Chemistry - Ionic Equilibrium Question with Solution | TestHub

Explain question :

Question is asking about concentration of ${\mathrm{CH}}_3\mathrm{COOH}$ in solution of ${\mathrm{CH}}_3{\mathrm{COONH}}_4$. Concept :

This question is based on salt hydrolysis×

Solution :-

CH₃COONH₄ is salt of WAWB 0.01 mole & 1 litre

⇒$10^{-2}$

${\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{NH}}_4^{+}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{NH}}_4\mathrm{OH}$

$C=10^{-2}$C 0 0

$C-CHC-CH$ $CH$ $CH$

$h=\sqrt{\frac{K_W}{K_a\times K_b}}$

So

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=C\sqrt{\frac{K_W}{K_a\times K_b}}\left(C=10^{-2}\right)$

$=10^{-2}\frac{10^{-14}}{{\left(1.8\times 10^{-5}\right)}_{}^2}=\frac{10^{-4}}{1.8}$

& Answer

$=\frac{10^{-4}}{1.8}\times 9\times 10^5=50$

The Final answer = $50$