Calculate the pH, and [H+] and [OH-] concentration

Question

Calculate the pH and [H+] and [OH-] concentration for a solution prepared by mixing 20.0mL of 0.25M acetic acid and 10.0mL of 0.25M sodium acetate.

Answer 1

Acetic acid = CH₃COOH

Sodium acetate = CH₃COONa

The mixture of these two compound forms a buffer because you have the weak acid and its conjugate base.

For such a buffer, we can use the Henderson Hasselbalch equation to find the pH but we first need to know the Ka or pKa for acetic acid. Looking it up I find the pKa to be 4.75.

pH = pKa + log [CH₃COONa] /[CH₃COOH] ... Henderson Hasselbalch equation

moles CH₃COOH = 20.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.005 mols

moles CH₃COONa = 10.0 ml x 1 L/1000 ml x 0.25 mol/L = 0.0025 mols

Total volume = 20 ml + 10 ml = 30 ml = 0.030 L

[CH₃COOH] = 0.005 mol / 0.030 L = 0.1667 M

[CH₃COONa] = 0.0025 mol / 0.030 L = 0.0833 M

pH = pKa + log [CH₃COONa] /[CH₃COOH]

pH = 4.75 + log (0.0833/0.1667) = 4.75 + (-0.301)

pH = 4.45

From the pH, we can easily find the [H⁺] and [OH^-]:

pH = -log[H⁺]

[H⁺] = 1x10^-pH = 1x10^-4.45

[H⁺] = 3.55x10^-5 M

[H⁺][OH^-] = 1x10^-14

[OH^-] = 1x10^-14 / 3.55x10^-5

[OH^-] = 2.82x10^-10 M