 1.00 L of water M 1 V 1 = M 2 V 2 (1.00 M)(1.0 ml) = M 2 (1001 ml) M 2 = 9.99 x 10-4 M = [H 3 O + ] ph =

6 6 Adding an Acid to a Buffer 31 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (after HCl, ph = 3.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph before = 4.68) 32 To play movie you must be in Slide Show Mode Solution to Part (b) Step 1 do the stoichiometry H 3 O + (from HCl) + OAc - (from buffer) HOAc (from buffer) The reaction occurs completely because K is very large. The stronger acid (H 3 O + ) will react with the conj. base of the weak acid (OAc - ) to make the weak acid (HOAc). Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (ph = 3.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) Solution to Part (b): Step 1 Stoichiometry [H 3 O + ] + [OAc - ] [HOAc] Before rxn Change After rxn Equil [] s mol mol mol mol mol mol L mol 1.00l L mol/l mol/l 33 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (ph = 3.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - [HOAc] [H 3 O + ] [OAc - ] Initial (M) mol/l mol/l Change (M) Equil. (M) -x x +x x +x x (re-established) 34 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (ph = 3.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (ph = 4.68) 35 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (ph = 3.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (ph = 4.68) 36 Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - [HOAc] [H 3 O + ] [OAc - ] Equilibrium x x x Because [H 3 O + ] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to and Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - ph = = 4.67 The ph has not changed much on adding HCl to the buffer!

7 7 Adding a Base to a Buffer Problem: What is the ph when 1.00 ml of 1.00 M NaOH is added to a) 1.00 L of pure water (before NaOH, ph = 7.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) (See slide 20) Solution to Part (a) Calc. [OH - ] after adding 1.00 ml of NaOH to 1.00 L of water M 1 V 1 = M 2 V 2 (1.00 M)(1.0 ml) = M 2 (1001 ml) M 2 = 1.00 x 10-3 M = [OH - ] poh = 3.00 ph = Adding an Base to a Buffer What is the ph when 1.00 ml of 1.00 M NaOH is added to a) 1.00 L of pure water (after NaOH, ph = 11.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph before = 4.68) Solution to Part (b) Step 1 do the stoichiometry OH - (from NaOH) + HOAc (from buffer) OAc - (from buffer and rxn) The reaction occurs completely because K is very large. The strong base (OH - ) will react with weak acid (HOAc) to make the conj. weak base (OAc - ). 38 Adding a Base to a Buffer What is the ph when 1.00 ml of 1.00 M NaOH is added to a) 1.00 L of pure water (ph = 11.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) Solution to Part (b): Step 1 Stoichiometry [OH - ] + [HOAc] [OAc - ] Before rxn Change After rxn Equil. [] s mol mol mol mol mol mol L mol L mol/l mol/l 39 Adding a Base to a Buffer What is the ph when 1.00 ml of 1.00 M NaOH is added to a) 1.00 L of pure water (ph = 11.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc - ] = M (ph = 4.68) Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - Initial(M) Change (M) Equil. (M) (re-established) [HOAc] [H 3 O + ] [OAc - ] mol/l mol/l -x +x +x x x x 40 Adding a Base to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to a) 1.00 L of pure water (ph = 11.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (ph = 4.68, poh = 9.32) Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - [HOAc] [H 3 O + ] [OAc - ] Equilibrium x x x Because [H 3 O + ] = 2.1 x 10-5 M BEFORE adding NaOH, we again neglect x relative to and Adding a Base to a Buffer What is the ph when 1.00 ml of 1.00 M NaOH is added to a) 1.00 L of pure water (ph = 11.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (ph = 4.68) Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - ph = = 4.67 The ph has not changed much on adding NaOH to the buffer! 42

8 8 Preparing a Buffer You want to buffer a solution at ph = This means [H 3 O + ] = 10 -ph = = 5.0 x 10-5 M It is best to choose an acid such that [H 3 O + ] is about equal to K a (or ph pk a ). then you get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base. [H 3 O + [Acid] ] = [Conj. base] x K a 43 Preparing a Buffer You want to buffer a solution at ph = 4.30 or [H 3 O + ] = 5.0 x 10-5 M POSSIBLE ACIDS/CONJ BASE PAIRS K a HSO 4- / SO x 10-2 HOAc / OAc x 10-5 HCN / CN x Best choice is acetic acid / acetate. 44 Preparing a Buffer 45 Preparing a Buffer 46 You want to buffer a solution at ph = 4.30 or [H 3 O + ] = 5.0 x 10-5 M [H 3 O + ] = 5.0 x 10-5 = [HOAc] (1.8 x 10-5 ) [OAc - ] Solve for [HOAc]/[OAc - ] ratio = Therefore, if you use mol of NaOAc and mol of HOAc, you will have ph = A final point CONCENTRATION of the acid and conjugate base are not as important as the RATIO OF THE NUMBER OF MOLES of each. Result: diluting a buffer solution does not change its ph Commercial Buffers 47 Preparing a Buffer Buffer prepared from 48 The solid acid and conjugate base in the packet are mixed with water to give the specified ph. Note that the quantity of water does not affect the ph of the buffer. 8.4 g NaHCO 3 (8.4g/84 g/mol) weak acid 16.0 g Na 2 CO 3 (16.0/106 g/mol) conjugate base HCO 3- + H 2 O H 3 O + + CO 3 What is the ph? HCO - 3 pka = 10.3 ph = log (1.51/1.0)=10.5

9 9 Buffer Example Phosphate buffers are used a lot in labs to simulate blood ph of As with any other buffer, you need an acid and its conjugate base. In this case the acid is NaH 2 PO 4 and the base is Na 2 HPO 4. So the equilibrium expression is: H 2 PO 4- (aq) + H 2 O(l) HPO 4 (aq) + H 3 O + (aq) A) If the pka of H 2 PO 4- is 7.21, what should the [HPO 4 ]/[H 2 PO 4- ] ratio be? B) If you weigh 6.0 g of NaH 2 PO 4 and make 500 ml of solution, how much Na 2 HPO 4 do you need add to make your ph 7.40 buffer? (assume the Na 2 HPO 4 doesn t affect the volume) 49 Buffer Example Part A H 2 PO 4- (aq) + H 2 O(l) HPO 4 (aq) + H 3 O + (aq) Can use H-H: ph = pka + log[hpo 4 ]/[H 2 PO 4- ] 7.40 = log[hpo 4 ]/[H 2 PO 4- ] log[hpo 4 ]/[H 2 PO 4- ] = 0.19 [HPO 4 ]/[H 2 PO 4- ] = = 1.55 Part B 6.0g NaH 2 PO 4 = mol in L = 0.10 M 120g/mol [HPO 4 ] = 1.55[H 2 PO 4- ] = M HPO 4 =0.155 mol/l * L * 142 g Na 2 HPO 4 /mol = 11.0 g Na 2 HPO 4 50 Preparing a buffer Practice Using the acetic acid (molar mass 60 g/mol) and sodium acetate (molar mass: 82 g/mol) buffer from before, how much sodium acetate should you weigh to make 1.0 L of a ph 5.00 buffer? Assume [HOAc] is 1.0 M in the final 1.0 L of buffer. (pka = 4.74) 51 Buffering Capacity Chapter 15.3 The ph of a buffered solution is determined by the ratio [A - ]/[HA]. As long as the ratio doesn t change much the ph won t change much. The more concentrated these two are the more H + and OH - the solution will be able to absorb. Larger concentrations bigger buffer capacity. 52 Buffer Capacity Calculate the change in ph that occurs when mol of HCl is added to 1.0L of each of the following: 5.00 M HOAc and 5.00 M NaOAc M HOAc and M NaOAc Ka= 1.8x10-5 (pka = 4.74) Note: Since [OAc-]=[HOAc] for both cases, the initial ph is Large Buffer System H 3 O + + OAc - HOAc + H 2 O Stoichiometry Calculation: H 3 O + OAc - HOAc Before Reaction (moles) After Reaction (moles)

