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A 0.1 molar solution of week base BOH is 1% dissociated. If 0.2 moles of bcl is added in 1 liter solution of BOH,the degree of dissociation of BOH will become (1){2}{10^-3} (2){5}{10^-5} (3)00.5 (4)0.02

A 0.1 molar solution of week base BOH is 1% dissociated. If 0.2 moles of bcl is added in 1 liter solution of BOH,the degree of dissociation of BOH will become
(1){2}{10^-3}
(2){5}{10^-5}
(3)00.5
(4)0.02

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2 Answers

vikas askiitian expert
509 Points
11 years ago

alfa(degree of dissotion) = (Kb/c)

1% means alfa = 0.01 , c = 0.1 so

Kb = 10-5                           .......................1(Kb is constant )

 

Kb = [B+][0H-]/[BOH]

finally 0.2mol of BCl is added so

B+ from BCl is added in the solution & here common ion effect is produced ...

now , Kb = [BCl][OH-]/[BOH]

OH- = C(alfa)                              (alfa is final degree of dissociation)

BOH = C

kb = [BCl]alfa

10-5 = 0.2(alfa)

alfa = 5 * 10-5

option b) is correct

vikas askiitian expert
509 Points
11 years ago

alfa(degree of dissotion) = (Kb/c)

1% means alfa = 0.01 , c = 0.1 so

Kb = 10-5                           .......................1(Kb is constant )

 

Kb = [B+][0H-]/[BOH]

finally 0.2mol of BCl is added so

B+ from BCl is added in the solution & here common ion effect is produced ...

now , Kb = [BCl][OH-]/[BOH]

OH- = C(alfa)                              (alfa is final degree of dissociation)

BOH = C

kb = [BCl]alfa

10-5 = 0.2(alfa)

alfa = 5 * 10-5

option b) is correct

 

approve my ans if u like

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