Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion 16.2

What is the pH of a solution of 0.10 M CH 3 COOH and 0.20 M CH 3 COONa? Initial (M) Change (M) Equilibrium (M) x-x +x+x x x+x x x x (0.20+x) x = 1.8 x Ka  Ka  0.20x 0.10 = 1.8 x Assume x << 0.1 < 0.2 so 0.10 – x  0.10 and 0.20+x  0.20 K a << 1 x = 9.0 x M = [H + ] Common ion CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) Acid and conjugate base are present initially !! K a = [CH 3 COOH] [H + ][CH 3 COO - ] = pH = -log [H + ] = - log (9.0 x ) = 5.05 => acidic Assumption was good!

What happens when a strong base or strong acid is added to a weak acid or weak base ? A solution initially contains 0.30 M CH 3 COOH and 0.20 M NaOH. What is the pH at equilibrium? => CH 3 COOH, Na +, OH -, H 2 O Weak acid K a =1.8 x Strong base Neutralization Reaction CH 3 COOH + OH - CH 3 COO - + H 2 O Start End 0.30 M 0.00 M0.20 M 0.10 M 0.20 M0.00 M K = 1/K b = K a / K w = 1.8 x 10 9 => Goes to completion

Initial (M) Change (M) Equilibrium (M) x-x +x+x x x+x x x x (0.20+x) x = 1.8 x Ka  Ka  0.20x 0.10 = 1.8 x Assume x << 0.1 < 0.2 so 0.10 – x  0.10 and 0.20+x  0.20 K a << 1 x = 9.0 x M = [H + ] CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) K a = [CH 3 COOH] [H + ][CH 3 COO - ] = pH = -log [H + ] = - log (9.0 x ) = 5.05 => acidic Problem now become the same as the last problem. => CH 3 COOH, Na +, CH 3 COO -, H 2 O

What happens if we now add mole HCl to 1.00 L of a mixed acid/conjugate base solution? => H +, Cl -, CH 3 COOH, CH 3 COO -, Na +, H 2 O Strong Acid Neutralization Reaction CH 3 COO - + H + CH 3 COOH Start End 0.20 M 0.00 M0.101 M M mol = 1.00 L M M K = 1/K a = 5.68 x 10 4 => Goes to completion => CH 3 COOH, CH 3 COO -, Cl -, Na +, H 2 O

Initial (M) Change (M) Equilibrium (M) x-x +x+x x x+x x x x (0.199+x) x = 1.8 x Ka  Ka  0.199x = 1.8 x Assume x << < so – x  and x  K a << 1 x = 9.1 x M = [H + ] CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) K a = [CH 3 COOH] [H + ][CH 3 COO - ] = pH = -log [H + ] = - log (9.0 x ) = 5.04 => ΔpH = 0.01 Problem now becomes similar to the the last problem. If M HCl was alone in a solution, what would the pH be? [H + ] = 1 x M => pH = 3

Buffers A solution that is resistant to change in pH upon addition of small amounts of acid or base Mixture of weak acid - HX and its conjugate base - X - Buffer has: acidic species - neutralize base basic species to neutralize acid OH - + HX X - + H 2 O [X - ] [HX]  constant => pH doesn’t change Examples: CH 3 COOH + CH 3 COO - CH 3 NH 2 + CH 3 NH 2 + H 2 PO HPO 4 -2 acid conj. base