Calculate the ratio of  $\left[HCO{O}^{-}\right]\&\left[F^{-}\right]$ in a mixture of $0.2MHCOOH(K_a=2\times {10}^{-4})$ &  $0.1MHF(K_a=6.6\times {10}^{-4})$

$$\begin{gathered} \underset{c_1-x}{HCOO{H}_{\left(aq\right)}}+H_2{O}_{\left(aq\right)}\rightleftharpoons \underset{x}{HCO{O^{-}}_{\left(aq\right)}}+\underset{x}{H_3{O^{\oplus }}_{\left(aq\right)}} \\ \underset{c_2-y}{H{F}_{\left(aq\right)}}+H_2{O}_{\left(l\right)}\rightleftharpoons \underset{y}{{F^{-}}_{\left(aq\right)}}+\underset{y}{H_3{O^{\oplus }}_{\left(aq\right)}} \end{gathered}$$

Due to the common ion effect, the concentration of H₃O⁺ will be x+y in both the equations. Since the value is acid constants for both compounds are very less, hence we can neglect x in case of c₁-x for HCOOH and y in case of c₂-y for HF.

$$\begin{gathered} K_a\left(HCOOH\right)=\frac{x(x+y)}{c_1};K_b\left(HF\right)=\frac{y(x+y)}{c_2} \\ \frac{x}{y}=\frac{K_a\left(HCOOH\right)c_1}{K_a\left(HF\right)c_2} \\ \frac{\left[HCO{O}^{-}\right]}{\left[F^{-}\right]}=\frac{x}{y} \\ \frac{2\times {10}^{-4}\times 0.2}{6.6\times {10}^{-4}\times 0.1}=\frac{2}{3.3} \end{gathered}$$