Question

Find the concentration of $H^{+}$ ions in an aqueous solution, which is saturated with $H_{2}S$ ($0.1M$) as well as $H_{2}C{O}_3$ ($0.2M$).

[Use data: $K_1={10}^{-7},K_2={10}^{-14}$ for $H_{2}S,K_1=4\times {10}^{-7},K_2=4\times {10}^{-11}$ for $H_{2}C{O}_3]$

• A. $3\times {10}^{-4}M$
• B. $3.83\times {10}^{-4}M$
• C. $2.83\times {10}^{-4}M$
• D. None of these

Answer 1

The correct option is A $3\times {10}^{-4}M$

Let the contribution of $\left[H^{+}\right]$ from $H_{2}S$ be $x$ and from $H_{2}C{O}_3$ be $y$. Since the second dissociation constants of those acids are very small, we consider only the first dissociation.

$H_{2}S\leftrightharpoons H^{+}+H{S}^{-}$

$x+y$ $x$

$H_{2}C{O}_3\leftrightharpoons H^{+}+HC{O}_3^{-}$

$x+y$ $y$

$K_{1\left(H_{2}S\right)}=\frac{\left[H^{+}\right]\left[H{S}^{-}\right]}{\left[H_{2}S\right]}=\frac{x(x+y)}{0.1}={10}^{-7}$

Given $K_1={10}^{-7}$

$\therefore x(x+y)={10}^{-8}$ --------- $\left(1\right)$

Similarly, $K_{1\left(H_{2}C{O}_3\right)}=\frac{y(x+y)}{0.2}=4\times {10}^{-7}$

$\therefore y(x+y)=8\times {10}^{-8}$ ------- $\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, $y=8x$

Subsitituting this in $\left(1\right)$, $x\left(9x\right)={10}^{-8}$

$⟹x^2=\frac{{10}^{-8}}{9}$

$\therefore x=\frac{1}{3}\times {10}^{-4}$ and $y=\frac{8}{3}\times {10}^{-4}$

Total $H^{+}$ concentration $=x+y=\left(\frac{1+8}{3}\right)\times {10}^{-4}$ $M=3\times {10}^{-4}$ $M$

Note: Hydrogen ion from 2nd dissociation is negligible in both the cases.