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Question

1a. Write a balanced equation and corresponding Ka expression for the ionization of the weak acid lactic acid (HC3H5O3, Ka= 1.62 x 10^-4) in water.

b. A buffer solution is prepared by dissolving together 0.020mol of lactic acid and 0.015mol of sodium lactate, NaC3H5O3, in enough water to make 250.0mL of solution. Determine the pH of this solution

Answer 1

Hello Divine

1a) HC3H5O3 (aq) + H2O (l) <=> H3O+ (aq) + C3H5O3- (aq)

Ka = [H3O+] [C3H5O3-] / [HC3H5O3] = 1.6 x 10^-4

b) Molarity of Lactic acid = 0.020 mol / 0.2500 L = 0.080 M

Molarity of Sodium lactate = 0.015 mol / 0.2500 L = 0.060 M

Use the Hendersen-Haaselbach equation to solve for the pH

pH = pKa + log ([C3H5O3-] / [HC3H5O3])

Ka = 1.62 x 10^-4

pKa = -log 1.62 x 10^-4 = 3.790

pH = 3.790 + log (0.060 / 0.080)

= 3.790 - 0.125

= 3.665