Chemistry - Redox Reactions Question with Solution | TestHub

$\mathrm{InH}\rightleftarrows \mathrm{I}{\mathrm{n}}^{-}+{\mathrm{H}}^{+}{K}_{\mathrm{In}}=10^{-5}$

$0.2\mathrm{M}{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_3\underset{0.1\mathrm{M}\mathrm{KOH}}{\overset{40\mathrm{ml}}{\to }}\mathrm{KHS}{\mathrm{O}}_3$ (4 oneq)

$\mathrm{pH}=\frac{3+6}{2}=4.5$

1ˢᵗ equation point

$10^{-5}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{In}}^{-}\right]}{[\mathrm{InH}]}$ for 66.6% ionization

$10^{-5}=\frac{\left[{\mathrm{H}}^{+}\right][66.6]}{[33.3]}$

$\Rightarrow \left({\mathrm{H}}^{+}\right)=\frac{10^{-5}}{2}\Rightarrow \mathrm{pH}=5.3$

The 1ˢᵗ equivalent point has a pH of 4.5. Hence, the titration should be continued but should not reach the $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_3$ state (as pH exceeds 7). We want only 5.3, which means we should reach the $\mathrm{HS}{\mathrm{O}}_3^{-}/\mathrm{S}{\mathrm{O}}_3^{2-}$ buffer of 5.3 pH.

$5.3=\mathrm{pKa}=\log \frac{\left({\mathrm{HSO}}_3^{-}\right)}{\left({\mathrm{SO}}_3^{2-}\right)}$

$5.3=6-\log \left(\frac{4-x}{x}\right)$

$\Rightarrow 0.7=\log \left(\frac{4-x}{x}\right)\Rightarrow \frac{4-x}{x}=5\Rightarrow 6x=4$