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Question: If 0.1M \(C{H_3}COOH\) is mixed with 0.1M \(C{H_2}ClCOOH\), [Given: \({K_a}{\text{ C}}{{\text{H}}_...

If 0.1M CH3COOHC{H_3}COOH is mixed with 0.1M CH2ClCOOHC{H_2}ClCOOH,
[Given: Ka CH3COOH=1.8×10−5{K_a}{\text{ C}}{{\text{H}}_3}COOH = 1.8 \times {10^{ - 5}} , Ka CH2ClCOOH=1.8×10−4{K_a}{\text{ C}}{{\text{H}}_2}ClCOOH = 1.8 \times {10^{ - 4}} ]
Find out total [H+][{H^ + }]
(A) 0.404×10−20.404 \times {10^{ - 2}}
(B) 1.44×10−31.44 \times {10^{ - 3}}
(C) 5.58×10−35.58 \times {10^{ - 3}}
(D) 8.44×10−48.44 \times {10^{ - 4}}

Explanation

Solution

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
Ka=[Product][Reactant]=[H+][A−][HA]{K_a} = \dfrac{{[{\text{Product]}}}}{{[{\text{Reactant]}}}} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}
Thus, for given acid CH3COOHC{H_3}COOH , dissociation reaction can be given as
CH3COOH→H++CH3COO−C{H_3}COOH \to {H^ + } + C{H_3}CO{O^ - }
we can give its dissociation constant as
Ka=[H+][CH3COO−][CH3COOH]{K_a} = \dfrac{{[{H^ + }][C{H_3}CO{O^ - }]}}{{[C{H_3}COOH]}}
We know that Ka=1.8×10−5{K_a} = 1.8 \times {10^{ - 5}}, [CH3COOH][C{H_3}COOH]= 0.1 M and suppose that [H+][{H^ + }] =[CH3COO−][C{H_3}CO{O^ - }] = x
So, we can write the above equation as
1.8×10−5=(x)(x)0.11.8 \times {10^{ - 5}} = \dfrac{{(x)(x)}}{{0.1}}
So, we can write that
x2=(0.1)(1.8×10−5)=1.8×10−6{x^2} = (0.1)(1.8 \times {10^{ - 5}}) = 1.8 \times {10^{ - 6}}
Thus, we can say that x=1.34×10−3x = 1.34 \times {10^{ - 3}}
So, we obtain that concentration of protons of CH3COOHC{H_3}COOH will be 1.34×10−3M1.34 \times {10^{ - 3}}M
Now, dissociation reaction of CH2ClCOOHC{H_2}ClCOOH can be given as
CH2ClCOOH→CH2ClCOO−+H+C{H_2}ClCOOH \to C{H_2}ClCO{O^ - } + {H^ + }
So, its dissociation constant can be given by the equation
Ka=[H+][CH2ClCOO−][CH2ClCOOH]{K_a} = \dfrac{{[{H^ + }][C{H_2}ClCO{O^ - }]}}{{[C{H_2}ClCOOH]}}
We already know that dissociation constant Ka{K_a} for CH2ClCOOHC{H_2}ClCOOH is given as 1.8×10−41.8 \times {10^{ - 4}} M. The concentration of CH2ClCOOHC{H_2}ClCOOH is given as 0.1M. Suppose that concentration of both H+{H^ + } and CH2ClCOO−C{H_2}ClCO{O^ - } is x M.
So, we can write that
1.8×10−4=(x)(x)0.11.8 \times {10^{ - 4}} = \dfrac{{(x)(x)}}{{0.1}}
So, we can write that
x2=(1.8×10−4)(0.1)=0.18×10−4{x^2} = (1.8 \times {10^{ - 4}})(0.1) = 0.18 \times {10^{ - 4}}
Thus, we obtained x=0.4242×10−2x = 0.4242 \times {10^{ - 2}} = 4.242×10−34.242 \times {10^{ - 3}} M.
So, we can say that the concentration of H+{H^ + } in the mixture will be equal to the sum of concentration of protons given by each acid.
So, [H+][{H^ + }] = 1.34×10−31.34 \times {10^{ - 3}} + 4.242×10−34.242 \times {10^{ - 3}} = 5.582×10−35.582 \times {10^{ - 3}} M

Thus, correct answer of the question is (C) 5.582×10−35.582 \times {10^{ - 3}} M

Note:
Remember that Acetic acid (CH3COOH)(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.