Not too sure how to answer 2 and 3, any help would be nice, thanks Show all your

Question

Not too sure how to answer 2 and 3, any help would be nice, thanks Show all your work 1. Write the hydrolysis equations (Bronsted-Lowry reactions) for the dissociation of the following monobasic bases in water: NH2OH, C4H&N; and CH.COO 2. What is the pH of a solution consisting of a mixture of 0.75 M NH2OH (Kb 1.07x10 C&H;&N; (Kb-2.51x104 ? and 0.75 M 3. H,so. has ??,-1000 and Ka2-12x10-2. What is Kb of the HS04. ion? would a solution of 1 M KHSO, be acidic, basic or pH neutral? Briefly explain.

Explanation / Answer

2) NH4OH(aq) <--->NH3(aq)+H2O(l)

Ammonia hydrolyses in solution as follows:

NH3(aq)+H2O(l) <--->NH4+(aq)+ OH-(aq)

kb=[NH4+][OH-]/[NH3]

ICE table will help calculate [OH-]

1.07*10^-8=x^2/(0.75-x)

but kb is very small so x is very small ,x<<0.75.

1.07*10^-8=x^2/(0.75)

or x=[OH-]=8.958*10^-5M

[H3O+]=kw/[OH-] where kw=1*10^-14=ionic product of water

[H3O+]=(1*10^-14M^2)/(8.958*10^-5M)=1.116*10^-10M........(1)

Also ,C4H4N hydrolyses in water as follows:

C4H4N +H2O <--->C4H4NH+ +OH-

kb=[C4H4NH+][OH-]/[C4H4N]=x^2/(0.75-x) [similarly like above]

2.51*10^-14=x^2/(0.75-x)

Again kb is very small,which means base is very weak,x<<<0.75

or , 2.51*10^-14=x^2/(0.75)

or, x=[OH-]=1.372*10^-7 M

[H3O+]=(1*10^-14)/(1.372*10^-7 )=7.289*10^-8M ...(2)

from (1) and (2),

total [H3O]=(7.289*10^-8M)+(1.116*10^-10M)=(728.9*10^-10M)+(1.116*10^-10M)=730.016*10^-10M

pH=-log [H3O+]total=-log (730.016 *10^-10)=7.1

pH=7.1

[NH3] [NH4+] [OH-] initial 0.75M 0 0 change -x +x +x equilibrium 0.75M-x x x

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