Question

At 25.0 mL sample of 0.100 M HClO (aq) is titrated with NaOH (aq). What is the pH of the solution after the addition of 25.00 mL of 0.100 M NaOH? Ka of HClO=3.5*10^-8.

The answer might be 4.70 but I don't know. Can anyone explain this problem to me? Full explanation please.

Answer 1

25.0 mL sample of 0.100 M HClO = 25.0 x 0.100 = 2.50 mmol of HClO

25.00 mL of 0.100 M NaOH = 25.00 x 0.100 = 2.50 mmol of NaOH

HClO + NaOH NaClO + H₂O

2.50 moles of NaOH will completely react with 2.50 moles of HClO to give 2.50 moles of NaClO.

Total volume of the solution after the reaction = (25 + 25) = 50 mL

Molarity of NaClO = 2.50/50 = 0.05 M

This NaClO is a salt of strong base and weak acid and therefore it will dissociate in water to give OH^-.

ClO^- + H₂O HClO + OH^-

Initial concentration (M)                                 0.05                         0           0

Change in concentration (M)                           - X                          X          X

Equilibrium concentration (M)                      (0.05 - X)                   X           X

Now, K_a of HClO=3.5 x10^-8

Therefore, K_b of ClO^- = K_w/K_a

                                   = 10^-14/(3.5 x10^-8)

                                   = 2.86 x 10^-7

Now,

K_b = [HClO][OH^-]/[ClO^-]

or, 2.86 x 10^-7 = (X)(X)/(0.05 - X)

or, 2.86 x 10^-7 = (X)(X)/0.05       [(0.05 - X) 0.05 as X << 0.05]

or, X² = 1.43 x 10^-8

or, X = 1.20 x 10^-4

or, [OH^-] = 1.20 x 10^-4

or, - log[OH^-] = - log(1.20 x 10^-4)    [Taking -log function on both sides]

or, pOH = 3.92

or, 14 - pH = 3.92      [As pH + pOH = 14]

or, pH = 10.08

Therefore, the pH of the solution = 10.08