Applications of Aqueous Equilibria

1 1 Applications of Aqueous Equilibria Acid-Base & Precipitation Reactions Chapter 15 Stomach Acidity & 2 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? a) Common ion effect and buffers b) Titrations Common Ion Effect: the shift in an equilibrium position caused by the addition or presence of an ion involved in the equilibrium reaction. 3 The Common Ion Effect 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 qe NH 4 (aq) OH - (aq) Here we are adding an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the. The ph will go. After all, NH 4 is an acid! 4 5 6 ph of Aqueous NH 3 ph of Aqueous NH 3 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 qe NH 4 (aq) OH - (aq) 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 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 qe NH 4 (aq) OH - (aq) K b 1.8 x -5 [NH 4 ][OH - ] [NH 3 ] x 2 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 11.33 for 0.25 M NH 3

ph of NH 3 /NH 4 Mixture 2 7 ph of NH 3 /NH 4 Mixture 8 Problem: What is the ph of a solution with 0. M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) H 2 O qe 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 Problem: What is the ph of a solution with 0. M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) H 2 O qe 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. 0 change -x x x equilib 0.25 - x 0. x x ph of NH 3 /NH 4 Mixture Problem: What is the ph of a solution with 0. M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) H 2 O qe NH 4 (aq) OH - (aq) K b 1.8 x -5 [NH 4 ][OH - ] [NH 3 ] x(0. x) 0.25 - x [OH - ] x (0.25 / 0.)K b 4.5 x -5 M This gives poh 4.35 and ph 9.65 ph drops from 11.33 to 9.65 on adding a common ion. 9 Example Calculate the [H ] and the % dissociation of HF in a solution containing 1.0 M HF (Ka7.2-4 ) and 1.0 M NaF. I C E 4 x 7.2 [H ] ph 3.14 % Dissociation [H ] 7.2 0 [HF] 1.0 HF(aq) 1.0 -x 1-x 0 0.0072 % K 7.2 [H ][F ] (x)(1.0 x) (x)(1.0) [HF] 1.0 x 1.0 4 a 0 4 H (aq) 0 x x F - (aq) 1.0 x 1x You can further compare this % dissociation with just straight 1.00 M HF which has a 2.7 % dissociation. Using Le Chatelier principle it should be easy to understand the large difference between the % dissociations. 11 12 HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4- and its conjugate base HPO 4 2-. 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. Buffering Capacity the ability of a buffered solution to absorb protons or hydroxide ions without significant change in ph; determined by the magnitudes of [HA] and [A - ] in the solution.

Buffer Composition Buffer Composition Weak Acid Conj. Base HOAc OAc - H 2 PO - 4 HPO 2-4 Weak Base Conj. Acid NH 3 NH 4 3 13 Consider HOAc/OAc - to see how buffers work ACID USES UP ADDED OH - We know that OAc - H 2 O qe 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 9 K reverse is VERY LARGE, so HOAc completely snarfs up OH -!!!! 14 15 16 Consider HOAc/OAc - to see how buffers work CONJ. BASE USES UP ADDED H HOAc H 2 O qe OAc - H 3 O has K a 1.8 x -5 Therefore, the reverse reaction of the WEAK BASE with added H has K reverse 1/ K a 5.6 x 4 K reverse is VERY LARGE, so OAc - completely snarfs up H! Problem: What is the ph of a buffer that has [HOAc] 0.700 M and [OAc - ] 0.600 M? HOAc H 2 O qe OAc - H 3 O K a 1.8 x -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 qe OAc - H 3 O K a 1.8 x -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 -5 [H 3 O ](0.600) 0.700 [H 3 O ] 2.1 x -5 and ph 4.68 17 Example A buffered solution contains 0.50 M acetic acid (K a 1.8-5 ) and 0.50 M sodium acetate. Calculate the ph of this solution. CH 3 COOH(aq) H (aq) CH 3 COO - (aq) I 0.50 0 0.50 C -x x x E 0.50-x x 0.50x K 1.8 a 5 [H ][CH3COO ] (x)(0.50 x) (x)(0.50) [CH COOH] 0.50 x 0.50 3 5 x 1.8 [H ] ph 4.74 18

Example 4 19 So what happens when 0.0 mols of NaOH is added to 1.0 L of this buffer? [CH 3 COOH] 0.49 M [CH 3 COO - ] 0.51 M I C E K 1.8 a CH 3 COOH(aq) 5 0.49 -x 0.49-x x 1.7 H (aq) 0 x x ph 4.76 [H ] CH 3 COO - (aq) 0.51 x 0.51x [H ][CH3COO ] (x)(0.51 x) (x)(0.51) [CH COOH] 0.49 x 0.49 3 5 Notice that the expression for calculating the H conc. of the buffer is [H 3 O Orig. conc. of HOAc ] Orig. conc. of OAc - K a [H 3 O ] [Acid] [Conj. base] K a [OH ] [Base] [Conj. acid] K b Notice that the H or OH - concs. depend on K and the ratio of acid and base concs. 20 Henderson-Hasselbalch Equation Calculates the ph of a Buffer [H 3 O ] [Acid] [Conj. base] K a Take the negative log of both sides of this equation [Acid] ph pk a - log [Conj. base] [Conj. base] ph pk a log [Acid] The ph is determined largely by the pk a of the acid and then adjusted by the ratio of acid and conjugate base. 21 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) 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 M 2 1.00 x -3 M ph 3.00 22 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) 23 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) 24 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. Solution to Part (b): Step 1 Stoichiometry [H 3 O ] [OAc - ] [HOAc] Before rxn Change After rxn 0.000-0.000 0 0.600-0.000 0.599 0.700 0.000 0.701

