More About Chemical Equilibria

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 (15.2 15.3) 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 15.1 5 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 0.25 0 0 change -x +x +x equilib 0.25 - x x x

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 ] 0.25 - x Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives poh = 2.67 and so ph = 14.00-2.67 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 = 11.33 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 0.25 0.10 0 change -x +x +x equilib 0.25 - x 0.10 + 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(0.10 + x) = [NH 3 ] 0.25 - 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 11.33 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 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 - 4 + HPO 4 NH + 4 + 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 10-10 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] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10-5 0.700 M HOAc has ph = 2.45 The ph of the buffer will have 1. ph < 2.45 2. ph > 2.45 3. ph = 2.45

4 Buffer Solutions 19 Buffer Solutions 20 Problem: What is the ph of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10-5 initial change equilib [HOAc] [OAc - ] [H 3 O + ] 0.700 0.600 0 -x +x +x 0.700 - x 0.600 + x x Problem: What is the ph of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x 10-5 [HOAc] [OAc - ] [H 3 O + ] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have K a = 1.8 x 10-5 = [H 3 O+ ](0.600) 0.700 [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 1.4 10-4. 24

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 0.10 - E 0.1x 0.10 + x x x(0.10 + x) = 1.4x10-4 0.12 - 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 = 3.77 25 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 0.12 0.10 ph = log (1.4 10-4 )+ log (0.10) (0.12) ph = 3.85 + ( 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 0.080 M ammonia and 0.045 M ammonium chloride. The K b for ammonia is 1.8 x 10-5. NH 3 (aq) + H 2 O(l) NH 4+ (aq) + OH - (aq) [base] [conj. acid] 0.080 0.045 27 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 = 4.74 + ( 0.25) poh = 4.49 ph = 14.00 4.49 = 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 4.00. (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] = 0.700 M and [OAc - ] = 0.600 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 = 3.00 30

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] = 0.700 M and [OAc - ] = 0.600 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] = 0.700 M and [OAc - ] = 0.600 M (ph = 4.68) Solution to Part (b): Step 1 Stoichiometry [H 3 O + ] + [OAc - ] [HOAc] Before rxn Change After rxn Equil [] s 0.00100 mol 0.600 mol 0.700 mol -0.00100-0.00100 +0.00100 0 0.599 mol 0.701 mol 0 0.599 mol 1.001 L 0.701 mol 1.00l L 0.598 mol/l 0.700 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] = 0.700 M and [OAc - ] = 0.600 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) 0.700 mol/l 0 0.598 mol/l Change (M) Equil. (M) -x 0.700-x +x x +x 0.598 + 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] = 0.700 M and [OAc-] = 0.600 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] = 0.700 M and [OAc-] = 0.600 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 0.700-x x 0.598+x Because [H 3 O + ] = 2.1 x 10-5 M BEFORE adding HCl, we again neglect x relative to 0.700 and 0.598. Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - ph = 4.74-0.07 = 4.67 The ph has not changed much on adding HCl to the buffer!

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] = 0.700 M and [OAc - ] = 0.600 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 = 11.00 37 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] = 0.700 M and [OAc - ] = 0.600 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] = 0.700 M and [OAc - ] = 0.600 M (ph = 4.68) Solution to Part (b): Step 1 Stoichiometry [OH - ] + [HOAc] [OAc - ] Before rxn Change After rxn Equil. [] s 0.00100 mol 0.700 mol 0.600 mol -0.00100-0.00100 +0.00100 0 0.699 mol 0.601 mol 0 0.699 mol 1.001 L 0.601 mol 1.001 L 0.698 mol/l 0.600 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] = 0.700 M and [OAc - ] = 0.600 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 - ] 0.698 mol/l 0 0.600 mol/l -x +x +x 0.698-x x 0.600 + 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] = 0.700 M and [OAc-] = 0.600 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 0.698-x x 0.600+x Because [H 3 O + ] = 2.1 x 10-5 M BEFORE adding NaOH, we again neglect x relative to 0.600 and 0.698. 41 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] = 0.700 M and [OAc-] = 0.600 M (ph = 4.68) Solution to Part (b): Step 2 Equilibrium HOAc + H 2 O H 3 O + + OAc - ph = 4.74 0.07 = 4.67 The ph has not changed much on adding NaOH to the buffer! 42

