For the reaction ${\mathrm{BaF}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Ba}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{HF}\left(\mathrm{aq}\right),K_c=2.56\mathrm{M}$ the solubility of ${\mathrm{BaF}}_2$

in a buffer solution of pH= 2.0 is 

Lets be the solubility of' ${\mathrm{BaF}}_2$ , in a buffer solution of pH= 2 (i.e. $\left[{\mathrm{H}}^{+}\right] = {10}^{-2}\mathrm{M})$. Then ${\mathrm{BaF}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to \underset{\mathrm{s}}{{\mathrm{Ba}}^{2+}(\mathrm{aq}})+\underset{2\mathrm{s}}{2\mathrm{HF}}(\mathrm{aq})$

$\begin{array}{l}{K}_c=\frac{\left[{\mathrm{Ba}}^{2+}\right][\mathrm{HF}{]}^2}{{\left[{\mathrm{H}}^{+}\right]}^2}=\frac{\left(s\right)(2s{)}^2}{{\left({10}^{-2}\mathrm{M}\right)}^2};4s^3={\left({10}^{-2}\mathrm{M}\right)}^2{K}_c={\left({10}^{-2}\mathrm{M}\right)}^2\left(2.56\mathrm{M}\right)\\ s={\left[\frac{\left({10}^{-4}{\mathrm{M}}^2\right)\left(2.56\mathrm{M}\right)}{4}\right]}^{1/3}={\left(64\times {10}^{-6}\mathrm{M}\right)}^{1/3}=4\times {10}^{-2}\mathrm{M}\end{array}$