Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{ \; \; \; \; \; \; \; \; \; \; \; }{\text{CH}}_3{\text{NH}}_2+\text{HCl}\rightleftharpoons {\text{CH}}_3{\text{NH}}_3^{+}{\text{Cl}}^{-}$

$\mathrm{Initial\; moles\; \; \; \; 0.1\; \; \; \; \; \; 0.08\; \; \; \; 0}$

Resulting solution contains [salt] = 0.08

[base] = 0.1 - 0.08 = 0.02

$\text{Applying pOH}={\text{pK}}_{\text{b}}+\text{log}\frac{\left[\text{salt}\right]}{\left[\text{base}\right]}$

pK_b= -log K_b= -log 5 × 10^-4= 4 - log 5 = 3.30

$\text{pOH}=\text{3.30}+\text{log}\frac{\text{0.08}}{\text{0.02}}=\text{3.30}+\text{0.60} =\text{3.90}$

pH + pOH = 14, pH = 14 - 3.902 = 10.1

Applying pH = -log [H⁺]

$10.1=-\log \left[{\text{H}}^{+}\right]$

So${\text{H}}^{+}=8\times {10}^{-11}$

[H⁺] ≫ 8 × 10^-11