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 and ph Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria 1 ph range where buffered solutions play important roles CHEMISTRY BIOLOGY MEDICINE 2

ph Box [H 3 O + ] = 10 -ph ph [H 3 O + ] ph = -log[h 3 O + ] ph + poh = 14 @ 25 C K w = 1 x 10-14 @ 25 C [OH - ] = 10 -poh poh [OH - ] poh = -log[oh - ] [H 3 O + ][OH - ] = 1 x 10-14 3 Common Ion Effect A shift in the equilibrium POSITION that occurs because of the addition of an ion already involved in the equilibrium reaction. HNO 2 + H 2 O H 3 O + + NO 2 - Adding NaNO 2 would shift the equilibrium to the left, decreasing the H 3 O + and NO 2 - produced by the reaction between HNO 2 and H 2 O. (Le Châtelier s principle) 4

Which of the following statements is true? 1. HF dissociates more readily in a solution of NaF than in pure water. 2. HF dissociates equally well in a solution of KCl and in pure water. 3. HF does not dissociate in water to any extent. 5 Like Example 8.1 - I Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution. Calculate the [H + ], the ph, and the percent dissociation of HNO 2 in a 1.0 M solution that is also 1.0 M in NaNO 2! HNO 2(aq) H + (aq) + NO 2 - (aq) [H + ] [NO - 2 ] K a = [HNO = 4.0 x 10-4 2 ] Initial Concentration (mol/l) Equilibrium Concentration (mol/l) [HNO 2 ] 0 = 1.0 M [HNO 2 ] = 1.0 x (from dissolved HNO 2 ) [NO - 2 ] 0 = 1.0 M [NO - 2 ] = 1.0 + x (from dissolved NaNO 2 ) [H + ] 0 0 [H + ] = x (neglect the contribution from water) 6

Like Example 8.1 - II [H K a = + ] [NO - 2 ] ( x ) ( 1.0 + x ) = = 4.0 x 10 [HNO -4 2 ] (1.0 x ) Assume (1.0 + x) 1.0 and (1.0 x) 1.0 : x (1.0) (1.0) = 4.0 x 10-4 or x = 4.0 x 10-4 = [H + ] Therefore ph = - log [H + ] = - log (4.0 x 10-4 ) = 3.40 The percent dissociation is: 4.0 x 10-4 x 100 = 0.040 % 1.0 Nitrous acid Nitrous acid alone + NaNO 2 [H + ] 2.0 x 10-2 4.0 x 10-4 ph 1.70 3.40 % Diss 2.0 0.040 7 What is a BUFFER? A buffer is a solution that contains concentrations of both components of a conjugate acid-base pair. When small quantities of H + or OH - are added to the buffer, they cause a small amount of one buffer component to convert into the other (HA A - or A - HA). Buffers resist a change in ph. As long as the amounts of H 3 O + and OH - are small (compared to the concentrations of the acid and base in the buffer), the added ions will have little effect on the ph since they are consumed by the buffer components. The ability of a buffer to resist a change of ph is due to the buffer capacity. ph equals pka IF the acid-to-base ratio in the buffer is 1:1 8

9 How Does a Buffer Work Original buffered solution ph (H + or OH - added) Modified ph Step 1: Do stoichiometric calculations to determine new concentrations. Assume reaction with H + /OH - goes to completion. Step 2: Do equilibrium calculations. 10

How Does a Buffer Work Addition of a strong base to a buffer has the following effect: OH - + HA A - + H 2 O Original buffer ph Final ph of buffer close to original K a = [H+ ] [A - ] [HA] Added OH - ions produce A - ions [H + ] = K a [HA] [A - ] 11 A buffer is made by placing 0.250 mol of acetic acid and 0.250 mol of sodium acetate in 1 L of solution. What is the ph of 100.0 ml of the buffer before and after 1.00 ml of concentrated HCl (12.0 M) is added? What is the ph of 300.00 ml of pure water before and after the same amount of acid is added? CH 3 COOH (aq) + H 2 O (l) How a Buffer Works I CH 3 COO - (aq) + H 3 O+ (aq) [CH [H 3 O + 3 COOH] (0.250) ] = K a x = 1.8 x 10-5 x = 1.8 x 10-5 [CH 3 COO - ] (0.250) ph = -log[h 3 O + ] = -log(1.8 x 10-5 ) = ph = 4.74 Before HCl added! Add 1.00 ml conc. HCl to 100.0 ml of buffer: 0.00100 L x 12.0 mol/l = 0.0120 mol H 3 O + 12

