Question

A solution of 0.357 M HClO (K_a = 3.5 x 10^-8) with volume 75. mL is mixed with 50. mL of a 0.800 M KOH solution. What is the final pH?

Answer 1

Moles of HClO = 0.357 M x 75. mL = 27 mmol

Moles of KOH = 0.800 M x 50. mL = 40 mmol

HClO + KOH KClO + H₂O

Theoretically, 1 mol of HClO gets completely neutralized by 1 mol of KOH.

Therefore, 27 mmol of HClO is completely neutralized by 27 mmol of KOH.

Hence, the unreacted KOH in the solution = (40 - 27) = 13 mmol

Total volume of the solution after mixing = (75. + 50.) mL =125 mL

Concentration of unreacted KOH = 13 mmol/125 mL

                                                      = 0.10 M

KOH is a strong base. Thus, 0.10 M of KOH completely dissociate in water to give 0.10 M of OH^- ions.

Now,

pOH = - log[OH-]

        = - log(0.10)

        = 1.0

And,

pH = 14 - pOH

      = 14 - 1.0

      = 13

Hence, the final pH of the solution = 13