Question

The pH of 0.1M monobasic acid solution is 2.Calculate degree of dissociation and dissociation constant.

Answer 1

general formula of monobasic acid is HB.

now, dissociation of HB is ....

$HB\Leftrightarrow H^++B^-$

Let degree of dissociation is $\alpha$

at equilibrium, $[H^+]=C\alpha$, $[B^-]=C\alpha$ and $[HB]=C(1-\alpha)$

where C is concentration of acid.

pH = -log[H^+]

2 = -log(0.1$\alpha$)

10^-2 = 10^-1$\alpha$

$\alpha$ = 0.1

now, dissociation constant, $K=\frac{[H^+][B^-]}{[HB]}$

= $\frac{C\alpha.C\alpha}{C(1-\alpha)}$

= $\frac{C\alpha^2}{(1-\alpha)}$

= 0.1×(0.1)²/(1 - 0.1)

= 0.1 × 0.01/0.9

= 0.01/9

= 1.111 × 10^-3 mol/L