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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=1.62x10^-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 the solution. Determine the pH of this solution.

Answer 1

a.  HC3H5O3 ==>H⁺ + C3H5O3^-

 

b. pH= pKa + log [lactate]/[lactic acid]

pKa = -log Ka = -log 1.62x10^-4 = 3.79

[lactate] = 0.015 mol/0.25 L = 0.06 M

[lactic acid] = 0.02 mol/0.25 L = 0.08 M

pH = 3.79 + log (0.06/0.08) = 3.79 + (-0.125)

pH = 3.79 - 0.125 = 3.67