Chemical and Ionic Equilibrium 2 Question 55
55. The equilibrium constant $K_{p}$ for this reaction at $298 \mathrm{~K}$, in terms of $\beta_{\text {equilibrium }}$ is
(2016 Adv.)
(a) $\frac{8 \beta^{2}{ }{\text {equilibrium }}}{2-\beta{\text {equilibrium }}}$
(b) $\frac{8 \beta^{2}{ }{\text {equilibrium }}}{4-\beta{\text {equilibrium }}^{2}}$
(c) $\frac{4 \beta^{2}{ }{\text {equilibrium }}}{2-\beta{\text {equilibrium }}}$
(d) $\frac{4 \beta^{2}{ }{\text {equilibrium }}}{4-\beta^{2}{ }{\text {equilibrium }}}$
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Solution:
- Partial pressure of $\mathrm{SO}_{2}$ in air $=10^{-5}$ atm
$$ \left[\mathrm{SO}{2}\right]{a q}=1.3653 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} $$
$\because \mathrm{p} K_{a}=1.92$ and concentration of $\mathrm{H}{2} \mathrm{SO}{3}$ is very low, it is almost completely ionised as
$$ \begin{gathered} \mathrm{H}{2} \mathrm{SO}{3} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HSO}_{3}^{-} \ {\left[\mathrm{H}^{+}\right]=1.3653 \times 10^{-5} \mathrm{M}} \ \mathrm{pH}=-\log \left(1.3653 \times 10^{-5}\right)=4.86 \end{gathered} $$
