pH of Weak Acids

1. pH of Weak Acids AP Chemistry Unit 9 Chapter 14

2. Strengths of Acids and Bases • “Strength” refers to how much an acid or base ionizes in a solution.

3. Relative Dissociation of Acids

4. Concentration vs- StrengthHA ↔ H+ + A- Ka = [H+] [A-] , therefore larger Ka values indicate stronger acids. [HA]

5. What is the pH of a 0.25M HF solution? • The Ka = at 25oC = 7.1 x 10-4 HF is a weak acid, so we can’t directly calculate the pH from the molarity. We need to set up a table as we did for other equilibrium problems • HF(aq) ⇄ H+(aq)+ F-(aq) Initial: 0.25M 0 M 0 M Equilibrium: 0.25-x x x • Ka = [H+][F-] = 7.1 x 10-4 • [HF] • 7.1 x 10-4 = x2 • (0.25-x)

6. 7.1 x 10-4 = x2 • (0.25-x) • and x2 + 7.1 x 10-4x -1.775 x 10-4 = 0 • Solving using the quadratic equation: • x = 1.30 x 10-2 or –1.37x10-2. • The negative solution is physically impossible because x was set to be the [H+], • so x = 1.30 x 10-2. • Substituting in this x into the original formula for Ka checks, so: • pH=-log(1.30 x 10-2) = 1.89

7. An easier method than using the quadratic equation is to make the assumption that the dissociation of HF is so insignificant that the initial concentration is not significantly changed at equilibrium. • Therefore, we assume that (0.25 – x) = 0.25 • So, 7.1 x 10-4 = x2 • 0.25 • 1.78 x 10-4 = x2 • therefore x = 1.33 x 10-2 • pH = - log (1.33 x 10-2) = 1.88 • Considering we need to round to 2 significant figures, our assumption is valid!

8. The 5% Rule • We can ignore the (-x) in the denominator if the acid dissociates less than 5% • Otherwise, we must use the quadratic equation. • % Ionization = [A-] • [HA]initial • Usually a very low Ka value means we can ignore the (-x) X 100