Question

Hydrofluoric acid is weak acid. At ${25}^{o}C$, the molar conductivity of $0.002M$ $HF$ is $176.2{\Omega }^{-1}$ ${cm}^2$ ${mol}^{-1}$. If its ${\Lambda }_m^o=405{\Omega }^{-1}$ ${cm}^2$ ${mol}^{-1}$. Equilibrium constant at the given concentration is:

• A. $6.7\times {10}^{-4}M$
• B. $3.2\times {10}^{-4}M$
• C. $6.7\times {10}^{-5}M$
• D. $3.2\times {10}^{-5}M$

Answer 1

The correct option is A $6.7\times {10}^{-4}M$

$\alpha =\frac{{\Lambda }_c}{{\Lambda }_m^o}=\frac{176.2}{405.2}=0.435$

$K=\frac{\left[H^{+}\right]\left[F^{-1}\right]}{\left[HF\right]}=\frac{C{\alpha }^2}{1-\alpha }$$=0.002M\times \frac{{0.435}^2}{1-0.435}=6.7\times {10}^{-4}M$

Hence, option $A$ is correct.