Question

The pH of a 0.01 M solution of the sodium salt of a monocarboxylic acid is 9. The pKa of the acid will be ………

Answer 1

Thus the pKa value of the acid is p K  a = 6

Explanation:

K  a H A  = K  w  ( 0.01 − x   /  x^2 )

[p K a  = l K  w  − log( 0.01 − x) + 2 log x ]     −−−(1)

pH = − log x = 9

⇒  x = 10 ^-9

Putting this in (1) and joining x in ( 0.01 − x )

as x << 0.01

p K  a   = p K  w  − log 0.01 + 2 log x

p K  a = 14 − 9 + 1

p K  a = 6

Thus the pKa value of the acid is p K  a = 6