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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 46 of 73
Marks: +1, -0
The ionization constant of acetic acid is 1.74 × 10−510^{-5}. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ions in the solution and its pH.
Solution:  
α = KaC\sqrt{\frac{K_a}{C}} = 1.74×10−55×10−2\sqrt{\frac{1.74 \times 10^{-5}}{5 \times 10^{-2}}} = 0.0186
CH3COOH\mathrm{CH_3COOH} ⇌ CH3COO−+H+\mathrm{CH_3COO^{-}}+\mathrm{H^{+}}
KaK_a = [CH3COO−][H+][CH3COOH]\frac{[\mathrm{CH_3COO^{-}}][\mathrm{H^{+}}]}{[\mathrm{CH_3COOH}]} = [H+]2[CH3COOH]\frac{[\mathrm{H^{+}}]^2}{[\mathrm{CH_3COOH}]}
or, [H+][\mathrm{H^{+}}] = Ka[CH3COOH]\sqrt{K_a[\mathrm{CH_3COOH}]} = (1.74×10−5)(5×10−2)\sqrt{(1.74 \times 10^{-5})(5 \times 10^{-2})} = 9.33 × 10−4 M10^{-4}\,\mathrm{M}
[CH3COO−][\mathrm{CH_3COO^{-}}] = [H+][\mathrm{H^{+}}] = 9.33 × 10−4 M10^{-4}\,\mathrm{M}
pH = - log (9.33 × 10−410^{-4}) = 4 - 0.9699 = 4 - 0.97 = 3.03
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