Question

$\left[H^{+}\right]$ in a solution containing $0.1M$ $HCOOH$ and $0.1M$ $HOCN$ is $X\times {10}^{-3}M$. Nearest integer to X is
$K_a$ for $HCOOH$ and $HOCN$ are $1.8\times {10}^{-4}$ and $3.3\times {10}^{-4}$ respectively

Answer 1

$\frac{aX}{0.1-a}=1.8\times {10}^{-4}$

$aX=1.8\times {10}^{-5}$......(1)

$\frac{bX}{0.1-b}=3.3\times {10}^{-4}$

$bX=3.3\times {10}^{-5}$......(2)

Divide equation (1) with equation (2).
$\frac{a}{b}=\frac{1.8\times {10}^{-5}}{3.3\times {10}^{-5}}=0.5455$

$a=0.5455b$......(3).

$bX=b(a+b)=3.3\times {10}^{-5}$

$bX=b\left(1.5455b\right)=3.3\times {10}^{-5}$

$b=4.626\times {10}^{-3}$

$a=0.5455b=2.524\times {10}^{-3}$

$X=a+b=7.15\times {10}^{-3}\simeq 7\times {10}^{-3}=x\times {10}^{-3}$

Thus, $x=7$