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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 59 of 73
Marks: +1, -0
The ionization constant of propanoic acid is 1.32 × 10−510^{-5}. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?
Solution:  
KaK_a = 1.32 × 10−510^{-5} ; C = 0.05 M ; α = KaC\sqrt{\frac{K_a}{C}} or α = 1.32×10−55×10−2\sqrt{\frac{1.32\times 10^{-5}}{5\times 10^{-2}}} = 1.62 × 10−210^{-2}
[H+][\mathrm{H}^{+}] = Cα = 0.05 × 1.62 × 10−210^{-2} = 8.1 × 10−410^{-4}
pH = – log(8.1 × 10−410^{-4}) = 3.09
In 0.01 M HCl, [H+][\mathrm{H}^{+}] = 0.01 M
C2H5COOH\mathrm{C_2H_5COOH} ⇌ C2H5COO−+H+\mathrm{C_2H_5COO}^{-} + \mathrm{H}^{+}
Initial molar conc.C00Eqm. molar conc.C(1−α)CαCα\begin{array}{l c c c} \text{Initial molar conc.} & C & 0 & 0 \\ \text{Eqm. molar conc.} & C(1-\alpha) & C\alpha & C\alpha \end{array}
Applying law of chemical eqm.
KaK_a = [C2H5COO−][H+][C2H5COOH]\frac{[\mathrm{C_2H_5COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{C_2H_5COOH}]} = Cα×0.01C(1−α)\frac{C\alpha \times 0.01}{C(1-\alpha)} ; KaK_a = Cα×0.01C\frac{C\alpha \times 0.01}{C}
1.32 × 10−510^{-5} = 0.01α ⇒ α = 1.32 × 10−310^{-3}
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