Question

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

Answer 1

Answer: (B)

$HA+NaOH \rightleftharpoons NaA+H_{2}O$
$K=10^{9}$
$\therefore $ For $NaA+H_{2}O \rightleftharpoons NaOH +HA; K=10^{-9}$
$\begin{matrix}\text{Initial}&0.1 M&&\\ \text{At eqm}&\left(0.1-\alpha\right)M&\alpha\,M&\alpha\,M\end{matrix}$
$\frac{\left[NaOH\right]\left[HA\right]}{\left[NaA\right]}=10^{-9}$
$\frac{\alpha^{2}}{0.1-\alpha}=10^{-9}$
As $\alpha$ is very small, $0.1 - \alpha =0.1$
$\alpha^{2}=10^{-9}\times0.1$
$\alpha=10^{-5}$
$\left[OH^{-}\right]=\left[NaOH\right]=10^{-5}M$
$PH=14-pOH=14-5=9$