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Question: At \(15^{\circ} \mathrm{C} , 0.05 \mathrm{N}\) solution of a weak monobasic acid is \(3.5 \%\) ionis...

At 15∘C,0.05N15^{\circ} \mathrm{C} , 0.05 \mathrm{N} solution of a weak monobasic acid is 3.5%3.5 \% ionised. Calculate the ionisation constant of acid.

Explanation

Solution

The ionised ions have a tendency to recombine back to the unionized electrolyte like, acid. Thus, a dynamic equilibrium exists between ionized and unionized electrolytes. In a binary electrolyte AB, AB⇌A++B−\mathrm{AB} \rightleftharpoons \mathrm{A}^{+}+\mathrm{B}^{-} Applying law of mass action, the ionisation constant K=[A+][B−][AB]\mathrm{K}=\dfrac{\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}

Complete step by step answer:
-According to Arrhenius theory of electrolytic dissociation, when an electrolyte is dissolved in water, it dissociates into positive and negative ions that move freely in the bulk of the solution. The number of positive ions is equal to the number of negative ions formed, hence the solution is electrically neutral.
-For a binary electrolyte, AB
-Applying law of mass action, the ionisation constant AB⇌A++B−\mathrm{AB} \rightleftharpoons \mathrm{A}^{+}+\mathrm{B}^{-} K=[A+][B−][AB]\mathrm{K}=\dfrac{\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]}{[\mathrm{AB}]}

-The fraction of electrolyte dissociated into ions at equilibrium out of the total number of molecules of electrolyte dissolved is known as the degree of dissociation or ionisation and is given by:
α=Number of molecules present as ionsTotal number of molecules of the electrolyte\alpha=\dfrac{\text {Number of molecules present as ions}}{\text {Total number of molecules of the electrolyte}}

Let the initial concentration of the electrolyte be c in molL−1\mathrm{mol} \mathrm{L}^{-1}
AB⇌A++B−\mathrm{AB} \rightleftharpoons \mathrm{A}^{+}+\mathrm{B}^{-}
[AB]=c(1−α)[\mathrm{AB}]=\mathrm{c}(1-\alpha)
[A+]=cα\left[\mathrm{A}^{+}\right]=\mathrm{c} \alpha
[B−]=cα\left[\mathrm{B}^{-}\right]=\mathrm{c} \alpha
Thus, the ionisation constant is Kc=α2c(1−α)\mathrm{K}_{\mathrm{c}}=\dfrac{\alpha^{2} \mathrm{c}}{(1-\alpha)}

-Now, let us approach the question.
Let HA be the monobasic acid.
Now, HA⇌H++A−\mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-}
α=3.5100\alpha=\dfrac{3.5}{100}
Now, α=0.035\alpha=0.035

c = 0.05N.
Thus, Kc=α2c(1−α)\mathrm{K}_{\mathrm{c}}=\dfrac{\alpha^{2} \mathrm{c}}{(1-\alpha)}
Kc=(0.035)2×0.051−0.035\mathrm{K}_{\mathrm{c}}=\dfrac{(0.035)^{2} \times 0.05}{1-0.035}
Kc=6.34×10−5\mathrm{K}_{\mathrm{c}}=6.34 \times 10^{-5}

Thus, the answer is 6.34×10−56.34 \times 10^{-5}.

Note: This law only holds good for weak electrolytes as strong electrolytes are completely ionised at all dilutions. This law is based on the fact that the electrolyte is partly ionised at ordinary dilution and completely ionised at infinite dilution. When the concentrations of ions are high, the chemical equilibrium is affected by the presence of these ions and the law of mass action does not hold.