Question

# A 0.1 molar solution of weak base BOH is 1% dissociated. If 0.2 mole of BCl is added in 1L solution of BOH, the degree of dissociation of BOH will become:                  (A) 0.02                  (B) 0.005                  (C) $5 \times {10^{ - 5}}$                   (D) $2 \times {10^{ - 3}}$

Hint: First find out the amount of ${B^ + }$ and $O{H^ - }$ present after dissociation. The addition of BCl causes a common ion effect due to an increase of ${B^ + }$ ions which results in backward reaction and increase in the concentration of BOH. Now find out the new amount of BOH that will be dissociated which will lead you to the new degree of dissociation.

-According to the question: weak base BOH has molarity 0.1 M and it is 1% dissociated. BOH dissociation can be written as:
$BOH \rightleftharpoons {B^ + } + O{H^ - }$
Where the initial concentration of BOH is 0.1 moles, of ${B^ + }$ is 0 and of $O{H^ - }$ is also 0.
We know that the degree of dissociation (α) is 1%. So, out of 0.1 moles of BOH, amount of it dissociated will be = degree of dissociation × concentration of BOH
= $0.1 \times \dfrac{1}{{100}}$ = ${10^{ - 3}}$ or 0.001 moles.
So, 0.001 moles of BOH will dissociate and form 0.001 moles of ${B^ + }$ and 0.001 moles of $O{H^ - }$also. This can be written as:
$BOH \rightleftharpoons {B^ + } + O{H^ - }$
At time t = 0: 0.1 0 0 (moles)
After dissociation: 0.1 – 0.001 , 0.001 , 0.001
= 0.099 moles
-Using these obtained concentrations we can find out the equilibrium constant (${K_b}$) for BOH: ${K_b} = \dfrac{{[{B^ + }][O{H^ - }]}}{{[BOH]}}$
= $\dfrac{{[0.001][0.001]}}{{0.099}}$
= $1.01 \times {10^{ - 5}}$
-Now 0.2 moles of BCl are added to this. BCl also dissociates as follows:
$BCl \rightleftharpoons {B^ + } + C{l^ - }$
After dissociation: 0 , 0.2 , 0.2

Total amount of ${B^ + }$ ions after addition of BCl will be = 0.001 + 0.2 = 0.201
Release of ${B^ + }$ ions here causes the common ion effect of ${B^ + }$ ions. So, due to increased concentration of ${B^ + }$ ions causes that dissociation reaction of BOH to move backwards and increase in the concentration of BOH. Now, the new concentration of all ions of BOH can be written as:
Let the amount of increase in BOH concentration be ‘x’ and so total amount of BOH will become = 0.099 + x
$BOH \rightleftharpoons {B^ + } + O{H^ - }$
(0.099 + x) (0.201 – x) (0.001 – x)
The dissociation constant for a base does not change with concentration, it remains constant. For BOH the dissociation constant we had calculated earlier to be = $1.01 \times {10^{ - 5}}$. Now using this value we will find out the value of ‘x’.
${K_b} = \dfrac{{[{B^ + }][O{H^ - }]}}{{[BOH]}}$
$1.01 \times {10^{ - 5}} = \dfrac{{[0.201][0.001 - x]}}{{0.099 + x}}$
By solving this equation we obtain x = 0.000995
-The new amount of BOH dissociated is = (0.001 – 0.000995) = 0.000005
We know that:
Amount of BOH dissociated = concentration of BOH × degree of dissociation (α)
0.000005 = 0.1 × α
α = $\dfrac{{0.000005}}{{0.1}}$
= 0.00005 = $5 \times {10^{ - 5}}$
So, the new degree of dissociation of BOH will be $5 \times {10^{ - 5}}$.
So, the correct answer is “Option C”.

Note: Common ion effect causes decrease in solubility of any ionic precipitate if we add a soluble compound with an ion in common with the precipitate. This happens by initiating a backward reaction to increase the concentration of the initial reactant and decreasing the concentration of the common ion.