Question

What will be the dissociation constant of $0.1M$ aqeous ammonia solution in terms of degree of dissociation constant '$\alpha$'? What will be the value if the concentration is $0.01N$?

Answer 1

$N{H}_{4}OH\rightleftharpoons {N{H}_4}^{+}{+OH}^{-}$

$0.1$ $0$ $0$

$0.1(1-\alpha )$ $0.1\alpha$ $0.1\alpha$

$K_a=\frac{\left[{N{H}_4}^{+}\right]\left[{OH}^{-}\right]}{\left[N{H}_{4}OH\right]}$

$K_a=\frac{0.1\alpha \times 0.1\alpha }{0.1(1-\alpha )}=0.1\frac{{\alpha }^2}{1-\alpha }$

If concentration is $0.01N$ then,

$K_a=\frac{0.1\alpha \times 0.1\alpha }{0.1(1-\alpha )}=0.01\frac{{\alpha }^2}{1-\alpha }$