10 10 Calculate ph via Henderson- Hasselbalch Equations At Equilibrium: HOAc + H 2 O H 3 O + + OAc X 4.99 [base] ph = pka + [acid] ph = -log (1.8 x 10-5) + log[ 4.99/5.01] ph = Since pka = 4.74, solution has gotten slightly more acidic 55 Small Buffer System H 3 O + + OAc - HOAc(aq) + H 2 O Stoichiometry Calculation: H + OAc - HOAc Before Reaction (moles) After Reaction (moles) Calculating ph for Small Buffer System HOAc H 3 O + + OAc - In table form (Equilibrium Re-established): [HOAc], M [H 3 O + ], M [OAc ], M Initially Change -x +x +x At Equilibrium 0.06-x x x 57 Calculate ph via Henderson- Hasselbalch Equations At Equilibrium: HOAc + H 2 O H 3 O + + OAc x X 0.04-x Ignore x s relative to 0.06 and 0.04 since they will be small. Calculate via H-H eqn: [base] ph = pka + [acid] ph = -log (1.8 x 10-5) + log[ 0.04/0.06] ph = = 4.56 So, solution is considerably more acidic than the ph 4.74 of the big buffer system. 58 Mixing Acids and Bases (Stoichiometry & Equilibrium) Chapter 15.4 If you mix 50.0 ml of 0.10 M HCl and 50.0 ml of 0.10 NaOH, what is the ph? Since it is a strong acid/strong base, what happens stoichiometrically? H + + OH - H 2 O(l) I(moles) After reaction Since you consume all acid and base and make water, what is the ph? 7 of course. 59 Mixing Acids and Bases (Stoichiometry & Equilibrium) Now, what if you mix 50.0 ml of 0.10 M HCl and 20.0 ml of 0.10 NaOH, what is the ph? H + + OH - H 2 O(l) I(moles) After reaction (mol) Total volume of solution is 70.0 ml ( L) So the [H + ] = mol/ L = M ph = -log[0.0429] =

11 11 Mixing Acids and Bases (Stoichiometry & Equilibrium) Now mix 50.0 ml of 0.10 M HCl and 70.0 ml of 0.10 NaOH, what is the ph? H + + OH - H 2 O(l) I(moles) After reaction (mol) Total volume of solution is ml ( L) Since there was more OH - added than H +, all of the H + is consumed and there is excess OH -. [OH-] = 0.002/0.120 L = M OH - poh = 1.77 so ph = Mixing Acids and Bases (Stoichiometry & Equilibrium) If you mix 50.0 ml of 0.10 M HOAc and 50.0 ml of 0.10 NaOH, what is the ph? Since it is a weak acid/strong base, what happens stoichiometrically? HOAc + OH - H 2 O(l) + OAc - I(moles) After reaction Since HOAc is a weak acid and OH- is a strong base, the OH- will still consume (take that H + ) from the HOAc. (The Stoichiometry part.) 62 Mixing Acids and Bases (Stoichiometry & Equilibrium) If you mix 50.0 ml of 0.10 M HOAc and 50.0 ml of 0.10 NaOH, what is the ph? Now, equilibrium will be re-established. H 2 O(l) + OAc - HOAc + OH - Need Conc: mol/0.100l I (conc) C -x +x +x E x x x You can think of this part as when we calculated ph s of salts of weak acids. It doesn t matter how you get to this point, the calculation is still the same! 63 Mixing Acids and Bases (Stoichiometry & Equilibrium) If you mix 50.0 ml of 0.10 M HOAc and 50.0 ml of 0.10 NaOH, what is the ph? Now, equilibrium will be re-established. H 2 O(l) + OAc - HOAc + OH - E x x x K rxn =K b = K w /K a = 1x10-14 /1.8x10-5 = 5.56x = 5.56x10-10 Usual assumption that x is small compared to so x = poh = -log[5.27x10-6 ] = 5.28 so ph = Mixing Acids and Bases (Stoichiometry & Equilibrium) If you mix 50.0 ml of 0.10 M HOAc and 20.0 ml of 0.10 NaOH, what is the ph? Since it is a weak acid/strong base, what happens stoichiometrically? HOAc + OH - H 2 O(l) + OAc - I(moles) Reaction After reaction Mixing Acids and Bases (Stoichiometry & Equilibrium) If you mix 50.0 ml of 0.10 M HOAc and 20.0 ml of 0.10 NaOH, what is the ph? Now, equilibrium will be re-established. Since you have an acid and its conjugate base, this is a buffer and you can use H-H eqn. H 2 O(l) + HOAc OAc - + H 3 O + E x ph = pka + log [OAc - ]/[HOAc] = log(0.002)/(0.003) = log(0.667) = =

12 ph Titrant volume, ml Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the ph increases very slowly Additional NaOH is added. ph rises as equivalence point is approached. Additional NaOH is added. ph increases and then levels off as NaOH is added beyond the equivalence point. QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is ph at half-way point? What is the ph at equivalence point? 71 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at the equivalence point? HBz + NaOH Na + + Bz - + H 2 O 72 Benzoic acid + NaOH ph of solution of benzoic acid, a weak acid C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

13 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at the equivalence point? HBz + NaOH Na + + Bz - + H 2 O The ph of the final solution will be 1. Less than 7 2. Equal to 7 3. Greater than 7 The product of the titration of benzoic acid is the benzoate ion, Bz -. Bz - is the conjugate base of a weak acid. Therefore, solution is basic at equivalence point. Bz - + H 2 O HBz + OH - + K b = 1.6 x QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. Benzoic acid + NaOH ph at equivalence point is basic 75 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at the equivalence point? Strategy find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate ph. This is a two-step problem 1. stoichiometry of acid-base reaction 2. equilibrium calculation QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at the equivalence point? QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at the equivalence point? STOICHIOMETRY PORTION 1. Calc. moles of NaOH req d (0.100 L HBz)(0.025 M) = mol HBz This requires mol NaOH 2. Calc. volume of NaOH req d mol (1 L / mol) = L 25 ml of NaOH req d STOICHIOMETRY PORTION 25 ml of NaOH req d 3. Moles of Bz - produced = moles HBz = mol 4. Calc. conc. of Bz - There are mol of Bz - in a TOTAL SOLUTION VOLUME of 125 ml [Bz - ] = mol / L = M

