Calculate the pH for each of the following cases in the titr

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.110 M HClO(aq) with 0.110 M KOH(aq). The ionization constant for HClO is 4.0*10-8.

(a) before addition of any KOH

(b) after addition of 25.0 mL of KOH

(c) after addition of 30.0 mL of KOH

(d) after addition of 50.0 mL of KOH

(e) after addition of 60.0 mL of KOH

Solution

(d) after addition of 50 ml KOH,

moles of KOH = 50 * 0.11 = 5.5 mmol

moles of HClO = 50 * 0.11 = 5.5 mmol

occurred hydrolysis solution,

pH = 0.5(14 + pKa + log [base conjugate])

pH = 0.5(14 + (- log (4 * 10^-8)) + log (5.5/(50 + 50)))

pH = 2.67