Question

$\alpha$ of $HA(K_e={10}^{-5})$ is $0.01$ at one molar concentration, what is $\alpha$ when concentration is $4M$?

• A. $0.05$
• B. $0.04$
• C. $0.005$
• D. $0.0025$

Answer 1

Answer: (A)

$0.05$

Given At $0.01=\alpha$ $C=1$ $M$ $K_e={10}^{-5}$

For $C=4$ $M$

$K_c=C{\alpha }^2⟹{10}^{-5}=4{\alpha }^2$

${\alpha }^2=\frac{{10}^{-5}}{4}=0.25\times {10}^{-5}⟹{\alpha }^2=25\times {10}^{-4}$

Hence $\alpha =5\times {10}^{-2}=0.05$