pH of acid solution
[H+] = √Ka×C
Ka = 1.35 × 10−3 , C = 0.1 M
[H+] = √1.35×10−3×0.1 = √1.35×10−4 = 1.16 × 10−2
pH = – log [H+] = – log(1.16 × 10–2) = 2 – 0.064 = 1.936
pH of 0.1 M sodium salt solution
Sodium salt of chloroacetic acid is salt of weak acid and strong base.
Hence,
pH =

1

2

[pKw+pKa+logC]
Ka = 1.35 × 10−3
pKa = - log Ka = - log (1.35 × 10−3) = - (0.1303 - 3) = 2.8697
pKw = - log 10−14 = 14
C = 0.1 M , log C = log (0.1) = - 1 ⇒ pH =

1

2

[14 + 2.8697 - 1] = 7.935