pH Calculating

Question

Calculate the pH of a solution prepared by dissolving 2.25 g lactic acid (90 g/mol) in 100 mL of 0.25 M NaOH solution. ( Ka (lactic acid)= 1.38x10-4 )

Answer 1

Lactic acid is a weak acid with the formula HC₃H₅O₃ where the bold H is the acid hydrogen. We can abbreviate such as weak acid simply as HA.

A buffer is formed when you have the weak acid and the salt of that weak acid. In this example, we have the weak acid present, and by adding NaOH, we are making the salt (lactate)

The reaction would then be ...

HA + NaOH ==> H₂O + NaA

Eliminating spectator ions, we have ...

HA + OH^- ==> H₂O + A^-

Moles HA present initially = 2.25 g x 1 mol/90 g = 0.025 mols lactic acid

Moles NaOH added = 100 ml x 1 L/1000 ml x 0.25 mol/L = 0.025 mols NaOH

Since moles NaOH = moles HA , we are at the equivalence point. This means ALL of the HA (lactic acid) has reacted with all of the NaOH to produce 0.025 moles of A^- (lactate; C₃H₅O₃^-). This is no longer a buffer since we no longer have any weak acid left.

To find the pH, we are now dealing with a solution of 0.025 moles of C₃H₅O₃^- in 100 mls water (0.25 M). We thus have the following equilibrium ...

C₃H₅O₃^- + H₂O ==> HC₃H₅O₃ + OH^- and C₃H₅O₃^- is acting as a base

Recall that KaKb = Kw so we can use the Ka value provided to find the Kb

Kb = Kw/Ka = 1x10^-14 / 1.38x10^-4 = 7.25x10^-11

Kb = 7.25x10^-11 = [HC₃H₅O₃][OH^-] / [C₃H₅O₃^-]

7.25x10^-11 = (x)(x) / 0.25

x² = 1.81x10^-11

x = 4.26x10^-6 M = [OH^-]

pOH = -log 4.26x10^-6 = 5.37

pH = 14 - 5.37

pH = 8.63