Calculate the pH of a 0.25 M NaOH solution. Note: Kw = [OH-] x [H3O+] = 1.0 × 10–14, pH = -log[H3O+]

Question

Calculate the pH of a 0.25 M NaOH solution. Note: Kw = [OH-] x [H3O+] = 1.0 × 10–14, pH = -log[H3O+]

Answer 1

0.25 M NaOH ==> 0.25 M Na+ + 0.25 M OH^-

Two ways to calculate pH:

1). [H₃O⁺][OH^-] = 1x10^-14

[H₃O⁺]0.25] = 1x10^-14

[H₃O⁺] = 4x10^-14

pH = -log [H₃O⁺] = -log 4x10^-14 = 13.40

2). pOH = -log [OH-] = -log 0.25 = 0.602

pH + pOH = 14

pH = 14 - 0.602

pH = 13.40