5 25 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 e H 3 O OAc - [HOAc] [H 3 O ] [OAc - ] Before rxn Change After rxn 0.701 0 0.599 -x x x 0.701-x x 0.599 x 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 e H 3 O OAc - [HOAc] [H 3 O ] [OAc - ] After rxn 0.701-x 0.599x x Because [H 3 O ] 2.1 x -5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. 26 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 - [H 3 O ] [HOAc] [OAc - ] K a 0.701 0.599 (1.8 x -5 ) [H 3 O ] 2.1 x -5 M ph 4.68 The ph has not changed on adding HCl to the buffer! 27 Buffer Capacity 28 The buffering capacity of a buffered solution represents the amount of protons or hydroxide ions the buffer can absorb without significant change in the ph. We know The ph of a buffer is determined by the ratio of [A - ] / [HA] The buffer capacity Is determined by the magnitudes of [HA] and [A - ] The most effective buffering will occur when [HA] [A - ] Therefore, the pka of the weak acid to be used in the buffer should be as close as possible to the desired ph. 29 Example 30 Calculate the change in ph that occurs when 0.0 mol gaseous HCl is added to 1.0 L of each of the following solutions. 5.00 M CH 3 COOH and 5.00 M NaCH 3 COO 0.050 M CH 3 COOH and 0.050 M NaCH 3 COO First determine the ph of the Buffers [A ] ph pka log [HA] since [A ] [HA] for both ph pk log 1 a ph - log (1.8-5 ) 4.74

First Solution 6 31 What will happen to the extra H? H CH 3 COO - CH 3 COOH Before Reaction: 0.0 M 5.00 M 5.00 M After Reaction: 0 M 4.99 M 5.01 M Second Solution What will happen to the extra H? H CH 3 COO - CH 3 COOH Before Reaction: 0.0 M 0.050 M 0.050 M After Reaction: 0 M 0.040 M 0.060 M 32 [A ] ph pka log [HA] [A ] ph pka log [HA] 4.99 ph 4.74 log 4.74 5.01 0.040 ph 4.74 log 4.56 0.060 So We can see The ph of a buffer is determined by the ratio of [A - ] / [HA] Is determined by the magnitudes of [HA] and [A - ] The most effective buffering will occur when [HA] [A - ] Therefore, the pka of the weak acid to be used in the buffer should be as close as possible to the desired ph. 33 Preparing a Buffer You want to buffer a solution at ph 4.30. This means [H 3 O ] -ph 5.0 x -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] K a 34 Preparing a Buffer You want to buffer a solution at ph 4.30 or [H 3 O ] 5.0 x -5 M POSSIBLE ACIDS K a HSO 4- / SO 4 2-1.2 x -2 HOAc / OAc - 1.8 x -5 HCN / CN - 4.0 x - Best choice is acetic acid / acetate. 35 Preparing a Buffer You want to buffer a solution at ph 4.30 or [H 3 O ] 5.0 x -5 M [H 3 O ] 5.0 x -5 [HOAc] [OAc - ] (1.8 x -5 ) Solve for [HOAc]/[OAc - ] ratio 2.78 1 Therefore, if you use 0.0 mol of NaOAc and 0.278 mol of HOAc, you will have ph 4.30. 36

Preparing a Buffer 7 37 Preparing a Buffer 38 A final point Buffer prepared from CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Result: diluting a buffer solution does not change its ph 8.4 g NaHCO 3 - weak acid 16.0 g Na 2 CO 2-3 conjugate base HCO 3- H 2 O qe H 3 O CO 2-3 What is the ph? 39 40 Titrations Acid-Base Titrations ph Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the ph increases very slowly. Titrant volume, ml 41 42 Acid-Base Titrations Acid-Base Titrations 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 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. ph at half-way point? 8 43 ph at equivalence point? Acid-Base Titration QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? HBz NaOH Na Bz - H 2 O 44 Benzoic acid NaOH ph of solution of benzoic acid, a weak acid C 6 H 5 CO 2 H HBz Benzoate ion Bz - Acid-Base Titrations The product of the titration of benzoic acid is the benzoate ion, Bz -. Bz - is the conjugate base of a weak acid. Therefore, final solution is basic. Bz - H 2 O HBz OH - K b 1.6 x - 45 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. ph at equivalence point is basic 46 Benzoic acid NaOH 47 48 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? 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 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? STOICHIOMETRY PORTION 1. Calc. moles of NaOH req d (0.0 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 / 0.0 mol) 0.025 L 25 ml of NaOH req d

9 49 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph of the final solution? 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 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 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 0.020 0 0 change - x x x equilib 0.020 - x x x 50 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 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 - 51 QUESTION: You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH to the equivalence point. What is the ph at half-way point? ph at half-way point? Equivalence point ph 8.25 52 K b 1.6 x - x 2 0.020 - x x [OH - ] 1.8 x -6 poh 5.75 -----> ph 8.25 53 Acetic acid titrated with NaOH 54 You titrate 0. ml of a 0.025 M solution of benzoic acid with 0.0 M NaOH. What is the ph at the half-way point? HBz H 2 O H 3 O Bz - K a 6.3 x -5 Both HBz and Bz - are present. This is a BUFFER! [H 3 O ] [HBz] [Bz - ] K a At the half-way point, [HBz] [Bz - ] Therefore, [H 3 O ] K a 6.3 x -5 ph 4.20 pk a of the acid Weak acid titrated with a strong base

56 Strong acid titrated with a strong base Weak diprotic acid (H 2 C 2 O 4 ) titrated with a strong base (NaOH) 57 58 Weak base (NH 3 ) titrated with a strong acid (HCl) 59 Indicators for Acid-Base Titrations