14 14 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at equivalence point? Equivalence Point Most important species in solution is benzoate ion, Bz -, the weak conjugate base of benzoic acid, HBz. Bz - + H 2 O HBz + OH - K b = 1.6 x [Bz - ] [HBz] [OH - ] initial change - x +x +x equilib x x x 79 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at equivalence point? Equivalence Point Most important species in solution is benzoate ion, Bz -, the weak conjugate base of benzoic acid, HBz. Bz - + H 2 O HBz + OH - K b = 1.6 x K b = 1.6 x x 2 = x x = [OH - ] = 1.8 x 10-6 poh = 5.75 and ph = QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at half-way point? 81 QUESTION: You titrate 100. ml of a M solution of benzoic acid with M NaOH to the equivalence point. What is the ph at half-way point? 82 ph at half-way point? 1. < 7 2. = 7 3. > 7 Equivalence point ph = 8.25 ph at halfway point Equivalence point ph = 8.25 You titrate 100. ml of a M solution of benzoic acid with M NaOH. What is the ph at the half-way point? HBz + H 2 O H 3 O + + Bz - K a = 6.3 x 10-5 Both HBz and Bz - are present. This is a BUFFER! [H 3 O + ] = [HBz] x K [Bz - a ] At the half-way point, [HBz] = [Bz - ] Therefore, [H 3 O + ] = K a = 6.3 x 10-5 ph = 4.20 = pk a of the acid 83 You titrate 100. ml of a M solution of benzoic acid with M NaOH. What was the ph when 6.0 ml of When doing these types of problems, NaOH was added? start from initial moles and go to the point you are considering. OH - + HBz H 2 O + Bz - Stoichiometry Before rxn, moles After rxn, mole After rxn, [ ] mol/0.106L / Equilbrium: HBz + H 2 O H 3 O + + Bz - pka = 4.20 Equilibrium Now can use H-H equation: ph = log (0.0057/0.018) = = 3.70 Remember: this is essentially a buffer solution since HBz and Bz - are both in the solution. 84

15 15 You titrate 100. ml of a M solution of benzoic acid with M NaOH. Practice: What is the ph when 20.0 ml of NaOH are added? OH - + HBz H 2 O + Bz - Stoichiometry Before rxn, moles After rxn, mole After rxn, [ ] mol/0.120L / Equilbrium: HBz + H 2 O H 2 O + Bz - pka = 4.20 Equilibrium Now can use H-H equation: ph = log (0.017/0.0042) = = Acetic acid titrated with NaOH Weak acid titrated with a strong base Strong acid titrated with a strong base Weak diprotic acid (H 2 C 2 O 4 ) titrated with a strong base (NaOH) See Figure 18.6 See Figure 18.4 ph Titration of a 1. Strong acid with strong base? 2. Weak acid with strong base? 3. Strong base with weak acid? 4. Weak base with strong acid? 5. Weak base with weak acid 6. Weak acid with weak base? Weak base (NH 3 ) titrated with a strong acid (HCl) Volume of titrating reagent added -->

16 16 Acid-Base Indicators Chapter 15.5 Weak acids that change color when they become bases. Since the Indicator is a weak acid, it has a K a. End point - when the indicator changes color. An indicator changes colors at ph=pk a 1, pk a is the acid dissoc. constant for the indicator Want an indicator where pk a ph at equiv. pt. 91 Acid-Base Indicators 92 Indicators for Acid-Base Titrations 93 Titration of a Base with an Acid 94 The ph at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice. Since the ph change is large near the equivalence point, you want an indicator that is one color before and one color after. Solubility and Complex Ion Equilibria Chapter Objectives Chapter 16 Define equilibrium constant for insoluble salts (K sp ) (16.1) Manipulate solubility by common ion effect (16.1) Precipitation Reactions comparing Q & K of a reaction (16.2) Define Complex ion & show how complex ions affect solubility (16.3) 96 Lead(II) iodide

17 17 Types of Chemical Reactions EXCHANGE REACTIONS: AB + CD AD + CB Acid-base: CH 3 CO 2 H + NaOH NaCH 3 CO 2 + H 2 O Gas forming: CaCO HCl CaCl 2 + CO 2 (g) + H 2 O Precipitation: Pb(NO 3 ) KI PbI 2 (s) + 2 KNO 3 OXIDATION REDUCTION (Redox Ch. 18) 4 Fe + 3 O 2 2 Fe 2 O 3 Apply equilibrium principles to acid-base and precipitation reactions. 97 Analysis of Silver Group Ag + Pb Hg 2 AgCl PbCl 2 Hg 2 Cl 2 All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions. 98 Ag + Pb Hg 2 AgCl PbCl 2 Hg 2 Cl 2 Analysis of Silver Group 99 Ag + Pb Hg 2 AgCl PbCl 2 Hg 2 Cl 2 Analysis of Silver Group 100 Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED. AgCl(s) Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10-5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? [Cl - ] = [Ag + ] = 1.67 x 10-5 M Ag + Pb Hg 2 AgCl PbCl 2 Hg 2 Cl 2 Analysis of Silver Group 101 Ag + Pb Hg 2 AgCl PbCl 2 Hg 2 Cl 2 Analysis of Silver Group 102 AgCl(s) Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x 10-5 M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67 x 10-5 )(1.67 x 10-5 ) = 2.79 x AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 2.79 x Because this is the product of solubilities, we call it K sp = solubility product constant See Table 16.1 and Appendix A5.4 Note: The property solubility is in mol/l. (Can also be expressed as g/l or mg/l!) Solubility Product has no units.

18 Solubility Products In general, for the equilibrium for the reaction is expressed as M m A a (s) mm + (aq) + aa - (aq) Some Values of K sp The K sp equilibrium expression is.. K sp =[M + ] m [A - ] a The higher the K sp, the more soluble the salt is. Lead(II) Chloride 105 Solubility of Lead(II) Iodide 106 PbCl 2 (s) Pb (aq) + 2 Cl - (aq) K sp = 1.9 x 10-5 = [Pb ][Cl ] 2 Consider PbI 2 dissolving in water PbI 2 (s) Pb (aq) + 2 I - (aq) Calculate K sp if solubility = M Solution 1. Solubility = [Pb ] = 1.30 x 10-3 M [I - ] =? [I - ] = 2 x [Pb ] = 2.60 x 10-3 M Solubility of Lead(II) Iodide 107 Solubility of Lead(II) Iodide 108 Consider PbI 2 dissolving in water PbI 2 (s) Pb (aq) + 2 I - (aq) Calculate K sp if solubility = M Solution 2. K sp = [Pb ] [I - ] 2 = [Pb ] {2 [Pb ]} 2 K sp = 4 [Pb ] 3 = 4 (solubility) 3 K sp = 4 (1.30 x 10-3 ) 3 = 8.79 x 10-9 Caveat 3. The value we just calculated by K sp = 4 (1.30 x 10-3 ) 3 = 8.79 x 10-9 solubility gives a ball-park value. The actual K sp of PbI 2 is 9.8 x Note: Calculating K sp from solubility gives approximate values that can be off by a few OOMs due to ionic interactions.

