For a weak monobasic acid, if ${\mathrm{pK}}_{\mathrm{a}}$ = 4, then at a concentration of 0.01 M of the acid solution, the Van’t Hoff factor is

Let a weak monobasic acid be HCN

Hence the degree of dissociation of HCN is:

$\mathrm{HCN} \to {\mathrm{H}}^{+}+ {\mathrm{CN}}^{-}$

$1 0 0$

$1-\alpha \alpha \alpha$

$a(1-\alpha ) a\alpha a\alpha$

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

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

Given, a=0.01

Again we know that Van't Hoff factor, 

$i=\frac{1+\alpha }{1}$

${\mathrm{H}}^{+}=\mathrm{a}\alpha$

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

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

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

Hence Van't Hoff factor,
$i=\frac{1+\alpha }{1}$

=1+0.01

=1.01

Hence, the correct option is A