Question

If the Ka of a weak monoprotic acid is 4* 10-6 and
its concentration is 0.2 M then pH of solution is
(1) 305
(2) 3.72
(3) 3.99
(4)4.62​

Answer 1

Answer:

See below.

Explanation:

We are given,

$\begin{aligned}K_{a}=4\times 10^{-6}\\ C=0.2\end{aligned}$

For a weak acid, we know,

$\begin{aligned}K_{a}=C\alpha ^{2}\\ \Rightarrow \alpha =\left( \dfrac {K_{a}}{c}\right) ^{\dfrac {1}{2}}\end{aligned}$

Calculating the degree of dissociation ( alpha ),

$\begin{aligned}\alpha ^{2}=\dfrac {4\times 10^{-6}}{0.2}=2\times 10^{-5}\\ \Rightarrow \alpha =0.447\times 10^{-2}\end{aligned}$

Calculating the concentration of H+ ions,

$\begin{aligned}\left[ H^{+}\right] =C\alpha =0.2\times 0.477\times 10^{-2}\\ =0.0954\times 10^{-2}\\ =9.54\times 10^{-4}\end{aligned}$

Calculating pH,

$\begin{aligned}pH=-\log \left[ H^{+}\right] \\ =-\log \left( 9.54\times 10^{-4}\right) \\ =-\left( 0.98-4\right) \\ =3.02\end{aligned}$

The correct option should be 3.05 ( nearest answer ).