Question

$\Lambda$ of a $\frac{M}{32}$ solution of a weak acid is $8\text{S}{\text{cm}}^2{\text{mol}}^{-1}$ and the limiting molar conductivity is $400\text{S}{\text{cm}}^2{\text{mol}}^{-1}$. $K_a$ of the acid is:

• A. $1.25\times {10}^{-6}$
• B. $6.25\times {10}^{-4}$
• C. $78.25$
• D. $1.25\times {10}^{-5}$

Answer 1

The correct option is D $1.25\times {10}^{-5}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{8}{400}=\frac{1}{50}$

$K_a=\frac{C{\alpha }^2}{1-\alpha }\alpha <<1(1-\alpha \approx 1)$

$K_a=C{\alpha }^2$

$K_a=\frac{1}{32}\times \frac{1}{50}\times \frac{1}{50}$

$=\frac{1}{800}\times {10}^{-2}=\frac{1000}{800}\times {10}^{-5}$

$K_a=1.25\times {10}^{-5}$