How a Buffer Works II updates in green After acid is added, all H 3 O + will react with available CH 3 COO - : Conc. (moles) CH 3 COOH (aq) + H 2 O (l) CH 3 COO - + H 3 O + Initial 0.025 ---- 0.025 0.012 Change +0.012 ---- -0.012-0.012 Equilibrium 0.037 ---- 0.013 0 Equilibrium must then be reestablished: total vol = 0.101 L Conc. (M) CH 3 COOH (aq) + H 2 O (l) CH 3 COO - + H 3 O + Initial 0.366 ---- 0.129 0 Change -x ---- +x +x Equilibrium 0.366 - x ---- 0.129 + x x Assuming: 0.366 - x = 0.366 and 0.129 + x = 0.129 [CH 3 COOH] [H 3 O + ] = K a x =1.8 x 10-5 (0.366) x = 5.11 x 10-5 [CH 3 COO - ] (0.129) ph = -log(5.11 x 10-5 ) = 4.29 After the acid is added! 13 How a Buffer Works III A buffer is made by placing 0.250 mol of acetic acid and 0.250 mol of sodium acetate in 1 L of solution. What is the ph of 100.0 ml of the buffer before and after 1.00 ml of concentrated HCl (12.0 M) is added? What is the ph of 300.0 ml of pure water after the same amount of acid is added? Add 1.00 ml conc. HCl: 0.00100 L x 12.0 mol/l = 0.0120 mol H 3 O + to 300.0 ml of water : 0.0120 mol H 3 O + = 0.0399 M H 3 O + 0.3010 L soln. ph = -log(0.0399 M) ph = 1.399 Without buffer! 14

How a Buffer Works IV updates in red Suppose we add 1.00 ml of a concentrated base instead of an acid. Add 1.00 ml of 12.0 M NaOH to 1 L of the buffer and to pure water, and let s see what the impact is: 1.00 ml x 12.0 mol/l OH - = 0.0120 mol OH - 0.0120 M INITIAL STOICHIOMETRY Moles CH 3 COOH (aq) + OH - (aq) CH 3 COO- + H 2 O Initial 0.250 0.0120 0.250 --- Change - 0.0120-0.0120 +0.0120 --- Equilibrium 0.238 0 0.262 --- REESTABLISH EQUILIBRIUM (in 1.001L) Conc. (M) CH 3 COOH (aq) + H 2 O (l) CH 3 COO - + H 3 O + Initial 0.238 ---- 0.262 0 Change - x ---- +x +x Equilibrium 0.238-x ---- 0.262+x x 15 How a Buffer Works V Making our normal assumptions: 0.238 - x = 0.238 and 0.262 + x = 0.262 [H 3 O + ] = 1.8 x 10-5 0.238 x = 1.64 x 10-5 0.262 ph = -log(1.64 x 10-5 ) = 4.79 After base is added! By adding 1.00 ml base to 300.0 ml of pure water we would get a [OH - ] concentration of: 0.0120 mol OH [OH - ] = - = 0.0399 M OH - 0.3010 L The [H 3 O + ] concentration is: K [H 3 O + w 1.0 x 10 ] = = -14 = 2.51 x 10-13 [OH - ] 0.0399 M ph = -log(2.5 x 10-13 ) = 12.60 Without buffer! 16

How a Buffer Works VI In summary: Buffer alone ph = 4.74 Buffer plus 1.0 ml acid ph = 4.29 Buffer plus 1.0 ml base ph = 4.79 Acid in water ph = 1.399 Base in water ph = 12.60 17 HENDERSON - HASSELBALCH EQUATION HA (aq) + H 2 O (l) H + (aq) + A- (aq) K a = [H+ ] [A - ] [HA] Solving for the hydronium ion concentration gives: [H + ] = K a x [HA] [A - ] Taking the negative logarithm of both sides: -log[h + ] = -log K a + - [HA] log( [A ) - ] ph = pk a - [HA] log( [A ) - ] 18

ph of a BUFFER (Henderson Hasselbalch eqn.) Generalizing for any conjugate acid-base pair : ph = pka + log ( [A - ] ) [HA] when [A - ] = [HA], ph= pka For acetic acid/sodium acetate solution: K a = 1.8 x 10-5 pka = 4.74 ph = 4.74 If [HA] and [A - ] are known, the ph can be calculated. If the ph is known, the ratio of [HA] and [A - ] can be calculated. NOTE: The ph is INDEPENDENT of [HA] and [A - ] only when [HA] = [A - ]. 19 The Effect of Added Acetate Ion on ph of Acetic Acid [CH 3 COOH] [CH 3 COO - ] added % Dissociation ph 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10 0.10 0.018 4.74 0.10 0.15 0.012 4.92 NOTE: Remember, sodium acetate solution is basic! Increasing concentration of acetate increases ph of solution! 20

