$0.010 \mathrm{M}$ solution an acid $H A$ freezes at $-0.0205^{\circ} \mathrm{C}$. If…

$0.010 \mathrm{M}$ solution an acid $H A$ freezes at $-0.0205^{\circ} \mathrm{C}$. If $\mathrm{K}_{\mathrm{f}}$ for water is $1.860 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, the ionization
constant of the conjugate base of the acid will be (assume $0.010 \mathrm{M}=0.010 \mathrm{~m}$ )
  1. $1.1 \times 10^{-4}$
  2. $1.1 \times 10^{-3}$
  3. $9.0 \times 10^{-11}$
  4. $9.0 \times 10^{-12}$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}$ (normal) $=\mathrm{K}_{\mathrm{f}} \mathrm{m}=1.86 \times 0.01=0.0186$;
$\mathrm{i}=\frac{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{obs})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{nor})}}=\frac{0.0205}{0.0186}=1.10=1+\alpha$
$\alpha=0.1$
$\mathrm{~K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}=\frac{0.01 \times 0.1^{2}}{1-0.1}=\frac{1}{9} \times 10^{-3}$
$\mathrm{~K}_{\mathrm{b}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}=1.0 \times 10^{-14} \times 9 \times 10^{3}=9 \times 10^{-11}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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