1. Finding the pH of Weak Acids

2. Strengths of Acids and Bases “Strength” refers to how much an acid or base ionizes in a solution. STRONGWEAK Ionize completely (~100%) Example: HCl  H + + Cl - NaOH  Na + + OH - Ionize partially (usually <10%) Example: HF ↔ H + + F - NH 3 + H 2 O ↔ NH 4 + + OH -

3. Weak acids will generally have a higher pH than a strong acid because the [H + ] is lower. However, you can make a weak acid more concentrated and that will lower the pH. A 1 Molar solution of a strong acid will have a pH of 0. This will not be the case for a 1M solution of a weak acid if only 5% dissociates. The [H+] will be less, so pH will be higher.

4. Concentration vs- Strength HA ↔ H + + A -

5. Finding the pH of Weak Acids Weak acids do not dissociate 100%. Therefore, we cannot assume that the concentration of the acid equals the concentration of H + ions like we did with the strong acids.

6. We must use equilibrium expressions to help us solve for the pH. Assume HA is a weak acid - HA  H + + A - Then Ka = [H + ] [A - ] [HA]

7. Generic Example- What is the pH of a 1 Molar solution of a weak acid HA? HA  H+ + A- Initial 1 M 0 0 At 1 - x x x Equilibrium * at equilibrium “some” of the HA will dissociate, which we call “x”. Therefore, [HA] will decrease by “x” and [H+] = [A-] = x

8. With a known Ka value and plugging in our equilibrium values we get Ka = (x)(x)= x 2 1 – x 1 – x Because the Ka is small, we can assume that “x” is very small in comparison to 1 M, so we can ignore the “x” in the (1 – x) value. So, Ka = x 2 x = [H + ] 1 from this we can get pH!!!!!

9. Real Example Problem What is the pH of a 0.5 Molar solution of benzoic acid, C 6 H 5 COOH, Ka = 6.6 x 10 -5 ? Write the equilibrium reaction and expression C 6 H 5 COOH  H + + C 6 H 5 COO- Initial 0.5 M 0 0 At Eq. 0.5 - x x x Ka = [H + ] [C 6 H 5 COO-] [C 6 H 5 COOH]

10. Ka = [H+] [C 6 H 5 COO-] [C 6 H 5 COOH] 6.6 x 10 -5 = x 2 0.5 - x We can ignore the ‘x’ as being very small compared to 0.5 in the denominator, so the equation becomes 6.6 x 10 -5 = x 2 0.5 Solve for ‘x’ = [H+] = 5.7 x 10 -3 Remember - solve for x 2 and then take square root of both sides to solve for x!!!!

11. Once you have the [H+], you can solve for pH pH = - log [H+] = - log (5.7 x 10 -3 ) = 2.24 Ka = _x 2 [HA] x = [H + ]