The ammonia-ammonium ion buffer has a ph of about 9.2 and can be used to keep solutions in the basic ph range. What mass of ammonium chloride must be added to 400.0 ml of a 3.00 M ammonia solution to prepare a buffer? K b (NH 3 ) = 1.8 x 10-5 1. 1.20 g 2. 20.44 g 3. 21.65 g 4. 62.94 g 5. 64.19 g 21 Preparing a Buffer Problem: The ammonia-ammonium ion buffer has a ph of about 9.2 and can be used to keep solutions in the basic ph range. What mass of ammonium chloride must be added to 400.0 ml of a 3.00 M ammonia solution to prepare a buffer? Plan: We want to add sufficient ammonium ion to equal the aqueous ammonia concentration. Solution: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) [NH + 4 ] [OH - ] K b = = 1.8 x 10-5 [NH 3 ] NH 3 = 3.00 mol x 0.4000 L = 1.20 mol same amount of NH + 4 needed L NH 4 Cl = 53.49 g/mol Therefore mass = 1.20 mol x 53.49 g/mol = 64.19 g NH 22 4 Cl

Instructions for making a buffer say to mix 60.0 ml of 0.100 M NH 3 with 40.0 ml of 0.100 M NH 4 Cl. What is the ph of this buffer? K b (NH 3 ) = 1.8 x 10-5 1. 2.7 x 10-5 2. 3.7 x 10-10 3. 4.57 4. 9.43 23 Like Example 8.3 I Problem: Instructions for making a buffer say to mix 60.0 ml of 0.100 M NH 3 with 40.0 ml of 0.100 M NH 4 Cl. What is the ph of this buffer? The combined volume is 60.0 ml + 40.0 ml = 100.0 ml Moles of ammonia = V x M NH = 0.060 L x 0.100 M = 0.0060 mol 3 NH3 Moles of ammonium ion = V NH 4Cl x M NH4Cl = 0.040 L x 0.100 M = 0.0040 mol 0.0060 mol [NH 3 ] = = 0.060 M ; [NH + 0.0040 mol 4 ] = = 0.040 M 0.100 L 0.100 L Concentration (M) NH 3 (aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) Starting 0.060 0.040 0 Change -x +x +x Equilibrium 0.060 x 0.040 + x x 24

Like Example 8.3 - II Substituting into the equation for K b : [NH K b = 4+ ] [OH - ] = 1.8 x 10-5 = [NH 3 ] (0.040 + x) (x) (0.060 x) Assume : 0.060 x = 0.060 and 0.040 + x = 0.040 0.040 (x) K b = 1.8 x 10-5 = x = 2.7 x 10 0.060-5 Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 0.000027 = 0.060 or 0.045% [OH - ] = 2.7 x 10-5 ; poh = - log[oh - ] = - log (2.7 x 10-5 ) = 4.57 ph = 14.00 poh = 14.00 4.57 = 9.43 25 OR, using the H-H equation: K K a = w K = 5.6 x 10-10 b Like Example 8.3 - III [A ph = pka + log ( - ] ) [HA] [NH ph = 9.25 + log ( 3 ]) [NH 4+ ] 0.060 ph = 9.25 + log ( ) 0.040 ph = 9.25 + log 1.5 = 9.43 pk a = -log 5.6 x 10-10 = 9.25 26

ph and Capacity of Buffered Solutions The ph of a buffered solution is determined by the ratio[a - ]/[HA]. The buffer capacity of a buffered solution is determined by the magnitudes of [HA] and [A - ] The HIGHER the concentration of buffer components the LARGER the buffering capacity. The more concentrated the buffer is the more it will resist ph change. 27 The Relation Between Buffer Capacity and ph Change 28

Example 8.5 - I Calculate the change in ph that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following acetic acid/acetate buffer solutions. For acetic acid, K a = 1.8 x 10-5 Solution A: 5.00 M HC 2 H 3 O 2 and 5.00 M NaC 2 H 3 O 2 Solution B: 0.050 M HC 2 H 3 O 2 and 0.0500 M NaC 2 H 3 O 2 Use the Henderson-Hasselbalch equation for initial ph: ph = pk a + log ( [C 2 H 3 O 2 - ] [H C 2 H 3 O 2 ] ph = pk a + log (1) = pk a = -log(1.8 x 10-5 ) = 4.74 ) Since [C 2 H 3 O 2 - ] = [H C 2 H 3 O 2 ] The equation becomes: Adding 0.010 mol of HCl will cause a shift in the equilibrium due to: H + (aq) + C 2 H 3 O 2 - (aq) HC 2 H 3 O 2 (aq) 29 Example 8.5 - II For Solution A: H + + C 2 H 3 O 2 - H C 2 H 3 O 2 Before reaction 0.010 M 5.00 M 5.00 M After reaction 0 4.99 M 5.01 M Calculate the new ph using the Henderson-Hasselbalch equation: ( ) ( ) [C ph = pk a + log 2 H 3 O - 2 ] 4.99 = 4.74 + log = 4.74 0.0017 [H C 2 H 3 O 2 ] 5.01 = 4.74 For Solution B: H + + C 2 H 3 O 2 - H C 2 H 3 O 2 Before reaction 0.010 M 0.050 M 0.050 M After reaction 0 0.040 M 0.060 M 0.040 The new ph is: ph = 4.74 + log = 4.74 0.18 = 4.56 0.060 ( ) 30

Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding base. They can involve a weak acid HA and its conjugate base A - or a weak base B and its conjugate acid BH +. When H + is added to a buffered solution, it reacts essentially to completion with the weak base present: H + + A - HA or H + + B BH + When OH - is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH - + HA A - + H 2 O or OH - + BH + B + H 2 O The ph of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the ph will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA/A - or B/BH + ) are large compared with the amounts of H + or OH - added. 31 Review TITRATIONS Volumetric analysis - technique by which one solution is used to analyze another solution Titrant is delivered from a buret. The sample and an indicator are in an Erlenmeyer flask. The indicator changes color at the equivalence point, the point at which the amounts of titrant and sample are stoichiometrically equal. The concentration of one of the solutions must be known exactly (in order to determine moles, millimoles (1 mmol = 10-3 mol), or equivalents added). mol = mmol ml * M = mmol L ml The concentration of an unknown sample is calculated from the volume and concentration of the titrant used to reach the equivalence point, along with the volume of the sample. 32

ACID-BASE TITRATIONS There are three types of titration curves: strong acid/strong base weak acid/strong base weak base/strong acid 33 STRONG ACID / STRONG BASE The curve is symmetrical. The equivalence point is at ph = 7.00 (think about effect of salts on ph/acidity). The range of ph values is proportional to original concentration of analyte at the starting point. A titration curve is characterized by starting point, equivalence point, and final point. 34

SA/SB titration curve: HCl titrated by NaOH Note: strong acid - strong base system neutral salt (NaCl) at equivalence point! HCl + NaOH HCl NaCl NaCl + H 2 O NaCl+NaOH ph calculations along the curve: 1)Stoichiometry with added titrant 2)pH based on what s left in the beaker 35 Calculating ph along the Titration Curve Initial ph: 100 ml 0.030 M HCl ph = -log (0.030 M) = 1.52 Region #1: Before the equivalence point 100 ml 0.030 M HCl + 25 ml 0.030 M NaOH ph determined by excess acid and new total volume Moles HCl (aq) + NaOH (aq) NaCl + H 2 O Initial 3.0 mmol 0.75 mmol 0 --- Change -0.75 mmol -0.75 mmol +0.75 mmol --- Equilibrium 2.25 mmol 0 0.75 mmol --- New total volume = 100 ml + 25 ml = 125 ml New [H+] = 2.25 mmol / 125 ml = 0.018 M ph = -log (0.018 M) = 1.74 36

Calculating ph along the Titration Curve Region #2: At the equivalence point 100 ml 0.030 M HCl + 100 ml 0.030 M NaOH ph determined by ions present at eq. pt. and new total volume Moles HCl (aq) + NaOH (aq) NaCl + H 2 O Initial 3.0 mmol 3.0 mmol 0 --- Change -3.0 mmol -3.0 mmol +3.0 mmol --- Equilibrium 0 mmol 0 mmol 3.0 mmol --- New total volume = 100 ml + 100 ml = 200 ml Only 0.015 M NaCl is present at the equivalence point ph = 7 at equivalence point for SA/SB 37 Calculating ph along the Titration Curve Region #3: After the equivalence point 100 ml 0.030 M HCl + 150 ml 0.030 M NaOH ph determined by excess base and new total volume Moles HCl (aq) + NaOH (aq) NaCl + H 2 O Initial 3.0 mmol 4.5 mmol 0 --- Change -3.0 mmol -3.0 mmol +3.0 mmol --- Equilibrium 0 mmol 1.5 mmol 3.0 mmol --- New total volume = 100 ml + 150 ml = 250 ml [OH-] = 1.5 mmol / 250 ml = 0.0060 M poh = -log (0.0060 M) = 2.22 ph = 14 - poh = 14-2.22 = 11.78 38

39 SB/SA TITRATION CURVE : titration of 100 ml of 0.5 M NaOH by 1.0 M HCl 40

Titration Curve Calculations for Weak Acid/Strong Base A stoichiometry problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined. An equilibrium problem. The position of the weak acid equilibrium is determined (I.C.E. table) and the ph is calculated. 41 Consider the salt that would be formed by the titration of a weak acid with a strong base. What do you expect the ph to be at the equivalence point? 1. < 7 2. > 7 3. = 7 42

Remember Effects of Salts on ph and Acidity WEAK ACIDS For any salt whose cation has neutral properties (cation from a strong acid - such as Na + or K + ) and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, etc. 43 TITRATION of WEAK ACID The curve is asymmetrical The equivalence point is NOT at ph = 7.00 (ph will be > 7.00) The range of ph values is proportional to original concentration of analyte at the starting point and the K a value of the titrated acid! A titration curve is characterized by starting point, buffer region, equivalence point, and final point. halfway point (or midpoint) in the buffer region Titration of 0.1 M acetic acid with 0.1 M NaOH 44

