Question

What will be the pH at the equivalence point in the titration of 45.00 mL of 0.115 M NaClO with 0.106 M HCl? ClO–(aq) + HCl(aq) → HClO(aq) + Cl–(aq) Ka(HClO)=3.0·10-8 You only need to provide the final answer (pH).

Answer 1

NaClO + HCl -------------------------> HClO + NaCl

both HCl and NaClO consumed in the reaction only HClO remains

HCl volume needed = 45 x 0.115 / 0.106

                                = 48.82 mL

[HOCl] = 45 x 0.115 / (45 + 48.82)

   C        = 0.0552 M

pKa = -log Ka

pKa = -log (3 x 10^-8)

pKa = 7.52

pH = 1/2 [pKa - log C]

pH = 1/2 [7.52 - log 0.0552]

pH = 4.39