19 19 Solubility/K sp Practice 1 Calculate the K sp if the solubility of CaC 2 O 4 is 6.1 mg/l. Note units here. What should they be in order to determine K sp? (Molar Mass = g/mol) CaC 2 O 4 (s) Ca (aq) + C 2 O 4 (aq) 109 Solubility/K sp Practice 2 Calculate the solubility of CaF 2 in g/l if the K sp is 3.9x (Molar Mass = 78.1 g/mol) 110 The Common Ion Effect Adding an ion common to an equilibrium causes the equilibrium to shift back to reactant. 111 Common Ion Effect PbCl 2 (s) Pb (aq) + 2 Cl - (aq) K sp = 1.9 x The Common Ion Effect 114 (a) BaSO 4 is a common mineral, appearing a white powder or colorless crystals. Barium Sulfate K sp = 1.1 x (b) BaSO 4 is opaque to x-rays. Drinking a BaSO 4 cocktail enables a physician to exam the intestines. Calculate the solubility of BaSO 4 in (a) pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x BaSO 4 (s) Ba (aq) + SO 4 (aq) Solution Solubility in pure water = [Ba ] = [SO 4 ] = x K sp = [Ba ] [SO 4 ] = x 2 x = (K sp ) 1/2 = 1.1 x 10-5 M Solubility in pure water = 1.1 x 10-5 mol/l

20 20 The Common Ion Effect 115 The Common Ion Effect 116 Calculate the solubility of BaSO 4 in (a) pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x BaSO 4 (s) Ba (aq) + SO 4 (aq) Calculate the solubility of BaSO 4 in (a) pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x BaSO 4 (s) Ba (aq) + SO 4 (aq) Solution Solubility in pure water = 1.1 x 10-5 mol/l. Now dissolve BaSO 4 in water already containing M Ba. Which way will the common ion shift the equilibrium? Will solubility of BaSO 4 be less than or greater than in pure water? Solution initial change equilib. [Ba ] [SO 4 ] y + y y y The Common Ion Effect Calculate the solubility of BaSO 4 in (a) pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x BaSO 4 (s) Ba (aq) + SO 4 (aq) Solution K sp = [Ba ] [SO 4 ] = ( y) (y) Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means y is about equal to Therefore, K sp = 1.1 x = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of added Ba ion. 117 The Common Ion Effect Calculate the solubility of BaSO 4 in (a) pure water and (b) in M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x BaSO 4 (s) Ba (aq) + SO 4 (aq) SUMMARY Solubility in pure water = x = 1.1 x 10-5 M Solubility in presence of added Ba = 1.1 x 10-8 M Le Chatelier s Principle is followed! 118 Common Ion Effect Practice Do you expect the solubility of AgI to be greater in pure water or in M AgNO 3? Explain your answer. Ksp(AgI) = 8.5 x AgI(s) Ag + (aq) + I - (aq) 119 Factors Affecting Solubility - ph ph If a substance has a basic anion (like OH - ), it is more soluble in a more acidic solution. Substances with acidic cations (Al 3+ ) are more soluble in more basic solutions. 120

21 21 Solubility in more Acidic Solution example: solid Mg(OH) 2 in water Equilibrium: Mg(OH) 2 Mg + 2OH - For K sp = 1.8 x 10-11, calculate ph: 1.8 x = x(2x) 2 x = 1.65 x 10-4 M [solubility of Mg(OH) 2 ] = [Mg ] in solution [OH - ] = 2 x solubility of Mg(OH) 2 poh = -log ( x 10-4 ) = 3.48 ph = Solubility in more Acidic Solution example: solid Mg(OH) 2 in ph=9 buffer Now put Mg(OH) 2 in a ph = 9 buffer solution Now [OH - ] = 1 x 10-5 (if ph =9 then poh = 5) [Mg ][OH - ] 2 = 1.8 x (K sp a function of Temp only) [Mg ] = 1.8x10-11 /(1x10-5 ) 2 = 0.18 M Considerably more soluble (~1000x) in more acidic solution Solubility in water was 1.65 x 10-4 M Essentially removed OH - by reacting with H Acidic solution example: Ag 3 PO 4 Ag 3 (PO 4 ) is more soluble in acid solution than neutral solution. Why? H + of acid solution reacts with PO 4 3- anion to produce weak acid HPO 4 Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3- (aq) Ksp = 1.8x10-18 H + (aq) + PO 4 3- (aq) HPO 4 (aq) K = 1/Ka = 2.8x10 12 Ag 3 PO 4 (s) + H + 3Ag + (aq) + HPO 42 (aq) K = 5.0 x 10-6 Addition of H + shifts equilibrium of 1 st rxn to right! PO 4 3- is actually a moderately strong base: PO H 2 O HPO 4 + OH - K b3 = 2.8x Acidic solution example: AgCl What about AgCl in an acidic solution, such as HNO 3 or H 2 SO 4. AgCl(s) Ag + (aq) + Cl - (aq) Ksp = 1.8x10-10 H + (aq) + Cl - (aq) HCl(aq) K = very small K of reaction remains very small, largely determined by the Ksp of AgCl. Note: Special case of HCl. HCl will affect solubility. How? Why? It would add chloride, actually make AgCl less soluble by Le Chatelier. 124 Solubility and Acidity Other conjugate base anions that will increase in solubility in an acidic solution: OH -, F -, S, CO 3, C 2 O 4 and CrO 4 Why? Because all form weak acids. Examples: PbS(s) + 2H + (aq) H 2 S(g) + Pb (aq) Ca(OH) 2 (s) + 2H + (aq) 2H 2 O(l) + Ca (aq) BaCO 3 (s) + 2H + (aq) CO 2 (g) + H 2 O(l) + Ba (aq) 125 Precipitation and Qualitative Analysis Section 16.2 Hg 2 Cl 2 (s) Hg 2 (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 ] [Cl - ] 2 If [Hg 2 ] = M, what [Cl - ] is req d to just begin the precipitation of Hg 2 Cl 2? That is, what is the maximum [Cl - ] that can be in solution with M Hg 2 without forming Hg 2 Cl 2? 126

22 22 Precipitation and Qualitative Analysis Hg 2 Cl 2 (s) Hg 2 (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 ] [Cl - ] 2 Recognize that K sp = product of maximum ion concs. Precip. begins when product of ion concentrations (Q) EXCEEDS the K sp. 127 Precipitation and Qualitative Analysis Hg 2 Cl 2 (s) Hg 2 (aq) + 2 Cl - (aq) K sp = 1.1 x = [Hg 2 ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 ] = M, [Cl - ] = K sp = 1.1 x 10-8 M If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate. 128 Precipitation and Qualitative Analysis Hg 2 Cl 2 (s) Hg 2 (aq) + 2 Cl - (aq) K sp = 1.1 x Now raise [Cl - ] to 1.0 M. What is the value of [Hg 2 ] at this point? Solution [Hg 2 ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x M The concentration of Hg 2 has been reduced by 10 16! 129 Precipitation and Qualitative Analysis Practice If a solution is M in Cl - ions, at what concentration of Pb does the PbCl 2 (K sp = 1.7 x 10-5 ) precipitate? 130 Separating Metal Ions Cu, Ag +, Pb K sp Values AgCl 1.8 x PbCl x 10-5 PbCrO x Separating Salts by Differences in K sp Add CrO 4 to solid PbCl 2. The less soluble salt, PbCrO 4, precipitates PbCl 2 (s) + CrO 4 PbCrO Cl- Salt Ksp 132 PbCl x 10-5 PbCrO x 10-14

23 23 Separating Salts by Differences in K sp PbCl 2 (s) + CrO 4 PbCrO Cl - Salt K sp PbCl x 10-5 PbCrO x PbCl 2 (s) Pb + 2 Cl - Pb + CrO 4 PbCrO 4 K net = (K 1 )(K 2 ) = 9.4 x 10 8 Net reaction is product-favored K 1 = K sp K 2 = 1/K sp 133 Separating Salts by Differences in K sp The color of the salt silver chromate, Ag 2 CrO 4, is red. A student adds a solution of Pb(NO 3 ) 2 to a test tube containing solid red Ag 2 CrO 4. After stirring the contents of the test tube, the student finds the contents changes from red to yellow. What is the yellow salt? Which one has a higher Ksp? 134 Separating Salts by Differences in K sp Solution 135 Separations by Difference in K sp 136 Qualitative Analysis 137 Equilibria Involving Complex Ions Chapter Figure 16.2 Group 1 Insoluble Chlorides Group 2 Sulfides Insoluble in acid Group 3 Sulfides insoluble in base Group 4 Insoluble Carbonates Group 5 Alkali Metal and ammonium ions The combination of metal ions (Lewis acids) with Lewis bases such as H 2 O and NH 3 leads to COMPLEX IONS