What is the ph of 20.00 ml of 0.250 M nitrous acid? (HNO 2 ; K a = 4.0 x 10-4 ) 1. 2.000 2. 2.849 3. 3.398 4. 4.000 5. 5.699 45 Calculating the ph During a Weak Acid-Strong Base Titration I (Like Ex 8.7) Problem: Calculate the ph during the titration of 20.00 ml of 0.250 M nitrous acid (HNO 2 ; K a = 4.0 x 10-4 ) after adding the following volumes of 0.150 M NaOH: (a) 0.00 ml (b) 15.00 ml (c) 20.00 ml (d) 33.33 ml, and (e) 40.00 ml. Solution: Stoichiometry HNO 2 (aq) + NaOH (aq) H 2 O (l) + NaNO 2 (aq) Weak Acid Dissoc. HNO 2 (aq) + H 2 O (l) H + (aq) + NO 2 - (aq) (a) [H + ] [NO - 2 ] x K a = = 2 = 4.00 x 10-4 [HNO 2 ] 0.250 M ph = -log(1.0 x 10-2 ) = 2.000 starting point x 2 = 1.00 x 10-4 x = 1.00 x 10-2 46

Calculating the ph During a WA/SB Titration II (b) 15.00 ml of 0.150 M NaOH is added; new total volume = 35.00 ml 15.00 ml x 0.150 mmol/ml = 2.25 mmol of OH - 20.00 ml x 0.250 mmol/ml = 5.00 mmol of HNO 2 Moles HNO 2 (aq) + OH - (aq) H 2 O (l) + NO 2 - (aq) Initial 5.00 mmol 2.25 mmol ---- 0 Change -2.25 mmol -2.25 mmol ---- +2.25 mmol Equilibrium 2.75 mmol 0 ---- 2.25 mmol Concentration (M) HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) Initial 0.0786 M ---- 0 0.0643 M Change -x ---- +x +x Equilibrium 0.0786 - x ---- x 0.0643 + x ph = pk a + log [NO ( - 2 ] [HNO 2 )] = 3.40 + log(0.0643/0.0786) ph = 3.40-0.0872 = 3.31 after 15.0 ml of NaOH added NOTE: this ph is actually before the halfway point of the titration! 47 Calculating the ph During a WA/SB Titration III (c) 20.00 ml of 0.150 M NaOH is added; new total volume = 40.00 ml 20.00 ml x 0.150 mmol/ml = 3.00 mmol of OH - 20.00 ml x 0.250 mmol/ml = 5.00 mmol of HNO 2 Moles HNO 2 (aq) + OH - (aq) H 2 O (l) + NO 2 - (aq) Initial 5.00 mmol 3.00 mmol ---- 0 Change -3.00 mmol -3.00 mmol ---- +3.00 mmol Equilibrium 2.00 mmol 0 ---- 3.00 mmol Concentration (M) HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) Initial 0.0500 M ---- 0 0.0750 M Change -x ---- +x +x Equilibrium 0.0500 - x ---- x 0.0750 + x ph = pk a + log [NO ( - 2 ] [HNO 2 )] = 3.40 + log(0.0750/0.0500) ph = 3.40 + 0.176 = 3.58 after 20.0 ml of NaOH added NOTE: this ph is actually after the halfway point of the titration! 48

Calculating the ph During a WA/SB Titration IV (d) 33.33 ml of 0.150 M NaOH is added; new total volume = 53.33 ml 33.33 ml x 0.150 mmol/ml = 5.00 mmol of OH - 20.00 ml x 0.250 mmol/ml = 5.00 mmol of HNO 2 Moles HNO 2 (aq) + OH - (aq) H 2 O (l) + NO 2 - (aq) Initial 5.00 mmol 5.00 mmol ---- 0 Change -5.00 mmol -5.00 mmol ---- +5.00 mmol Equilibrium 0 0 ---- 5.00 mmol Weak Base Dissociation!! Concentration (M) NO - 2 (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial 0.0938 M ---- 0 0 Change -x ---- +x +x Equilibrium 0.0938 - x ---- x x [OH - ] [HNO 2 ] x K b = = 2 = 2.50 x 10-11 x 2 = 2.34 x 10-12 [NO - 2 ] 0.0938 M x = 1.53 x 10-6 ph = 14 poh = 14 5.815 = 8.185 after 33.33 ml of NaOH added 49 NOTE: this ph is AT the equivalence point of the titration! Calculating the ph During a WA/SB Titration V (e) 40.00 ml of 0.150 M NaOH is added; new total volume = 60.00 ml 40.00 ml x 0.150 mmol/ml = 6.00 mmol of OH - 20.00 ml x 0.250 mmol/ml = 5.00 mmol of HNO 2 Moles HNO 2 (aq) + OH - (aq) H 2 O (l) + NO 2 - (aq) Initial 5.00 mmol 6.00 mmol ---- 0 Change -5.00 mmol -5.00 mmol ---- +5.00 mmol Equilibrium 0 1.00 mmol ---- 5.00 mmol Excess of 1.00 mmol of NaOH which will control the ph. [OH-] = 1.00 mmol / 60.00 ml = 0.0167 M poh = -log (0.0167 M) = 1.778 ph = 14 - poh = 14 1.778 = 12.222 when all of the acid is neutralized and there is an excess of NaOH 50

