Question

Limiting molar conductance of $H^{+}$ and ${CH}_3{COO}^{-}$ ions are $344$ and $40$ respectively. Molar conductance of $0.008M{CH}_{3}COOH$ is $48$. What will be the value of $K_a$ for ${CH}_{3}COOH$:-

• A. $1.4\times {10}^{-5}$
• B. $1.2\times {10}^{-5}$
• C. $1.4\times {10}^{-4}$
• D. $1\times {10}^{-5}$

Answer 1

The correct option is C $1.4\times {10}^{-4}$

${\lambda }_{m\left(C{H}_{3}COOH\right)}^{\infty }={\lambda }_m^{\infty }\left(H^{+}\right)+{\lambda }_{m\left(C{H}_{3}CO{O}^{-}\right)}^{\infty }$

$=344+40$

$=384$

${\lambda }_{mc}=48$

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

$=\frac{48}{384}$

$=0.125$

Now, $C{H}_{3}COOH⟷C{H}_{3}CO{O}^{-}+H^{+}$

C. 0 0

$C(1-\alpha )$ $C_{\alpha }$ $C_{\alpha }$

$K_a=\frac{\left(C_{\alpha }\right)\left(C_{\alpha }\right)}{C(1-\alpha )}$

$=\frac{C_{\alpha 2}}{1-\alpha C}$

$=0.008M$

So, $K_a=0.008\times \frac{\left(0.125\right)2}{1-0.125}$

$=\frac{0.000125}{0.875}$

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