8 Preparing a Buffer You want to buffer a solution at ph = 4.30. This means [H 3 O + ] = 10 -ph =10-4.30 = 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 4 1.2 x 10-2 HOAc / OAc - 1.8 x 10-5 HCN / CN - 4.0 x 10-10 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 = 2.78 1 Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have ph = 4.30. 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 = 10.3 + log (1.51/1.0)=10.5

9 Buffer Example Phosphate buffers are used a lot in labs to simulate blood ph of 7.40. 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 = 7.21 + log[hpo 4 ]/[H 2 PO 4- ] log[hpo 4 ]/[H 2 PO 4- ] = 0.19 [HPO 4 ]/[H 2 PO 4- ] = 10 0.19 = 1.55 Part B 6.0g NaH 2 PO 4 = 0.050 mol in 0.500 L = 0.10 M 120g/mol [HPO 4 ] = 1.55[H 2 PO 4- ] = 0.155 M HPO 4 =0.155 mol/l * 0.500 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 0.010 mol of HCl is added to 1.0L of each of the following: 5.00 M HOAc and 5.00 M NaOAc 0.050 M HOAc and 0.050 M NaOAc Ka= 1.8x10-5 (pka = 4.74) Note: Since [OAc-]=[HOAc] for both cases, the initial ph is 4.74. 53 Large Buffer System H 3 O + + OAc - HOAc + H 2 O Stoichiometry Calculation: H 3 O + OAc - HOAc Before Reaction (moles) 0.01 5.00 5.00 After Reaction (moles) 0 4.99 5.01 54

10 Calculate ph via Henderson- Hasselbalch Equations At Equilibrium: HOAc + H 2 O H 3 O + + OAc - 0.501 X 4.99 [base] ph = pka + [acid] ph = -log (1.8 x 10-5) + log[ 4.99/5.01] ph = 4.738 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) 0.01 0.05 0.05 After Reaction (moles) 0 0.04 0.06 56 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 0.06 0 0.04 Change -x +x +x At Equilibrium 0.06-x x 0.04 + x 57 Calculate ph via Henderson- Hasselbalch Equations At Equilibrium: HOAc + H 2 O H 3 O + + OAc - 0.06-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 = 4.74 0.18 = 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) 0.005 0.005 - After reaction 0 0 - 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) 0.005 0.002 - After reaction (mol) 0.003 0 - Total volume of solution is 70.0 ml (0.0700 L) So the [H + ] = 0.003 mol/0.0700 L = 0.0429 M ph = -log[0.0429] = 1.37 60

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) 0.005 0.007 - After reaction (mol) - 0.002 - Total volume of solution is 120.0 ml (0.1200 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 = 0.017 M OH - poh = 1.77 so ph = 12.23 61 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) 0.005 0.005 - After reaction 0 0-0.005 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: 0.005 mol/0.100l I (conc) 0.05 0 0 C -x +x +x E 0.050-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 0.050-x x x K rxn =K b = K w /K a = 1x10-14 /1.8x10-5 = 5.56x10-10. = 5.56x10-10 Usual assumption that x is small compared to 0.050 so x = 0.050 5.56 10 5.27 10 poh = -log[5.27x10-6 ] = 5.28 so ph = 8.72 64 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) 0.0050 0.0020 0 Reaction -0.0020-0.0020 +0.0020 After reaction 0.0030 0 0.0020 65 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 0.0030 0.0020 x ph = pka + log [OAc - ]/[HOAc] = 4.74 + log(0.002)/(0.003) = 4.74 + log(0.667) = 4.74 0.18 = 4.56 66

12 67 68 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. 69 70 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 0.025 M solution of benzoic acid with 0.100 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 0.025 M solution of benzoic acid with 0.100 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 73 74 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 10-10 + QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. Benzoic acid + NaOH ph at equivalence point is basic 75 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 76 77 78 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the ph at the equivalence point? QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2. Calc. volume of NaOH req d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 ml of NaOH req d STOICHIOMETRY PORTION 25 ml of NaOH req d 3. Moles of Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of 125 ml [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M

14 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 10-10 [Bz - ] [HBz] [OH - ] initial 0.020 0 0 change - x +x +x equilib 0.020 - x x x 79 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 10-10 K b = 1.6 x 10-10 x 2 = 0.020 - x x = [OH - ] = 1.8 x 10-6 poh = 5.75 and ph = 8.25 80 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the ph at half-way point? 81 QUESTION: You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 0.025 M solution of benzoic acid with 0.100 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 0.025 M solution of benzoic acid with 0.100 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 0.0006 0.0025 -- After rxn, mole 0 0.0019 0.0006 After rxn, [ ] 0 0.0019mol/0.106L 0.0006/0.106 0 0.018 0.0057 Equilbrium: HBz + H 2 O H 3 O + + Bz - pka = 4.20 Equilibrium Now can use H-H equation: ph = 4.20 + log (0.0057/0.018) = 4.20.50 = 3.70 Remember: this is essentially a buffer solution since HBz and Bz - are both in the solution. 84

