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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 47 of 73
Marks: +1, -0
It has been found that the pH of 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.
Solution:  
pH = - log [H+][\mathrm{H}^{+}]
∴ [H+][\mathrm{H}^{+}] = antilog (- 4.15) = 7.08 × 10−5M10^{-5}\mathrm{M}
HA(aq)\mathrm{HA}_{(aq)} ⇌ H(aq)++A(aq)−\mathrm{H}^{+}_{(aq)} + \mathrm{A}^{-}_{(aq)}
∴ [H+][\mathrm{H}^{+}] = [A−][\mathrm{A}^{-}] = 7.08 × 10−5 M10^{-5}\,\mathrm{M}
and α = [H+]X\frac{[\mathrm{H}^{+}]}{X} = 7.08×10−50.01\frac{7.08 \times 10^{-5}}{0.01} = 7.08 × 10−310^{-3}
KaK_a = Cα21−α\frac{C\alpha^2}{1-\alpha} = 0.01(7.08×10−3)21−7.08×10−3\frac{0.01(7.08 \times 10^{-3})^2}{1-7.08 \times 10^{-3}} = 5.05 × 10−710^{-7}
pKapK_a = - log KaK_a = - log (5.05 × 10−710^{-7}) = 6.29
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