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

Given:

pH = 9

Concentration = 0.01 M

To Find:

The pKa of the acid.

Calculation:

- Let the acid be HA

 A⁻  +  H2O  →  HA  +  OH⁻

0.01                   0         0

(0.01-x)               x         x

K = x²/(0.01-x)

⇒ Kw/Ka = x²/(0.01-x)

⇒ Ka = Kw × (0.01-x)/ x²

⇒ pKa = pKw + 2 log x - log (0.01-x)

- pH = - log x

⇒ 9 = - log x

⇒ x = 10⁻⁹

- Putting in the eqn of pKa and as x << 1, we neglect it in (0.01-x), we get:

pKa = pKw + 2 log 10⁻⁹ - log 0.01

⇒ pKa = 14 - 18 + 2

⇒ pKa = -2

- The pKa value of the acid will be -2.