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(A) 0.02

(B) 0.005

(C) $5 \times {10^{ - 5}}$

(D) $2 \times {10^{ - 3}}$

Answer
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-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 $5 \times {10^{ - 5}}$.