Answer
424.5k+ views
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.
Complete answer:
-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.
Complete answer:
-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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)