Question

Titration: Calculate the pH of each solution listed below. You are starting with a solution of NaClO

and titrating that solution with HCl.

A) What is the pH of 25.0 mL of 0.035 M NaClO? The Kb of ClO- is 2.86 x 10-7?

B) What is the pH when you add 12.5mL of 0.035M HCl?

C) What is the pH when you add 25.0 mL of 0.035 M HCl?

Answer 1

A) NaClO is salt of wak acid and strong base and weak acid.

for such salts pH = 1/2 [pKw + pKa + logC]

pKb = - log Kb = - log [2.86 x 10^-7] = 6.55

pka = 14 - 6.55 = 7.45

pH = 1/2 [14 + 7.45 + log 0.035]

pH = 10.0

B) millimoles of NaClO = 25 x 0.035 = 0.875

millimoles of HCl added = 12.5 x 0.035 = 0.4375

means 50% of NaClO convert to HCLO

so now in solution

[NaClO] = [HClO]

pH = pKa + log [NaClO] /[HClO]

[NaClO] = [HClO] so

pH =pKa

pH = 7.45

C) millimoles of HCl added = 25 x 0.035 = 0.875

means all NaCLO becomes HOCl

pH = 1/2 [pKa - logC]

C = [HClO] = 0.035 / 2 = 0.0175 M

pH = 1/2 [7.45 - log 0.0175]

pH = 4.60