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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
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
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
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
approve my ans if u like
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