Determine the degree of ionizationand pH of a 0.05 M of ammonia solution .The ionization constant of ammonia can be taken fromK_b=1.77xx10^(-5) . Also, calculate the ionizationconstant of the conjugate acid of ammonia .

Determine the degree of ionizationand pH of a 0.05 M of ammonia soluti

1.	Determine the degree of ionizationand pH of a 0.05 M of ammonia solution .The ionization constant of ammonia can be taken fromK_b=1.77xx10^(-5) . Also, calculate the ionizationconstant of the conjugate acid of ammonia .
Answer» Solution :Dissociated degree of weak base `NH_3= alpha` `THEREFORE` Ionised `[NH_3]=C alpha =0.05 alphaM` [Ionised `[NH_3]`] = `0.05 alpha` = [ Formed `[NH_4^+]` ] = [Formed `[OH^-]` ] And at equilibrium `[NH_3]=(0.05 - 0.05 alpha)` `=0.05 (1-alpha)M` `{:(,NH_(3(aq)) + H_2O hArr , NH_(4(aq))^(+)+, OH_((aq))^(-)),("Equili.M:",0.05(1-alpha),0.05 alpha,0.05 alpha):}` `therefore K_b=([NH_4^+][OH^-])/([NH_3])=1.77xx10^(-5)` `therefore ((0.05alpha)(0.05 alpha))/(0.05 (1-alpha))=1.77xx10^(-5)` `therefore (0.05 alpha^2)/(1-alpha)=1.77xx10^(-5)` `therefore alpha^2=(1.77xx10^(-5))/0.05 = 3.54xx10^(-4)` (`because 1 gt gt gt alpha`, So `(1-alpha)=1`) `therefore alpha=(3.54xx10^(-4))^(1/2)=1.8814xx10^(-2)` `[OH^-]=alpha=0.05xx1.88xx10^(-2) = 9.4xx10^(-4)` M pOH=-log `(9.4xx10^(-4))` = 3.0269 and pH=14-3.0269 = 10.97 `NH_4^+` is CONJUGATE acid of `NH_3`. `K_w=K_a xx K_b` `therefore K_a = K_w/K_b = (1.0xx10^(-14))/(1.77xx10^(-5))=5.65xx10^(-10)`