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Question: At 298K a 0.1 M \(C{{H}_{3}}COOH\) solution is 1.34% ionised. The ionisation constant \({{K}_{a}}\) ...

At 298K a 0.1 M CH3COOHC{{H}_{3}}COOH solution is 1.34% ionised. The ionisation constant Ka{{K}_{a}} for acetic acid will be?
[A] 1.82×10−51.82\times {{10}^{-5}}
[B] 18.2×10−518.2\times {{10}^{-5}}
[C] 0.182×10−50.182\times {{10}^{-5}}
[D] None of these

Explanation

Solution

To solve this, firstly write down the dissociation reaction for acetic acid and write down Ka{{K}_{a}} for the reaction following the law of chemical equilibrium. From the percentage of ionisation, find out the number of moles of acetic acid ionised and then find out the number of moles of dissociated ions at equilibrium. Putting the values in the Ka{{K}_{a}} equation will give you the correct answer.

Complete step by step solution:
Here, the acid given to us is acetic acid. Firstly let’s discuss what ionisation is to solve this question.
We know that ionisation is the process by which atom or a molecule gains or loses an electron and obtains a negative or a positive charge. The degree of ionization (or dissociation) is basically a way of representing the strength of an acid.
Now, let us write down the ionisation reaction for acetic acid-
CH3COOH⇌CH3COO−+H+C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
Therefore, according to the law of chemical equilibrium, we can write that-
Ka=[CH3COO−][H+][CH3COOH]{{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}
The initial concentration of acetic acid is given to us as 0.1M and it is 1.34% ionised. It means that the amount of acetic acid ionised = 1.34100×0.1\dfrac{1.34}{100}\times 0.1 = 0.00134 moles
Therefore, at equilibrium, [CH3COOH]\left[ C{{H}_{3}}COOH \right]= 0.1 – 0.00134 = 0.9866 moles,
[CH3COO−]=[H+]\left[ C{{H}_{3}}CO{{O}^{-}} \right]=\left[ {{H}^{+}} \right] = 0.00134 moles
Thus, substituting the values in the Ka{{K}_{a}} equation, we will get -
Ka=0.00134×0.001340.9866=1.82×10−5{{K}_{a}}=\dfrac{0.00134\times 0.00134}{0.9866}=1.82\times {{10}^{-5}}

Therefore, the correct answer is option [A] 1.82×10−51.82\times {{10}^{-5}}.

Note: On ionisation with water the acids which produce a single hydronium ion i.e. readily ionises only one proton, are known as monobasic acids. Here, acetic acid is also a monobasic acid. While solving such questions remember to take into account the nature of the acid i.e. whether it is monobasic, dibasic or tribasic as it affects the concentration of the ions in the solution.