24 24 Reaction of NH 3 with Cu (aq) 139 Solubility and Complex Ions Consider the formation of Ag(NH 3 ) : Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) (aq) The Ag(NH 3 ) is called a complex ion. A Complex Ion is a central transition metal ion that is bonded to a group of surrounding molecules or ions and carries a net charge. It happens a lot with transition metals because they have available d- orbitals. Bases, like ammonia, are attracted to that. (See Chapter 21 for details.) 140 Solubility and Complex Ions 141 Solubility and Complex Ions 142 Note that the equilibrium expression is opposite to what we usually write; the individual species are on the left and the new substance on the right. This leads to K f s that are very LARGE. Dissolving Precipitates by forming Complex Ions Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion. 143 Solubility and Complex Ions Complex Ions The formation of these complex ions increase the solubility of these salts. 144 AgCl(s) + 2 NH 3 Ag(NH 3 ) Cl -

25 25 Solubility and Complex Ions Example Determine the Keq of a reaction if the solution contains AgBr and NH 3. In water, solid AgBr exists in equilibrium with its ions according to AgBr Ag + + Br - K sp = 5 x Also, Ag + make complex ions with ammonia expressed as Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.7 x Solubility and Complex Ions Example Therefore, the equilibrium expressed by both reactions is AgBr Ag + + Br - K sp = 5 x Ag + + 2NH 3 Ag(NH 3 ) + 2 K f = 1.7 x 10 7 AgBr + 2NH 3 Ag(NH 3 ) + Br - K eq = 8.5 x 10-6 K eq = K sp x K f So, this shows overall that the equilibrium is more to the right (aqueous ions) than the dissolution of the AgBr itself in pure water. In other words, NH 3 makes AgBr more soluble AP Exam Practice Complex A/B Problems 2011 AP Exam #1 2006B AP Exam # AP Exam #1 2002B AP Exam # AP Exam #1 Ksp Problems 2011B AP Exam # AP Exam # AP Exam # AP Exam # AP Exam #1

Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College

Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College The Common Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq)

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium Sample Exercise 17.1 (p. 726) What is the ph of a 0.30 M solution of acetic acid? Be sure to use a RICE table, even though you may not need it. (2.63) What

More information

Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA

Operational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA APPLICATIONS OF AQUEOUS EQUILIBRIA Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution

More information

Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline

Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline Name: The Common-Ion Effect Suppose we have a weak acid and a soluble salt of that acid. CH 3 COOH NaCH 3 COO CH 3 COOH CH 3 COO + H + Since NaCH

More information

Chapter 15 - Applications of Aqueous Equilibria

Chapter 15 - Applications of Aqueous Equilibria Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) SA-SB rxn goes to completion (one-way ) Write ionic and net ionic

More information

Chapter 17. Additional Aspects of Equilibrium

Chapter 17. Additional Aspects of Equilibrium Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak

More information

Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2

Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution

More information

Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107

Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107 Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107 Chapter 15 Applications of Aqueous Equilibria (mainly acid/base & solubility)

More information

Try this one Calculate the ph of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite.

Try this one Calculate the ph of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite. Chapter 17 Applying equilibrium 17.1 The Common Ion Effect When the salt with the anion of a is added to that acid, it reverses the dissociation of the acid. Lowers the of the acid. The same principle

More information

Chemistry 102 Chapter 17 COMMON ION EFFECT

Chemistry 102 Chapter 17 COMMON ION EFFECT COMMON ION EFFECT Common ion effect is the shift in equilibrium caused by the addition of an ion that takes part in the equilibrium. For example, consider the effect of adding HCl to a solution of acetic

More information

Chapter 17: Additional Aspects of Aqueous equilibria. Common-ion effect

Chapter 17: Additional Aspects of Aqueous equilibria. Common-ion effect Chapter 17: Additional Aspects of Aqueous equilibria Learning goals and key skills: Describe the common ion effect. Explain how a buffer functions. Calculate the ph of a buffer solution. Calculate the

More information

ACID-BASE REACTIONS. Titrations Acid-Base Titrations

ACID-BASE REACTIONS. Titrations Acid-Base Titrations Page III-b-1 / Chapter Fourteen Part II Lecture Notes ACID-BASE REACTIONS Chapter (Part II A Weak Acid + Strong Base Titration Titrations In this technique a known concentration of base (or acid is slowly

More information

CHM 112 Dr. Kevin Moore

CHM 112 Dr. Kevin Moore CHM 112 Dr. Kevin Moore Reaction of an acid with a known concentration of base to determine the exact amount of the acid Requires that the equilibrium of the reaction be significantly to the right Determination

More information

AP Chemistry. CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect Buffered Solutions. Composition and Action of Buffered Solutions

AP Chemistry. CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect Buffered Solutions. Composition and Action of Buffered Solutions AP Chemistry CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak electrolyte.

More information

Chapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc.

Chapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc. Chapter 17 Additional Aspects of Aqueous Equilibria 蘇正寬 chengkuan@mail.ntou.edu.tw Additional Aspects of Aqueous Equilibria 17.1 The Common-Ion Effect 17.2 Buffers 17.3 Acid Base Titrations 17.4 Solubility

More information

Chapter 8: Applications of Aqueous Equilibria

Chapter 8: Applications of Aqueous Equilibria Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations

More information

Chapter 18. Solubility and Complex- Ionic Equilibria

Chapter 18. Solubility and Complex- Ionic Equilibria Chapter 18 Solubility and Complex- Ionic Equilibria 1 The common ion effect Le Chatelier Why is AgCl less soluble in sea water than in fresh water? AgCl(s) Ag + + Cl Seawater contains NaCl 2 Problem: The

More information

Flashback - Aqueous Salts! PRECIPITATION REACTIONS Chapter 15. Analysis of Silver Group. Solubility of a Salt. Analysis of Silver Group

Flashback - Aqueous Salts! PRECIPITATION REACTIONS Chapter 15. Analysis of Silver Group. Solubility of a Salt. Analysis of Silver Group Page III-15-1 / Chapter Fifteen Lecture Notes Flashback - Aqueous Salts! If one ion from the Soluble Compd. list is present in a compound, the compound is water soluble. PRECIPITATION REACTIONS Chapter

More information

Chap 17 Additional Aspects of Aqueous Equilibria. Hsu Fu Yin

Chap 17 Additional Aspects of Aqueous Equilibria. Hsu Fu Yin Chap 17 Additional Aspects of Aqueous Equilibria Hsu Fu Yin 1 17.1 The Common-Ion Effect Acetic acid is a weak acid: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Sodium acetate is a strong electrolyte: NaCH

More information

Solubility and Complex-ion Equilibria

Solubility and Complex-ion Equilibria Solubility and Complex-ion Equilibria Contents and Concepts Solubility Equilibria 1. The Solubility Product Constant 2. Solubility and the Common-Ion Effect 3. Precipitation Calculations 4. Effect of ph

