Question

If equilibrium constant of the reaction of weak acid HA with $NaOH$ is $10^9$, then pH of $0.1MNaA$ 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 $NaA+H_{2}O\rightleftharpoons NaOH+HA;K=10^{-9}$
$\begin{array}{cccc}\text{Initial}& 0.1M& & \\ \text{At eqm}& \left(0.1-\alpha \right)M& \alpha M& \alpha M\end{array}$
$\frac{\left[NaOH\right]\left[HA\right]}{\left[NaA\right]}=10^{-9}$
$\frac{{\alpha }^2}{0.1-\alpha }=10^{-9}$
As $\alpha$ is very small, $0.1-\alpha =0.1$
${\alpha }^2=10^{-9}\times 0.1$
$\alpha =10^{-5}$
$\left[O{H}^{-}\right]=\left[NaOH\right]=10^{-5}M$
$PH=14-pOH=14-5=9$