Question

17. The pH of 0.2 M solution of ammonia is 10.78. Calculate its
a) OH ion concentration
b) The degree of dissociation and
c) The dissociation constant.​

Answer 1

Answer:

The dissociation constant.

Answer 2

Given: Ammonia solution of

pH = 10.78

C = 0.2 M (Concentration)

To find:

a) OH ion concentration = $OH^{-}$

b) The degree of dissociation = α

c) The dissociation constant = $K_{b}$

Solution:

a) OH ion concentration = $OH^{-}$

First of all, let us find pOH of the given solution.

pH + pOH = 14

∴ 10.78 + pOH = 14

∴ pOH = 3.22

Now, the pOH can be calculated as follows:

$pOH = -log_{10} [OH^{-}]$

$3.22 = -log_{10}[OH^{-}]$

$-3.22 = log_{10}[OH^{-}]$

Now, by the definition of log, we will get,

$[OH^{-}]=10^{-3.22}$

$[OH^{-}]=0.0006=6\times 10^{-4}$

b) The degree of dissociation = α

α = $=\frac{[OH^{-}]}{C}$

α = $\frac{0.0006}{0.2}$

α = 0.003

c) The dissociation constant = $K_{b}$

$K_{b}=\frac{C\:\alpha^{2} }{1-\alpha }$

$K_{b}= \frac{0.2\times (0.003)^{2}}{1-0.003}$

$K_{b}=1.8\times10^{-6}$

Answer:

Thus, the answers are

a) OH ion concentration = $6\times 10^{-4}$

b) The degree of dissociation = 0.003

c) The dissociation constant = $1.8\times10^{-6}$