Question

What is the final pH of the solution when 25mL of 0.131 mol/L NaOH(aq) and 45mL of 0.191 mol/L CH3COOH(aq) are mixed? Assume the temperature is 25°C. (For CH3COOH, pKa = 4.74 at 25°C)

Answer 1

K_a = 10^-pKa = 1.82*10^-5

CH₃COOH(aq) + NaOH(aq) -----> CH₃COONa(aq) + H₂O(l)

moles of NaOH present = molarity*volume of solution in litres = 0.131*0.025 = 0.0033

Moles of CH₃COOH present = 0.191*0.045 = 0.0086

Thus, moles of CH₃COOH unreacted = 0.0053

Molarity of unreacted CH₃COOH = moles/volume of solution in litres = 0.0053/0.07 = 0.076 M

Now,

CH₃COOH(aq) <-------> CH₃COO^-(aq) + H⁺(aq)

Initially, [CH₃COOH] = 0.076 M; [CH₃COO^-] = 0 = [H⁺]

Let at eqb., [CH₃COOH] = (0.076-x) M; [CH₃COO^-] = x = [H⁺]

K_a = {[CH₃COO^-]*[H⁺]}/[CH₃COOH]

or, 1.82*10^-5 = x²/(0.076-x)

or, x = [H⁺] = 0.00116 M

Thus, pH = -log[H⁺] = 2.94