Question

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

• A. $1:6.6$
• B. $1:3.3$
• C. $2:3.3$
• D. $3.3:2$

Answer 1

The correct option is C $2:3.3$

$\begin{array}{ccccccc}HCOOH(aq.)& +& H_{2}O\left(l\right)& \rightleftharpoons & HCO{O}^{-}(aq.)& +& H_3{O}^{+}\\ C_1-x& & & & x& & +x+y\end{array}$
$\begin{array}{ccccccc}HF(aq.)& +& H_{2}O\left(l\right)& \rightleftharpoons & F^{-}(aq.)& +& H_3{O}^{+}(aq.)\\ C_2-y& & & & y& & x+y\end{array}$
$\because \alpha$ is very less for both acid so
$K_{a\left(HCOOH\right)}=\frac{x(x+y)}{C_1};$
$K_{a\left(HF\right)}=\frac{y(x+y)}{C_2}$
$\therefore \frac{x}{y}=\frac{K_{a\left(HCOOH\right)}{C}_1}{K_{a\left(HF\right)}{C}_2}$
$\frac{\left[HCO{O}^{-}\right]}{\left[F^{-}\right]}=\frac{x}{y}$
$\Rightarrow \frac{2\times {10}^{-4}\times 0.2}{6.6\times {10}^{-4}\times 0.1}=\frac{2}{3.3}$