Question

For a weak monobasic acid, if $p{K}_4=4$, then at a concentration of $0.01M$ of the acid solution, the van't Hoff factor is:

• A. $1.01$
• B. $1.02$
• C. $1.10$
• D. $1.20$

Answer 1

The correct option is A $1.01$

Let a weak monobasic acid is $HCN$

Hence the degree of dissociation of $HCN$ is:

$HCN\to H^{+}+C{N}^{-}1001-\alpha \alpha \alpha a(1-\alpha )a\alpha a\alpha$

We know that-$pH=-lo{g}_{10}\left[H^{+}\right]$

${\left[H\right]}^{+}={10}^{-pH}$

Given, $a=0.01$

Again we know that vant Hoff factor,$i=\frac{1+\alpha }{1}$

$H^{+}=a\alpha$

${10}^{-pH}=0.01\alpha$

${10}^{-4}=0.01\alpha \alpha =\frac{{10}^{-4}}{0.01}=0.01$

Hence vant Hoff factor,$i=\frac{1+\alpha }{1}$

$=1+0.01=1.01$

Hence, the correct option is $A$