Question

Limiting molar conductance of ${\mathrm{H}}^{+}$ and ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$are $344$and $40$ respectively. Molar conductance of $0.008\mathrm{M}$ ${\mathrm{CH}}_3\mathrm{COOH}$ is $48$. Find the value of $\mathrm{Ka}$ for ${\mathrm{CH}}_3\mathrm{COOH}$.

Answer 1

Step 1: Collecting Given data

Limiting molar conductance of ${\mathrm{H}}^{+}=344$

Limiting molar conductance of ${\mathrm{CH}}_3{\mathrm{COO}}^{-}=40$

Molar conductance of $0.008\mathrm{M}$ ${\mathrm{CH}}_3\mathrm{COOH}=48$

Step 2: Molar conductance of ${\mathrm{CH}}_3\mathrm{COOH}$

${\mathrm{\lambda }}_{\mathrm{m}}^{\infty }\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\mathrm{\lambda }}_{\mathrm{m}}^{\infty }\left({\mathrm{H}}^{+}\right)+{\mathrm{\lambda }}_{\mathrm{m}}^{\infty }\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)$

$=344+40\Rightarrow 384$

${\mathrm{\lambda }}_{\mathrm{mc}}$ of ${\mathrm{CH}}_3\mathrm{COOH}=48$

Now, degree of dissociation$\left(\mathrm{\alpha }\right)=\frac{{\mathrm{\lambda }}_{\mathrm{mc}}}{{\mathrm{\lambda }}_{\mathrm{m}}^{\infty }}$

$\mathrm{\alpha }=\frac{48}{384}\Rightarrow 0.125$

Step 3: Calculating $\mathrm{Ka}$ for ${\mathrm{CH}}_3\mathrm{COOH}$

$\underset{\mathrm{Acetic}\mathrm{acid}}{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{l}\right)}\rightleftharpoons \underset{\mathrm{Acetate}\mathrm{ion}}{{\mathrm{CH}}_3{\mathrm{COO}}^{-}}+\underset{\mathrm{Hydrogen}\mathrm{ion}}{{\mathrm{H}}^{+}}$

Let, Initial concentration of acetic acid$=\mathrm{C}$

The initial concentration of Acetate and hydrogen ion$=0$

The concentration of acetic acid after some time$=\mathrm{C}(1-\mathrm{\alpha })$

The concentration of acetate and hydrogen ion after some time$={\mathrm{C}}_{\mathrm{\alpha }}$

$\mathrm{Ka}=\frac{\left({\mathrm{C}}_{\mathrm{\alpha }}\right)\left({\mathrm{C}}_{\mathrm{\alpha }}\right)}{\mathrm{C}(1-\mathrm{\alpha })}$

$=\frac{{\mathrm{C}}_{{\mathrm{\alpha }}^2}}{1-\mathrm{\alpha C}}$$\Rightarrow 0.008\times \frac{{(0.125)}^2}{1-0.125}\Rightarrow \frac{0.000125}{0.875}$

$=1.4\times {10}^{-4}$

Therefore, $\mathrm{Ka}$ for ${\mathrm{CH}}_3\mathrm{COOH}=1.4\times {10}^{-4}$.