Question

The degree of dissociation of acetic acid in a 0.1 N solution is $1.32\times {10}^{-2}$. At what concentration of nitrous its, degree of dissociation will be same as that of acetic acid? $\left\{K_a\left(HN{O}_2\right)=4\times {10}^{-4}\right\}$

Answer 1

Since degree of dissociation is same.

$\therefore \alpha =1.32\times {10}^{-2}$

$HN{O}_2\rightleftharpoons H^{+}+N{O}_2^{-},Ka=4\times {10}^{-4}$

$x$ $0$ $0$ ----- Initially

$x(1-\alpha )$ $x\alpha$ $x\alpha$ ----- At eqbm.

$Ka=\frac{\left[H^{+}\right]\left[N{O}_2^{-}\right]}{\left[HN{O}_2\right]}=4\times {10}^{-4}$

$\Rightarrow \frac{(1.32\times {10}^{-2}x{)}^2}{x-1.32\times {10}^{-2}x}=4\times {10}^{-4}$

$1-1.32\times {10}^{-2}=\frac{(1.32{)}^2\times {10}^{-4}}{4\times {10}^{-4}}xx$

$(1-1.32\times {10}^{-2})\times 4\times {10}^{-4}=(1.32{)}^2\times {10}^{-4}x$

$x=2.29\times (1-0.0132)=2.27M$