More information

Unit 3: Solubility Equilibrium

Unit 3: Solubility Equilibrium Unit 3: Chem 11 Review Preparation for Chem 11 Review Preparation for It is expected that the student understands the concept of: 1. Strong electrolytes, 2. Weak electrolytes and 3. Nonelectrolytes. CHEM

More information

SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two)

SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two) SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two) ADEng. PRGORAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore

More information

Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria

Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Solutions of Acids or Bases Containing a Common Ion A common ion often refers to an ion that is added by two or more species. For

More information

AP Chemistry Table of Contents: Ksp & Solubility Products Click on the topic to go to that section

AP Chemistry Table of Contents: Ksp & Solubility Products Click on the topic to go to that section Slide 1 / 91 Slide 2 / 91 AP Chemistry Aqueous Equilibria II: Ksp & Solubility Products Table of Contents: K sp & Solubility Products Slide 3 / 91 Click on the topic to go to that section Introduction

More information

Chapter 17 Additional Aspects of

Chapter 17 Additional Aspects of Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville,

More information

Chapter 15 Additional Aspects of

Chapter 15 Additional Aspects of Chemistry, The Central Science Chapter 15 Additional Aspects of Buffers: Solution that resists change in ph when a small amount of acid or base is added or when the solution is diluted. A buffer solution

More information

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.3 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00

More information

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and Solubility Equilibria ACIDS-BASES COMMON ION EFFECT SOLUBILITY OF SALTS Acid-Base Equilibria and Solubility Equilibria Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 The common

More information

Unit 3: Solubility Equilibrium

Unit 3: Solubility Equilibrium Unit 3: Chem 11 Review Preparation for Chem 11 Review Preparation for It is expected that the student understands the concept of: 1. Strong electrolytes, 2. Weak electrolytes and 3. Nonelectrolytes. CHEM

More information

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.5 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H (aq) CH 3 COO (aq) Initial (): 0.40 0.00 0.00

More information

Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy.

Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy. Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy Chapter 17 Additional Aspects of Aqueous Equilibria Ahmad Aqel Ifseisi Assistant

More information

ACID-BASE EQUILIBRIA. Chapter 14 Big Idea Six

ACID-BASE EQUILIBRIA. Chapter 14 Big Idea Six ACID-BASE EQUILIBRIA Chapter 14 Big Idea Six Acid-Base Equilibria Common Ion Effect in Acids and Bases Buffer SoluDons for Controlling ph Buffer Capacity ph-titradon Curves Acid-Base TitraDon Indicators

More information

Lecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc.

Lecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc. Lecture Presentation Chapter 16 Aqueous Ionic Equilibrium Sherril Soman Grand Valley State University The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze.

More information

Aqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry

Aqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry 2012 Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry The Common-Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) If

More information

Acid-Base Equilibria and Solubility Equilibria

Acid-Base Equilibria and Solubility Equilibria Acid-Base Equilibria and Solubility Equilibria Acid-Base Equilibria and Solubility Equilibria Homogeneous versus Heterogeneous Solution Equilibria (17.1) Buffer Solutions (17.2) A Closer Look at Acid-Base

More information

CHAPTER 7.0: IONIC EQUILIBRIA

CHAPTER 7.0: IONIC EQUILIBRIA Acids and Bases 1 CHAPTER 7.0: IONIC EQUILIBRIA 7.1: Acids and bases Learning outcomes: At the end of this lesson, students should be able to: Define acid and base according to Arrhenius, Bronsted- Lowry

More information

Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations

Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations Acid-Base Equilibria And the beat goes on Buffer solutions Titrations 1 Common Ion Effect The shift in equilibrium due to addition of a compound having an ion in common with the dissolved substance. 2

More information

Saturated vs. Unsaturated

Saturated vs. Unsaturated Solubility Equilibria in Aqueous Systems K sp (Equilibria of Slightly Soluble Salts, Ionic Compounds) Factors that Affect Solubility (Common Ion Effect, AcidBase Chemistry) Applications of Ionic Equilibria

More information

Flashback - Aqueous Salts! PRECIPITATION REACTIONS Chapter 15. Analysis of Silver Group. Solubility of a Salt. Analysis of Silver Group

Flashback - Aqueous Salts! PRECIPITATION REACTIONS Chapter 15. Analysis of Silver Group. Solubility of a Salt. Analysis of Silver Group Page III-15-1 / Chapter Fifteen Lecture Notes Flashback - Aqueous Salts! If one ion from the Soluble Compd. list is present in a compound, the compound is water soluble. PRECIPITATION REACTIONS Chapter

More information

Modified Dr. Cheng-Yu Lai

Modified Dr. Cheng-Yu Lai Ch16 Aqueous Ionic Equilibrium Solubility and Complex Ion Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate Modified Dr. Cheng-Yu Lai Solubility-product constant

More information

Equilibri acido-base ed equilibri di solubilità. Capitolo 16

Equilibri acido-base ed equilibri di solubilità. Capitolo 16 Equilibri acido-base ed equilibri di solubilità Capitolo 16 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

More information

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT Lecture Presentation Chapter 17 Additional Aspects of James F. Kirby Quinnipiac University Hamden, CT Effect of Acetate on the Acetic Acid Equilibrium Acetic acid is a weak acid: CH 3 COOH(aq) H + (aq)

More information

Advanced Placement Chemistry Chapters Syllabus

Advanced Placement Chemistry Chapters Syllabus As you work through the chapter, you should be able to: Advanced Placement Chemistry Chapters 14 16 Syllabus Chapter 14 Acids and Bases 1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and

More information

Learning Objectives. Solubility and Complex-ion Equilibria. Contents and Concepts. 3. Precipitation Calculations. 4. Effect of ph on Solubility

Learning Objectives. Solubility and Complex-ion Equilibria. Contents and Concepts. 3. Precipitation Calculations. 4. Effect of ph on Solubility Solubility and Comple-ion Equilibria. Solubility and the Common-Ion Effect a. Eplain how the solubility of a salt is affected by another salt that has the same cation or anion. (common ion) b. Calculate

More information

Chemistry Lab Equilibrium Practice Test

Chemistry Lab Equilibrium Practice Test Chemistry Lab Equilibrium Practice Test Basic Concepts of Equilibrium and Le Chatelier s Principle 1. Which statement is correct about a system at equilibrium? (A) The forward and reverse reactions occur

More information

Chem 112, Fall 05 Exam 3A

Chem 112, Fall 05 Exam 3A Before you begin, make sure that your exam has all 10 pages. There are 32 required problems (3 points each, unless noted otherwise) and two extra credit problems (3 points each). Stay focused on your exam.

More information

HW 16-10: Review from textbook (p.725 #84, 87, 88(mod), 89, 95, 98, 101, 102, 110, 113, 115, 118, 120, SG#23,A)

HW 16-10: Review from textbook (p.725 #84, 87, 88(mod), 89, 95, 98, 101, 102, 110, 113, 115, 118, 120, SG#23,A) HW 6: Review from textbook (p.75 #84, 87, 88(mod), 89, 95, 98,,,, 3, 5, 8,, SG#3,A) 6.84 The pk a of the indicator methyl orange is 3.46. Over what ph range does this indicator change from 9 percent HIn

More information

What is the ph of a 0.25 M solution of acetic acid (K a = 1.8 x 10-5 )?

