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Question: Ionisation constant of \(C{{H}_{3}}COOH\) is \(1.7\times {{10}^{-5}}\) and concentration of \({{H}^{...

Ionisation constant of CH3COOHC{{H}_{3}}COOH is 1.7×10−51.7\times {{10}^{-5}} and concentration of H+{{H}^{+}} ions is 3.4×10−43.4\times {{10}^{-4}}. Then find out the initial concentration of CH3COOHC{{H}_{3}}COOH molecule.
A)3.4×10−43.4\times {{10}^{-4}}
B) 3.4×10−33.4\times {{10}^{-3}}
C) 6.8×10−46.8\times {{10}^{-4}}
D) 6.8×10−36.8\times {{10}^{-3}}

Explanation

Solution

The answer here relies upon the basic concept of physical chemistry that deals with the dissociation of acetic acid and calculation of acid dissociation constant Ka{{K}_{a}} which gives the correct answer.

Complete answer:
We have studied the physical chemistry concept that deals with the dissociation constants of acid and base at initial and final time and also change in their concentration with the time.
We have also come across the factors on which the dissociation constant depends.
We shall now see the dissociation constant of the acid that is acetic acid which is as shown below,
CH3COOH(aq)CH3COO−+H+C{{H}_{3}}COOH(aq)C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
Here, one mole of acetic acid dissociates to give one more of acetate ion and one mole of hydrogen ion.
Thus, the concentration change of these molecules at two time intervals can be depicted as,
CH3COOH(aq)CH3COO−+H+C{{H}_{3}}COOH(aq)C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}

At time t=0Conc.= C00
At time t=tConc.= C-xxx

Now, the dissociation constant is given by,
Ka=x×xC−x=x2C−x{{K}_{a}}=\dfrac{x\times x}{C-x}=\dfrac{{{x}^{2}}}{C-x}
At larger concentration that is C >>> x, we can write the above equation as,
Ka=x2C{{K}_{a}}=\dfrac{{{x}^{2}}}{C}
Rearranging the above equation we get,
C=x2KaC=\dfrac{{{x}^{2}}}{{{K}_{a}}}
From the given data, we have = 1.7×10−51.7\times {{10}^{-5}} and x = 3.4×10−43.4\times {{10}^{-4}}
Substituting these values in the above equation,
C=(3.4×10−4)21.7×10−5=6.8×10−3MC=\dfrac{{{(3.4\times {{10}^{-4}})}^{2}}}{1.7\times {{10}^{-5}}}=6.8\times {{10}^{-3}}M
Thus, we can say that the initial concentration of acetic acid molecule that is CH3COOHC{{H}_{3}}COOH is 6.8×10−3M6.8\times {{10}^{-3}}M in a solution.

Therefore, the correct answer is option D) 6.8×10−36.8\times {{10}^{-3}}.

Note:
Note that the dissociation constants of the acids and bases are the measure of the strength of acids and bases in the solution and this degree of dissociation depends on the concentration of the electrolyte.