Question

Determine the degree of dissociation (in %) of 0.05M $N{H_3}$ at 298K in a solution of pH = 11.

Answer 1

Hint: First find out pOH and from that calculate the concentration of $O{H^ - }$ ions. Now use the formula $\alpha = \dfrac{{[O{H^ - }]}}{C}$ to calculate the degree of dissociation.
Formulas used:
-Concentration of $O{H^ - }$ ions:
 $\left[ {O{H^ - }} \right] = {10^{ - pOH}}$

-Degree of dissociation:
$\alpha = \dfrac{{[O{H^ - }]}}{C}$

Where, C = initial concentration.

Complete step by step answer:
-First we will see what degree of dissociation is. The degree of dissociation of any substance is basically the fraction of its molecules that are being dissociated at a given time.
-To find out the degree of dissociation first we need to find out the concentration of $O{H^ - }$ ions.
-The question gives us the pH of the solution formed after dissociation to be = 11
Since we all know that: pH + pOH = 14.
From the above given formula we can find out the pOH of the solution. It will be:
pOH = 14 – pH
= 14 – 11
pOH = 3

Also, $pOH = - \log [O{H^ - }]$
And hence, $\left[ {O{H^ - }} \right] = {10^{ - pOH}}$

From the above given formula we can calculate the concentration of $O{H^ - }$ ions.
 $\left[ {O{H^ - }} \right] = {10^{ - 3}}$

-Now we will calculate the degree of dissociation :
 $\alpha = \dfrac{{[O{H^ - }]}}{C}$
 = $\dfrac{{{{10}^{ - 3}}}}{{0.05}}$
 = $2 \times {10^{ - 2}}$

In the form of percentage the degree of dissociation will be = $(2 \times {10^{ - 2}}) \times 100$ = 2%
Hence the degree of dissociation will be 2%.

Note: pOH is basically the measure of concentration of hydroxide ions. It is an expression of the alkalinity of the solution like pH is the expression of acidity of the solution. At 25$^ \circ C$ if a solution has a pOH of less than 7, then it is alkaline; if pOH is greater than 7, then solution is acidic and pOH equal to 7, the solution is neutral.