What is the ph of a 0.25 M solution of acetic acid (K a = 1.8 x 10-5 )? 1 of 17 After completing this chapter, you should, at a minimum, be able to do the following. This information can be found in my lecture notes for this and other chapters and also in your text. Correctly

More information

Acid-Base Equilibria and Solubility Equilibria Chapter 17

Acid-Base Equilibria and Solubility Equilibria Chapter 17 PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 The common ion effect is the shift in equilibrium caused by the addition

More information

AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS

AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS THE COMMON ION EFFECT The common ion effect occurs when the addition of an ion already present in the system causes the equilibrium to shift away

More information

Secondary Topics in Equilibrium

Secondary Topics in Equilibrium Secondary Topics in Equilibrium Outline 1. Common Ions 2. Buffers 3. Titrations Review 1. Common Ions Include the common ion into the equilibrium expression Calculate the molar solubility in mol L -1 when

More information

Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation

Chem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation Chem 106 3--011 Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation 3//011 1 The net ionic equation for the reaction of KOH(aq) and

More information

2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect

2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento,

More information

APPLICATIONS OF AQUEOUS EQUILIBRIA REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS

APPLICATIONS OF AQUEOUS EQUILIBRIA REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS APPLICATIONS OF AQUEOUS EQUILIBRIA REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS COMMON IONS Common ion effect- The addition of an ion already present(common) in a system causes equilibrium

More information

Solubility Equilibria

Solubility Equilibria Chapter 17 SOLUBILITY EQUILIBRIA (Part II) Dr. Al Saadi 1 Solubility Equilibria The concept of chemical equilibrium helps to predict how much of a specific ionic compound (salt) will dissolve in water.

More information

Consider a normal weak acid equilibrium: Which direction will the reaction shift if more A is added? What happens to the % ionization of HA?

Consider a normal weak acid equilibrium: Which direction will the reaction shift if more A is added? What happens to the % ionization of HA? ch16blank Page 1 Chapter 16: Aqueous ionic equilibrium Topics in this chapter: 1. Buffers 2. Titrations and ph curves 3. Solubility equilibria Buffersresist changes to the ph of a solution. Consider a

More information

CHEM 12 Unit 3 Review package (solubility)

CHEM 12 Unit 3 Review package (solubility) CHEM 12 Unit 3 Review package (solubility) 1. Which of the following combinations would form an ionic solid? A. Metalloid - metal B. Metal non-metal C. Metalloid metalloid D. Non-metal non-metal 2. Which

More information

Chapter 4. Reactions in Aqueous Solution. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO

Chapter 4. Reactions in Aqueous Solution. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation Chapter 4 in Solution 2012 Pearson Education, Inc. John D. Bookstaver St. Charles Community College Cottleville, MO Properties of Solutions Solute: substance in lesser quantity in

More information

Ionic Equilibria in Aqueous Systems

Ionic Equilibria in Aqueous Systems Ionic Equilibria in Aqueous Systems Chapter Nineteen AP Chemistry There are buffers in our blood that keep the ph of our blood at a constant level. The foods that we eat are often acidic or basic. This

More information

SOLUBILITY REVIEW QUESTIONS

SOLUBILITY REVIEW QUESTIONS Solubility Problem Set 1 SOLUBILITY REVIEW QUESTIONS 1. What is the solubility of calcium sulphate in M, g/l, and g/100 ml? 2. What is the solubility of silver chromate? In a saturated solution of silver

More information

CH 4 AP. Reactions in Aqueous Solutions

CH 4 AP. Reactions in Aqueous Solutions CH 4 AP Reactions in Aqueous Solutions Water Aqueous means dissolved in H 2 O Moderates the Earth s temperature because of high specific heat H-bonds cause strong cohesive and adhesive properties Polar,

More information

CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA

CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA The introduction refers to the formation of stalactites and stalagmites as an example of the material in this chapter: CO 2

More information

Chapter 4. Reactions In Aqueous Solution

Chapter 4. Reactions In Aqueous Solution Chapter 4 Reactions In Aqueous Solution I) General Properties of Aqueous Solutions Homogeneous mixture on a molecular level - prop. same throughout - separable by physical means - variable composition

More information

Chapter 6. Types of Chemical Reactions and Solution Stoichiometry

Chapter 6. Types of Chemical Reactions and Solution Stoichiometry Chapter 6 Types of Chemical Reactions and Solution Stoichiometry Chapter 6 Table of Contents (6.1) (6.2) (6.3) (6.4) (6.5) (6.6) (6.7) (6.8) Water, the common solvent The nature of aqueous solutions: Strong

More information

Chemical Equilibrium

Chemical Equilibrium Chemical Equilibrium THE NATURE OF CHEMICAL EQUILIBRIUM Reversible Reactions In theory, every reaction can continue in two directions, forward and reverse Reversible reaction! chemical reaction in which

More information

Solubility and Complex-ion Equilibria

Solubility and Complex-ion Equilibria Solubility and Complex-ion Equilibria Solubility Equilibria Many natural processes depend on the precipitation or dissolving of a slightly soluble salt. In the next section, we look at the equilibria of

More information

AP Chapter 15 & 16: Acid-Base Equilibria Name

AP Chapter 15 & 16: Acid-Base Equilibria Name AP Chapter 15 & 16: Acid-Base Equilibria Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 15 & 16: Acid-Base Equilibria 2 Warm-Ups (Show

More information

Chapter 4. Reactions in Aqueous Solution

Chapter 4. Reactions in Aqueous Solution Chapter 4. Reactions in Aqueous Solution 4.1 General Properties of Aqueous Solutions A solution is a homogeneous mixture of two or more substances. A solution is made when one substance (the solute) is

More information

Ionic Equilibria. weak acids and bases. salts of weak acids and bases. buffer solutions. solubility of slightly soluble salts

Ionic Equilibria. weak acids and bases. salts of weak acids and bases. buffer solutions. solubility of slightly soluble salts Ionic Equilibria weak acids and bases salts of weak acids and bases buffer solutions solubility of slightly soluble salts Arrhenius Definitions produce H + ions in the solution strong acids ionize completely

More information

Chapter 8: Phenomena. Chapter 8: Applications of Aqueous Equilibrium

Chapter 8: Phenomena. Chapter 8: Applications of Aqueous Equilibrium Chapter 8: Phenomena ph ph ph ph 14 12 10 8 6 4 2 0 14 12 10 8 6 4 2 0 Phenomena: Buffers are sometimes defined as: a solution that resists changes in ph when an acid base is added to it. This definition

More information

TYPES OF CHEMICAL REACTIONS

TYPES OF CHEMICAL REACTIONS TYPES OF CHEMICAL REACTIONS Precipitation Reactions Compounds Soluble Ionic Compounds 1. Group 1A cations and NH 4 + 2. Nitrates (NO 3 ) Acetates (CH 3 COO ) Chlorates (ClO 3 ) Perchlorates (ClO 4 ) Solubility

More information

Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria Chapter 17 Additional Aspects of Aqueous Equilibria Water is a common solvent. Dissolved materials can be involved in different types of chemical equilibria. 17.1 The Common Ion Effect Metal ions or salts

More information

Acid - Base Equilibria 3

Acid - Base Equilibria 3 Acid - Base Equilibria 3 Reading: Ch 15 sections 8 9 Ch 16 sections 1 7 * = important homework question Homework: Chapter 15: 97, 103, 107, Chapter 16: 29*, 33*, 35, 37*, 39*, 41, 43*, 49, 55, 57, 61,