51 Influence of K a value on shape of titration curve. Buffer region contains halfway point (midpoint) 52

53 Consider the salt that would be formed by the titration of a strong acid with a weak base. What do you expect the ph to be at the equivalence point? 1. < 7 2. > 7 3. = 7 54

Remember Effects of Salts on ph and Acidity WEAK BASES A salt whose cation is the conjugate acid of a weak base produces an acidic solution when dissolved in water. Examples: NH 4 Cl, AlCl 3, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. 55 Figure 8.5: ph curve for 100 ml of 0.05 M NH 3 and 0.1 M HCl B + H + BH + Like the weak acid titration, the curve is asymmetrical. Buffer region (halfway point, midpoint) The equivalence point is NOT at ph = 7.00 (ph will be < 7.00) 56

NH 4 Cl ph = 5.27 at equivalence point 57 The Color Change of the Indicator Bromthymol Blue 58

Figure 8.8 : Useful ph ranges for indicators (pka ± 1) 59 For titration of strong acid with strong base EITHER phenolphtalein OR methyl red can be used Color change will be sharp Several suitable indicators available 60

For titration of acetic acid (weak acid) by a strong base phenolphtalein is OK, methyl red is not!! NaOH + acetate Na acetate The weaker the acid, the smaller the vertical area around the equivalence point Look for useful ph range with midpoint as close to ph at equivalence point of titration Much less flexibility compared to SA/SB titrations Acetic acid Buffer point 61 SOLUBILITY and SOLUBILITY PRODUCT When a solution becomes saturated (no more solute will dissolve), a precipitate forms and we can calculate the quantity of material that remains in solution. We are working with the: Solubility Product The equilibrium constant that is used for these calculations is called the solubility product constant: K sp Example : Lead chromate PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) K sp = [Pb 2+ ][CrO 4 2- ] [PbCrO 4 ] Since the activity of a solid is 1.0, we can move it to the other side of the equal sign and combine it with the constant yielding the solubility product constant K sp K sp = [Pb2+ ][CrO 4 2- ] 62

Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds Problem: Write the ion-product expression for (a) silver bromide; (b) strontium phosphate; (c) aluminum carbonate; (d) nickel(iii) sulfide. Plan: Write the equation for a saturated solution, then write the expression for the solubility product. Solution: (a) silver bromide: AgBr (s) Ag + (aq) + Br - (aq) K sp = [Ag + ] [Br - ] (b) strontium phosphate: Sr 3 (PO 4 ) 2(s) 3 Sr 2+ (aq) + 2 PO 4 3- (aq) K sp = [Sr 2+ ] 3 [PO 4 3- ] 2 (c) aluminum carbonate: Al 2 (CO 3 ) 3 (s) 2 Al 3+ (aq) + 3 CO 3 2- (aq) K sp = [Al 3+ ] 2 [CO 3 2- ] 3 (d) nickel(iii) sulfide: Ni 2 S 3 (s) + 3 H 2 O (l) 2 Ni 3+ (aq) + 3 HS - (aq) + 3 OH- (aq) K sp =[Ni 3+ ] 2 [HS - ] 3 [OH - ] 3 63 64

65 Determining K sp from Solubility Problem: Lead chromate is an insoluble compound that at one time was used as the pigment in the yellow stripes on highways. It s solubility is 4.33 x 10-6 g/100 ml water. What is the K sp? Plan: We write an equation for the dissolution of the compound to see the number of ions formed, then write the ion-product expression. Solution: PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) 4.33 x 10 Molar solubility of PbCrO 4 = -6 g 1000 ml x x 100 ml 1 L 1mol PbCrO 4 323.2 g = 1.34 x 10-8 M PbCrO 4 1 Mole PbCrO 4 = 1 mole Pb 2+ and 1 mole CrO 4 2- Therefore [Pb 2+ ] = [CrO 4 2- ] = 1.34 x 10-8 M K sp = [Pb 2+ ] [CrO 4 2- ] = (1.34 x 10-8 M) 2 = 1.54 x 10-16 66

