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6-2.Equilibrium-II (Ionic Equilibrium)
easy
If $pK_a =\, -\,log K_a=4$ for a weak acid $HX$ and $K_a= C\alpha ^2$ then Van't Haff factor when $C = 0.01\,M$ is
A
$1.01$
B
$1.02$
C
$1.10$
D
$0.20$
Solution
$\mathrm{K}_{\mathrm{a}}=\mathrm{c} \alpha^{2}$
$\alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}} $ $=\sqrt{\frac{10^{-4}}{10^{-2}}}$
$=10^{-1}=0.1$
$\mathrm{i}=1-\alpha+\mathrm{n} \alpha=1-0.1+2 \times 0.1=1.1$
Standard 11
Chemistry
