What is the PH at equivalence point when 25ml of 0.100M HClO is titrated with 0.200M KOH ..ka HClO=2.9*10^-8

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SOLUTION:
At the equivalence point, the number of moles of HClO and KOH are the same. Hence,
HClO + OH- ↔ ClO- + H2O
Initial mole 0.0025 0.0025 ----
Change -0.0025 -0.0025 +0.0025
Equilibrium mole ------------ ---------- +0.0025
Equilibrium concentration ------------ ---------- 0.0025 mol/0.0375 L
= 0.0667 M
NOTE:
Computing for the volume of KOH required to reach the equivalence point:
Volume = 0.0025 mol / 0.2 M
Volume = 0.0125 L = 12.5 mL
Hence,
Total volume = 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

At the equivalence point, the pH of the solution is dependent on the hydrolysis of ClO-:
ClO- + H2O ↔ HClO + OH-
Initial 0.0667 ----- ----
Change -x +x +x
Equilibrium 0.0667 - x x x

Kb = [HClO][OH-] / [ClO-]
Kb= Kw / Ka = 1 x 10-14 / 2.9 x 10-8 = x2 / 0.0667 - x
3.45 x 10-7 = x2 / 0.0667 - x
Assume x as negligible:
3.45 x 10-7 = x2 / 0.0667
x2 = 2.3 x 10-8
x = 1.517 x 10-4 = [OH-]

pOH = -log [OH-]
pOH = - log 1.517 x 10-4
pOH = 3.819

pH = 14 - pOH
pH = 14 - 3.819
pH = 10.181

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