What is the solubility of PbCrO 4 in water if the K sp is equal to 2.00 x 10-16? 1. 1.41 x 10-16 M 2. 2.00 x 10-16 M 3. 1.41 x 10-8 M 4. 2.00 x 10-8 M 5. 2.82 x 10-8 M 6. 4.00 x 10-8 M 67 Determining Solubility from K sp Problem: Calculate the solubility of PbCrO 4 in water if the K sp is equal to 2.00 x 10-16. Plan: We write the dissolution equation and the ion-product expression. Solution: PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) K sp = 2.00 x 10-16 =[Pb 2+ ][CrO 4 2 ] Concentration (M) PbCrO 4 Pb 2+ CrO 4 2- Initial ---------- 0 0 Change ---------- +x +x Equilibrium ---------- x x K sp = [Pb 2+ ] [CrO 4 2- ] = (x)(x) = 2.00 x 10-16 x = 1.41 x 10-8 Therefore the solubility of PbCrO 4 in water is 1.41 x 10-8 M 68

Like Example 8.12 The K sp value for the mineral fluorite, CaF 2 is 3.4 x 10-11. Calculate the solubility of fluorite in units of grams per liter. Concentration (M) CaF 2 (s) Ca 2+ (aq) + 2 F- (aq) Initial 0 0 Change +x +2x Equilibrium x 2x Substituting into K sp : [Ca 2+ ][F - ] 2 = K sp (x) (2x) 2 = 3.4 x 10-11 4x 3 = 3.4 x 10-11 x = 3 x = 2.0 x 10-4 3.4 x 10-11 4 The solubility is 2.0 x 10-4 moles CaF 2 per liter of water. To get mass we must multiply by the molar mass of CaF 2 (78.1 g/mol). 78.1 g CaF 2.0 x 10-4 2 mol CaF 2 x 1 mol CaF = 1.6 x 10-2 g CaF 2 per L 2 69 Relationship Between K sp and Solubility at 25 C No. of Ions Formula Cation:Anion K sp Solubility (M) 2 PbSO 4 1:1 1.6 x 10-8 1.3 x 10-4 2 MgCO 3 1:1 3.5 x 10-8 1.9 x 10-4 2 BaCrO 4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH) 2 1:2 6.5 x 10-6 1.2 x 10-2 3 BaF 2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF 2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag 2 CrO 4 2:1 2.6 x 10-12 8.7 x 10-5 70

A solution of silver nitrate is added to a solution of the slightly soluble salt, silver chromate (K sp = 2.6 x 10-12 ). The concentration of chromate ion in solution will. 1. Increase 2. Decrease 3. Stay the same 71 Calculating the Effect of a Common Ion on Solubility Problem: What is the solubility of silver chromate in 0.0600 M silver nitrate solution? K sp = 2.6 x 10-12 Plan: From the equation and the ion product expression for Ag 2 CrO 4, we predict that the addition of silver ion will decrease the solubility. Solution: Writing the equation and ion-product expression: Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) K sp = [Ag+ ] 2 [CrO 4 2- ] Concentration (M) Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) Initial --------- 0.0600 0 Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x Assuming that K sp is small, 0.0600 M + 2x 0.600 M K sp = 2.6 x 10-12 = (0.0600) 2 (x) x = 7.22 x 10-10 M Therefore, the solubility of silver chromate here is 7.22 x 10-10 M 72

Predicting the Effect on Solubility of Adding Strong Acid Problem: Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of: (a) iron(ii) cyanide, (b) potassium bromide, and (c) aluminum hydroxide Plan: Write the balanced dissolution equation and note the anion. Anions of weak acids react with H 3 O + and shift the equilibrium position toward more dissolution. Strong acid anions do not react, so added acid has no effect. Solution: (a) Fe(CN) 2 (s) Fe 2+ (aq) + 2 CN- (aq) Increases solubility CN - ion reacts with H 3 O + to form the weak acid HCN, so CN - would be removed from the product side of the equation; shifts rxn to the right. (b) KBr (s) K + (aq) + Br - (aq) No effect Br - is the anion of a strong acid and K + is the cation of a strong base. (c) Al(OH) 3 (s) Al 3+ (aq) + 3 OH- (aq) Increases solubility OH - is the anion of water, a very weak acid, so it reacts with the added acid to produce water in a simple acid-base reaction; shifts 73 rxn to the right. Predicting the Formation of a Precipitate: Q sp vs. K sp The solubility product constant, K sp, can be compared to the ion product constant, Q sp, to understand the characteristics of a solution with respect to forming a precipitate. Q sp = K sp : When a solution becomes saturated, no more solute will dissolve. No changes will occur. Q sp > K sp : Precipitates will form until the solution becomes saturated. Q sp < K sp : Solution is unsaturated; no precipitate will form. 74

