Question

Calculate $\left[H^{+}\right]$ in a solution containing $0.1$M $HCOOH$ and $0.1$M $HOCN$. $K_a$ for $HCOOH$ and $HOCN$ are $1.8\times {10}^{-4}$ and $3.3\times {10}^{-4}$ respectively.

Answer 1

Given that $K_a$ for $HCOOH=1.8\times {10}^{-4}$

$K_a$ for $HOCN=3.3\times {10}^{-4}$

$\frac{\left[H^{+}\right]\left[HCO{O}^{-}\right]}{\left[HCOOH\right]}=1.8\times {10}^{-4};\frac{\left[H^{+}\right]\left[OC{N}^{-}\right]}{\left[HOCN\right]}=3.3\times {10}^{-4}$

$\underset{\downarrow }{\frac{(X+Y)X}{1-X}=1.8\times {10}^{-4}};\underset{\downarrow }{\frac{(X+Y)Y}{1-Y}=3.3\times {10}^{-4}}$

$\left(1\right)$ $\left(2\right)$

$\frac{\left(1\right)}{\left(2\right)}=\frac{X}{Y}=\frac{1.8}{3.3}(\because 1-X=1,1-Y=1)$

$Y=\frac{3.3\times X}{1.8};X=$

Substitute in (1)

$\frac{5.1}{1.8}{X}^2=1.8\times {10}^{-4}\Rightarrow X=7.97\times {10}^{-3}$

$Y=14.612\times {10}^{-3}$

$X+Y=22.582\times {10}^{-3}$

$\left[H^{+}\right]=0.2582$