Browse Questions

# If equilibrium constant of the reaction of weak acid $HA$ with $NaOH$ is $10^9$, then $pH$ of $0.1\; M\; NaA$ is

(a) 5

(b) 9

(c) 7

(d) 8

$HA + NaOH \rightleftharpoons NaA + H_2O$

$K = 10^9$

$\therefore$ For $NaA + H_2O \rightleftharpoons NaOH + HA; K = 10^9$
Initialliy, [NaA] = 0.1 M
At equilibrium, $[NaA] = (0.1-\alpha) M, [NaOH] =\alpha M, [HA] = \alpha M$

$\frac{[NaOH][HA]}{[NaH]} = 10^{-9}$
$\frac{\alpha^2}{0.1 - \alpha} = 10^{-9}$

As $\alpha$ is very small, $0.1 - \alpha \approx 0.1$
$\therefore \alpha^2 = 10^{-9} \times 0.1$
$\Rightarrow \alpha = 10^{-5}$

$[OH^-] = [NaOH] = 10^{-5} M$

$pH = 14 - pOH = 14 - (- log 10^{-5}) = 14 - 5 = 9$