Will a precipitate form when a concentrated solution containing barium nitrate is added to a concentrated solution of sodium chromate? 1. Yes 2. No 3. Not enough info 75 Predicting Whether a Precipitate Will Form I Problem: Will a precipitate form when 0.100 L of a solution containing 0.55 M barium nitrate is added to 200.0 ml of a 0.100 M solution of sodium chromate? Plan: Determine if the solutions will yield soluble ions. Calculate the new concentrations, adding the two volumes together to get the total volume of the solution. Calculate the ion product constant (Q sp ) and compare it to the solubility product constant to see if a precipitate will form. Solution: Both Na 2 CrO 4 and Ba(NO 3 ) 2 are soluble, so we will have Na +, CrO 4 2-, Ba 2+ and NO 3 - ions present in 0.300 L of solution. After referencing the table of solubility rules (Ch 4), and we find that BaCrO 4 is insoluble, so we calculate it s ion-product constant and compare it to the solubility product constant of 2.1 x 10-10 : For Ba 2+ : (0.100 L Ba(NO 3 ) 2 ) x (0.55 M) = 0.055 mol Ba 2+ [Ba 2+ 0.055 mol Ba ] = 2+ = 0.183 M in Ba 2+ 0.300 L 76

Predicting Whether a Precipitate Will Form II For CrO 4 2- : (0.100 M Na 2 CrO 4 )(0.200 L) = 0.0200 mol CrO 4 2-0.0200 mol CrO 2- [CrO 2-4 4 ] = = 0.667 M in CrO 2-4 0.300 L Q sp = [Ba 2+ ] [CrO 4 2- ] =(0.183 M Ba 2+ )(0.667 M CrO 4 2- ) = 0.121 Since K sp = 2.1 x 10-10 and Q sp = 0.121, Q sp >> K sp so a precipitate will form. 77 FYI Figure 8.12 : Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ 78

FYI The General Procedure for Separating Ions in Qualitative Analysis 79 Separating Ions by Selective Precipitation I Problem: A solution consists of 0.10 M AgNO 3 and 0.15 M CuNO 3. Calculate the [I - ] that would separate the metal ions as their iodides. K sp of AgI = 8.3 x 10-17 ; K sp of CuI = 1.0 x 10-12 Plan: Since the two iodides have the same formula type (1:1), we compare their K sp values and we see that CuI is about 100,000 times more soluble than AgI. Therefore, AgI precipitates first, and we solve for [I - ] that will give a saturated solution of AgI. Solution: Writing chemical equations and ion product expressions: H AgI 2 O (s) Ag + (aq) + I - (aq) K sp = [Ag+ ][I - ] H CuI 2 O (s) Cu + (aq) + I - (aq) K sp = [Cu+ ][I - ] Calculating the quantity of iodide needed to give a saturated solution of CuI: K sp [I - 1.0 x 10 ] = = -12 = 6.7 x 10-11 M [Cu + ] 0.15 M 80

Separating Ions by Selective Precipitation II Thus, the concentration of iodide ion that will give a saturated solution of copper(i) iodide is 6.7 x 10-11 M, and this will not precipitate the copper(i) ion, but should remove most of the silver ion. Calculating the quantity of silver ion remaining in solution we get: K sp [Ag + 8.3 x 10 ] = = -17 = 1.2 x 10-6 M [I - ] 6.7 x 10-11 Since the initial silver ion was 0.10 M, most of it has been removed, and essentially none of the copper(i) was removed, so the separation was quite complete. If the iodide was added as sodium iodide, you would have to add only a few nanograms of NaI to remove nearly all of the silver from solution: 1 mol NaI 6.7 x 10-11 mol I - x x 149.9 g NaI = 10.0 ng NaI mol I - mol NaI 81 Complex ion: A charged species consisting of a metal ion surrounded by ligands Ligand: A neutral molecule or ion having a lone pair that 82 can be used to form a covalent bond to a metal ion

The Stepwise Exchange of NH 3 for H 2 O in M(H 2 O) 4 2+ Formation (stability) constants: equilibrium constants for the addition of ligands to metal ions 83 Formation Constants (K f ) of Some Complex Ions at 25 o C Complex Ion K f Complex Ion K f Ag(CN) - 2 3.0 x 10 20 Ag(NH 3 ) + 2 1.7 x 10 7 Ag(S 2 O 3 ) 3-2 4.7 x 10 13 AlF 3-6 4 x 10 19 Al(OH) - 4 3 x 10 33 Be(OH) 2-4 4 x 10 18 CdI 2-4 1 x 10 6 Co(OH) 2-4 5 x 10 9 Cr(OH) - 4 8.0 x 10 29 Cu(NH 3 ) 2+ 4 5.6 x 10 11 Fe(CN) 4-6 3 x 10 35 Fe(CN) 3-6 4.0 x 10 43 Hg(CN) 2-4 9.3 x 10 38 Ni(OH) 2-4 2 x 10 28 Pb(OH) - 3 8 x 10 13 Sn(OH) - 3 3 x 10 25 Zn(CN) 2-4 4.2 x 10 19 Zn(NH 3 ) 2+ 4 7.8 x 10 8 Zn(OH) 2-4 3 x 10 15 84