More information

CHEM134- Fall 2018 Dr. Al-Qaisi Chapter 4b: Chemical Quantities and Aqueous Rxns So far we ve used grams (mass), In lab: What about using volume in lab? Solution Concentration and Solution Stoichiometry

More information

Review 7: Solubility Equilibria

Review 7: Solubility Equilibria Review 7: Solubility Equilibria Objectives: 1. Be able to write dissociation equations for ionic compounds dissolving in water. 2. Given Ksp, be able to determine the solubility of a substance in both

More information

Chapter 16. Equilibria in Aqueous Systems

Chapter 16. Equilibria in Aqueous Systems Chapter 16 Equilibria in Aqueous Systems Buffers! buffers are solutions that resist changes in ph when an acid or base is added! they act by neutralizing the added acid or base! but just like everything

More information

Applications of Aqueous Equilibria. Chapter 18

Applications of Aqueous Equilibria. Chapter 18 Applications of Aqueous Equilibria Chapter 18 What we learn from Chap 18 This chapter is the third in the three-chapter sequence about equilibrium, this one building upon the core principles raised in

More information

Exam 2 Sections Covered: 14.6, 14.8, 14.9, 14.10, 14.11, Useful Info to be provided on exam: K K [A ] [HA] [A ] [B] [BH ] [H ]=

Exam 2 Sections Covered: 14.6, 14.8, 14.9, 14.10, 14.11, Useful Info to be provided on exam: K K [A ] [HA] [A ] [B] [BH ] [H ]= Chem 101B Study Questions Name: Chapters 14,15,16 Review Tuesday 3/21/2017 Due on Exam Thursday 3/23/2017 (Exam 3 Date) This is a homework assignment. Please show your work for full credit. If you do work

More information

Chapter 17: Additional Aspects of Aqueous Equilibria

Chapter 17: Additional Aspects of Aqueous Equilibria Chapter 17: Additional Aspects of Aqueous Equilibria -Buffer Problems -Titrations -Precipitation Rx Common Ion Effect- The effect that a common ion (from two different sources) has on an equilibrium. (LeChatelier's

More information

Chapter 17 Additional Aspects of Aqueous Equilibria (Part A)

Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) Often, there are many equilibria going on in an aqueous solution. So, we must determine the dominant equilibrium (i.e. the equilibrium reaction

More information

SOLUBILITY EQUILIBRIA (THE SOLUBILITY PRODUCT)

SOLUBILITY EQUILIBRIA (THE SOLUBILITY PRODUCT) SOLUBILITY EQUILIBRIA (THE SOLUBILITY PRODUCT) Saturated solutions of salts are another type of chemical equilibria. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and

More information

Chemical Equilibrium

Chemical Equilibrium Chemical Equilibrium Many reactions are reversible, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product

More information

Acids, Bases and Buffers

Acids, Bases and Buffers 1 Acids, Bases and Buffers Strong vs weak acids and bases Equilibrium as it relates to acids and bases ph scale: [H+(aq)] to ph, poh, etc ph of weak acids ph of strong acids Conceptual about oxides (for

More information

III.1 SOLUBILITY CONCEPT REVIEW

III.1 SOLUBILITY CONCEPT REVIEW III.1 SOLUBILITY CONCEPT REVIEW Read Hebden p. 73 76 and review basic solubility definitions. Soluble means Insoluble means The Dissolving Process IONIC Solutions MOLECULAR Solutions (Covalent compounds)

More information

AP Chemistry. Chapter 4

AP Chemistry. Chapter 4 AP Chemistry Chapter 4 1 Properties of Aqueous Solution Solutions Definition: Any substance (solid, liquid or gas) EVENLY distributed throughout another substance. Solutions have 2 parts: 1) Solvent the

More information

Chapter 4. Reactions in Aqueous Solution

Chapter 4. Reactions in Aqueous Solution Chapter 4 Reactions in Aqueous Solution Topics General properties of aqueous solutions Precipitation reactions Acid base reactions Oxidation reduction reactions Concentration of solutions Aqueous reactions

More information

ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA Acids- taste sour Bases(alkali)- taste bitter and feel slippery Arrhenius concept- acids produce hydrogen ions in aqueous solution while

More information

Solubility Equilibria. Dissolving a salt... Chem 30S Review Solubility Rules. Solubility Equilibrium: Dissociation = Crystalization

Solubility Equilibria. Dissolving a salt... Chem 30S Review Solubility Rules. Solubility Equilibrium: Dissociation = Crystalization Chem 30S Review Solubility Rules Solubility Equilibria Salts are generally more soluble in HOT water(gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li 3

More information

Part One: Solubility Equilibria. Insoluble and slightly soluble compounds are important in nature and commercially.

Part One: Solubility Equilibria. Insoluble and slightly soluble compounds are important in nature and commercially. CHAPTER 17: SOLUBILITY AND COMPLEX ION EQUILIBRIA Part One: Solubility Equilibria A. Ksp, the Solubility Product Constant. (Section 17.1) 1. Review the solubility rules. (Table 4.1) 2. Insoluble and slightly

More information

Chapter 15. Acid-Base Equilibria

Chapter 15. Acid-Base Equilibria Chapter 15 Acid-Base Equilibria The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion already involved in the equilibrium

More information

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO

Chapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO Lecture Presentation Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville, MO The Common-Ion Effect Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l)

More information

SI session Grue 207A

SI session Grue 207A Chem 105 Wednesday 21 Sept 2011 1. Precipitation and Solubility 2. Solubility Rules 3. Precipitation reaction equations 4. Net ionic equations 5. OWL 6. Acids and bases SI session Grue 207A TR, 12:001:30

More information

Chemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B

Chemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B Chemical Equilibrium Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product formation,

More information

UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12)

UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12) I. Multiple Choice UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12) 1) Which one of the following would form an ionic solution when dissolved in water? A. I 2 C. Ca(NO 3 ) 2 B. CH 3 OH D.

More information

Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria

Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria This is a PRACTICE TEST. Complete ALL questions. Answers will be provided so that you may check your work. I strongly

More information

116 PLTL Activity sheet / Solubility Equilibrium Set 11

116 PLTL Activity sheet / Solubility Equilibrium Set 11 Predicting Solubility Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a slightly soluble salt and the equilibrium constant for the reaction is the solubility product

More information

Chapter 4. Concentration of Solutions. Given the molarity and the volume, the moles of solute can be determined.

Chapter 4. Concentration of Solutions. Given the molarity and the volume, the moles of solute can be determined. Molarity Chapter 4 Concentration of Solutions Molarity (M) = moles of solute liters of solution Given the molarity and the volume, the moles of solute can be determined. Given the molarity and the moles

More information

REVIEW QUESTIONS Chapter 17

REVIEW QUESTIONS Chapter 17 Chemistry 102 REVIEW QUESTIONS Chapter 17 1. A buffer is prepared by adding 20.0 g of acetic acid (HC 2 H 3 O 2 ) and 20.0 g of sodium acetate (NaC 2 H 3 O 2 ) in enough water to prepare 2.00 L of solution.

More information

SOLUBILITY AND PRECIPITATION EQUILIBRIA

SOLUBILITY AND PRECIPITATION EQUILIBRIA 16 CHAPTER SOLUBILITY AND PRECIPITATION EQUILIBRIA 16.1 The Nature of Solubility Equilibria 16.2 Ionic Equilibria between Solids and Solutions 16.3 Precipitation and the Solubility Product 16.4 The Effects

More information