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) = (K_b/c)

1% means alfa = 0.01 , c = 0.1 so

K_b = 10^-5                           .......................1(K_b is constant )

K_b = [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 , K_b = [BCl][OH^-]/[BOH]

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

BOH = C

k_b = [BCl]alfa

10^-5 = 0.2(alfa)

alfa = 5 * 10^-5

option b) is correct

Approved

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

1% means alfa = 0.01 , c = 0.1 so

K_b = 10^-5                           .......................1(K_b is constant )

K_b = [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 , K_b = [BCl][OH^-]/[BOH]

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

BOH = C

k_b = [BCl]alfa

10^-5 = 0.2(alfa)

alfa = 5 * 10^-5

option b) is correct

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