Question

Calculate the pH of a solution that is composed of 90.0 mL of 0.345 M

sodium hydroxide, NaOH, and 50.0 mL of 0.123 M lactic acid,
CH3COHCOOH.
(Ka of lactic acid = 1.38x104)

Answer 1

To solve this problem, we need to use the equation for the ionization of lactic acid:

CH3COHCOOH + H2O ⇌ CH3COHCOO- + H3O+

The equilibrium constant expression for this reaction is:

Ka = [CH3COHCOO-][H3O+] / [CH3COHCOOH]

We can assume that the concentration of [H3O+] is the same as the concentration of [OH-] because NaOH is a strong base and completely dissociates in water:

[OH-] = 0.345 M x 90.0 mL / 1000 mL = 0.031 M

Now we can use the equilibrium constant expression to calculate [H3O+]:

1.38x10^-4 = [CH3COO-][H3O+] / [CH3COHCOOH]

[CH3COO-] = 0.123 M x 50.0 mL / 1000 mL = 0.00615 M

[CH3COOH] = 0 (since it is completely consumed in the reaction)

[H3O+] = Ka x [CH3COHCOOH] / [CH3COO-] = 1.38x10^-4 x 0 / 0.00615 = 0

pH = -log[H3O+] = -log(0) = undefined

Therefore, the pH of the solution cannot be calculated, as it is not acidic or basic.