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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 53 of 73
Marks: +1, -0
Calculate the degree of ionization of 0.05 M acetic acid if its pKa\mathrm{p}K_a value is 4.74. How is the degree of dissociation affected when its solution also contains
(a) 0.01 M
(b) 0.1 M HCl?
Solution:  
pKa\mathrm{p}K_a = 4.74
–log KaK_a = 4.74
KaK_a = antilog(–4.74) = 1.820 × 10−510^{-5}
Cα2C\alpha^2 = 1.82 × 10−510^{-5}
⇒ α2\alpha^2 = 1.82×10−5C\frac{1.82 \times 10^{-5}}{C} = 1.82×10−50.05\frac{1.82 \times 10^{-5}}{0.05} = 3.64 × 10−410^{-4} ⇒ α = 1.9 × 10−210^{-2}
(a) CH3COOH\mathrm{CH_3COOH} ⇌ CH3COO−+H+\mathrm{CH_3COO}^{-} + \mathrm{H}^{+}
Initial molar conc.C00Eqm. molar conc.C(1−α)CαCα\begin{array}{llll} \text{Initial molar conc.} & C & 0 & 0 \\ \text{Eqm. molar conc.} & C(1-\alpha) & C\alpha & C\alpha \end{array}
KaK_a = [CH3COO−][H+]C\frac{[\mathrm{CH_3COO}^{-}][\mathrm{H}^{+}]}{C}
[H+][\mathrm{H}^{+}] = 0.01 M
1.82 × 10−510^{-5} = CαC\frac{C\alpha}{C} × 0.01 ⇒ α = 1.82×10−50.01\frac{1.82 \times 10^{-5}}{0.01} = 1.82 × 10−310^{-3}
(b) [H+][\mathrm{H}^{+}] = 0.1 M
1.82 × 10−510^{-5} = CαC\frac{C\alpha}{C} × 0.1 = α × 0.1 ⇒ α = 1.82 × 10−410^{-4}
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