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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 44 of 73
Marks: +1, -0
The ionization constant of phenol is 1.0 × 10−1010^{-10}. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?
Solution:  
C6H5OH\mathrm{C_6H_5OH} ⇌ C6H5O−+H+\mathrm{C_6H_5O^{-}} + \mathrm{H}^{+}
Initial concentration0.05 M00After dissociation0.05−xxx\begin{array}{lccc} \text{Initial concentration} & 0.05\,\text{M} & 0 & 0 \\ \text{After dissociation} & 0.05 - x & x & x \end{array}
∴ Ka_{a} = x×x0.05−x\frac{x \times x}{0.05 - x} = 1.0 × 10−1010^{-10} (Given)
or x20.05\frac{x^2}{0.05} = 1.0 × 10−1010^{-10} [Since 0.05 - x ≈ 0.05]
or x2x^2 = 5 × 10−1210^{-12} or, x = 2.33 × 10−6 M10^{-6}\,\text{M}
In presence of 0.01 M C6H5ONa\mathrm{C_6H_5ONa}, suppose y is the amount of phenol dissociated, then at equilibrium
[C6H5OH][\mathrm{C_6H_5OH}] = 0.05 – y ≈ 0.05 M
[C6H5O−][\mathrm{C_6H_5O^{-}}] = 0.01 + y ≈ 0.01 M, [H+][\mathrm{H}^{+}] = y M
∴ KaK_{a} = (0.01)(y)0.05\frac{(0.01)(y)}{0.05} = 1.0 × 101010^{10} (Given) or y = 5 × 10−1010^{-10}
∴ α = yC\frac{y}{C} = 5×10−100.05\frac{5 \times 10^{-10}}{0.05} = 1 × 10−810^{-8}
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