Question

Calculate the concentration of $H^{+}$ ions in a solution that consists of a mixture of $0.01M$ $HCl$ and $0.01M$ $CHC{l}_{2}COOH$?

$[K_{a\left(CHC{l}_{2}COOH\right)}=5\times {10}^{-2}]$

• A. $0.0174M$
• B. $0.174M$
• C. $0.107M$
• D. $1.740M$

Answer 1

The correct option is A $0.0174M$

Given that the solution is $0.01MHCl$ and $0.01MCHC{l}_{2}COOH$.

The dissociation constant $\left(K_a\right)$ for $CHC{l}_{2}COOH=5\times {10}^{-2}$

The dissociation reaction is as follows:

$CHC{l}_{2}COOH\rightleftharpoons CHC{l}_{2}CO{O}^{⊝}+H^{\oplus }$

$C$ $0$ $0.01$ (initial)
$C(1-\alpha )$ $C\alpha$ $C\alpha +0.01$ (final)

Here, $C$ is the concentration of acid and $\alpha$ is the degree of dissociation.

$\therefore K_a=\frac{\left[CHC{l}_{2}CO{O}^{-}\right]\left[H^{+}\right]}{\left[CHC{l}_{2}COOH\right]}=\frac{C\alpha \times (C\alpha +0.01)}{C(1-\alpha )}=\frac{\alpha (0.01\alpha +0.01)}{(1-\alpha )}=5\times {10}^{-2}$

or $\frac{0.01\alpha (1+\alpha )}{(1-\alpha )}=5\times {10}^{-2}$ or ${\alpha }^2+6\alpha -5=0$

$\therefore \alpha =0.7416$

$\therefore \left[CHC{l}_{2}CO{O}^{⊝}\right]=0.01\times 0.7416=7.416\times {10}^{-3}M$

$\left[H^{+}\right]=7.416\times {10}^{-3}+0.01=0.0174M$

Hence, the concentration of $H^{+}$ is $0.0174M$.

Hence, the correct answer is option $\text{A}$.