Question

If 0.1M $C{H_3}COOH$ is mixed with 0.1M $C{H_2}ClCOOH$,
[Given: ${K_a}{\text{ C}}{{\text{H}}_3}COOH = 1.8 \times {10^{ - 5}}$ , ${K_a}{\text{ C}}{{\text{H}}_2}ClCOOH = 1.8 \times {10^{ - 4}}$ ]
Find out total $[{H^ + }]$
(A) $0.404 \times {10^{ - 2}}$
(B) $1.44 \times {10^{ - 3}}$
(C) $5.58 \times {10^{ - 3}}$
(D) $8.44 \times {10^{ - 4}}$

Answer 1

Hint:Find the concentration of protons given by each acid after dissociation using dissociation constant and then add the concentration of protons given by both the acids to get the final concentration of protons in the solution.

Complete answer:
We are given the dissociation constants of two weak acids. We will find the protons given by each acid in order to find the concentration of total protons present in the solution.
- We know that dissociation constant shows the ability of the acid to dissociate into ionic form. The equation for acid HA can be given as
     \[{K_a} = \dfrac{{[{\text{Product]}}}}{{[{\text{Reactant]}}}} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}\]
Thus, for given acid $C{H_3}COOH$ , dissociation reaction can be given as
     \[C{H_3}COOH \to {H^ + } + C{H_3}CO{O^ - }\]
we can give its dissociation constant as
     \[{K_a} = \dfrac{{[{H^ + }][C{H_3}CO{O^ - }]}}{{[C{H_3}COOH]}}\]
We know that ${K_a} = 1.8 \times {10^{ - 5}}$, $[C{H_3}COOH]$= 0.1 M and suppose that $[{H^ + }]$ =$[C{H_3}CO{O^ - }]$ = x
So, we can write the above equation as
     \[1.8 \times {10^{ - 5}} = \dfrac{{(x)(x)}}{{0.1}}\]
So, we can write that
     \[{x^2} = (0.1)(1.8 \times {10^{ - 5}}) = 1.8 \times {10^{ - 6}}\]
 Thus, we can say that $x = 1.34 \times {10^{ - 3}}$
So, we obtain that concentration of protons of $C{H_3}COOH$ will be $1.34 \times {10^{ - 3}}M$
Now, dissociation reaction of $C{H_2}ClCOOH$ can be given as
     \[C{H_2}ClCOOH \to C{H_2}ClCO{O^ - } + {H^ + }\]
So, its dissociation constant can be given by the equation
     \[{K_a} = \dfrac{{[{H^ + }][C{H_2}ClCO{O^ - }]}}{{[C{H_2}ClCOOH]}}\]
We already know that dissociation constant ${K_a}$ for $C{H_2}ClCOOH$ is given as $1.8 \times {10^{ - 4}}$ M. The concentration of $C{H_2}ClCOOH$ is given as 0.1M. Suppose that concentration of both ${H^ + }$ and $C{H_2}ClCO{O^ - }$ is x M.
So, we can write that
     \[1.8 \times {10^{ - 4}} = \dfrac{{(x)(x)}}{{0.1}}\]
So, we can write that
     \[{x^2} = (1.8 \times {10^{ - 4}})(0.1) = 0.18 \times {10^{ - 4}}\]
Thus, we obtained $x = 0.4242 \times {10^{ - 2}}$ = $4.242 \times {10^{ - 3}}$ M.
So, we can say that the concentration of ${H^ + }$ in the mixture will be equal to the sum of concentration of protons given by each acid.
So, $[{H^ + }]$ = $1.34 \times {10^{ - 3}}$ + $4.242 \times {10^{ - 3}}$ = $5.582 \times {10^{ - 3}}$ M

Thus, correct answer of the question is (C) $5.582 \times {10^{ - 3}}$ M

Note:
Remember that Acetic acid $(C{H_3}COOH)$ and its derivatives are weak acids. So, they do not dissociate completely into ionic form in aqueous medium. We can say that higher the dissociation constant of an acid, higher its dissociation will be and the acid will be stronger.