15 You titrate 100. ml of a 0.025 M solution of benzoic acid with 0.100 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 0.0020 0.0025 -- After rxn, mole 0 0.0005 0.0020 After rxn, [ ] 0 0.0005mol/0.120L 0.0020/0.120 0 0.0042 0.017 Equilbrium: HBz + H 2 O H 2 O + Bz - pka = 4.20 Equilibrium Now can use H-H equation: ph = 4.20 + log (0.017/0.0042) = 4.20 + 0.61 = 4.81 85 Acetic acid titrated with NaOH Weak acid titrated with a strong base 86 87 88 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? 89 90 Weak base (NH 3 ) titrated with a strong acid (HCl) Volume of titrating reagent added -->

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 16 95 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 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 3 + 2 HCl CaCl 2 + CO 2 (g) + H 2 O Precipitation: Pb(NO 3 ) 2 + 2 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 10-10 AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 2.79 x 10-10 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.

18 103 104 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 = 0.00130 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 = 0.00130 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 10-9. Note: Calculating K sp from solubility gives approximate values that can be off by a few OOMs due to ionic interactions.

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 = 128.1 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 3.9x10-11. (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 10-5 112 113 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 10-10 (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 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10-10 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 The Common Ion Effect 115 The Common Ion Effect 116 Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10-10 BaSO 4 (s) Ba (aq) + SO 4 (aq) Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10-10 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 0.010 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 ] 0.010 0 + y + y 0.010 + y y The Common Ion Effect Calculate the solubility of BaSO 4 in (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10-10 BaSO 4 (s) Ba (aq) + SO 4 (aq) Solution K sp = [Ba ] [SO 4 ] = (0.010 + y) (y) Because y < 1.1 x 10-5 M (= x, the solubility in pure water), this means 0.010 + y is about equal to 0.010. Therefore, K sp = 1.1 x 10-10 = (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 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10-10 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 0.020 M AgNO 3? Explain your answer. Ksp(AgI) = 8.5 x 10-17 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 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 10-11 = 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 (2 1.65 x 10-4 ) = 3.48 ph = 10.52 121 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 10-11 (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 + 122 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 4 3- + H 2 O HPO 4 + OH - K b3 = 2.8x10-2 123 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 10-18 = [Hg 2 ] [Cl - ] 2 If [Hg 2 ] = 0.010 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 0.010 M Hg 2 without forming Hg 2 Cl 2? 126

22 Precipitation and Qualitative Analysis Hg 2 Cl 2 (s) Hg 2 (aq) + 2 Cl - (aq) K sp = 1.1 x 10-18 = [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 10-18 = [Hg 2 ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 ] = 0.010 M, [Cl - ] = K sp 0.010 = 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 10-18 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 10-18 M The concentration of Hg 2 has been reduced by 10 16! 129 Precipitation and Qualitative Analysis Practice If a solution is 0.020 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 10-10 PbCl 2 1.7 x 10-5 PbCrO 4 1.8 x 10-14 131 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 4 + 2 Cl- Salt Ksp 132 PbCl 2 1.7 x 10-5 PbCrO 4 1.8 x 10-14

23 Separating Salts by Differences in K sp PbCl 2 (s) + CrO 4 PbCrO 4 + 2 Cl - Salt K sp PbCl 2 1.7 x 10-5 PbCrO 4 1.8 x 10-14 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 16.3 138 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 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 ) 2 + + Cl -

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 10-13 Also, Ag + make complex ions with ammonia expressed as Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.7 x 10 7 145 Solubility and Complex Ions Example Therefore, the equilibrium expressed by both reactions is AgBr Ag + + Br - K sp = 5 x 10-13 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. 146 147 AP Exam Practice Complex A/B Problems 2011 AP Exam #1 2006B AP Exam #1 2003 AP Exam #1 2002B AP Exam #1 2002 AP Exam #1 Ksp Problems 2011B AP Exam #1 2010 AP Exam #1 2006 AP Exam #1 2004 AP Exam #1 2001 AP Exam #1