Question

Problem 4.6

·         Predict the position of equilibrium and calculate the equilibrium constant, , for each acid-base reaction.

1. Methylamine + acetic acid ⇄ methylammonium ion + acetate ion

2. Ethoxide ion + ammonia ⇄ ethanol + amide ion

Answer 1

methylamine Kb = 4.4*10^-4

acetic acid Ka = 1.8*10^-5

the reactions:

CH3NH2 + H2O <---> CH3NH3+ + OH- Kb =  4.4*10^-4

CH3COOH + H2O <-> CH3COO- + H3O+ Ka =  1.8*10^-5

add all:

CH3NH2 + H2O + CH3COOH + H2O <--> CH3NH3+ + OH- + CH3COO- + H3O+ Kt = Kb*Ka = (4.4*10^-4)(1.8*10^-5) = 7.92*10^-9

we need to add hydrolysis of watetr

2H2O(l) <--> H3O+(aq) + OH-(aq) Kw = 10^-14

CH3NH2 + H2O + CH3COOH + H2O <--> CH3NH3+ + OH- + CH3COO- + H3O+ Kt = Kb*Ka = 7.92*10^-9

H3O+(aq) + OH-(aq) <--> 2H2O(l) Kw = 1/(10^-14)

reaction

CH3NH2 + CH3COOH <--> CH3NH3+ + CH3COO-

K = (7.92*10^-9)(1/(10^-14))= 792000

This is very likely to happen

since it is a weak base + weak acid = neutralization + ions in solution

b)

Ethoxide ion CH3CH2O- + NH3(aq)

CH3CH2O⁻ + NH3(aq) <--> CH3CH2OH + NH2-(aq)

NH3(aq) <--> NH2- + H K = 10^-35 approx

CH3CH2OH <--> H+ CH3CH2O- K = 10^-15.9

then..

H+ CH3CH2O- + NH3(aq) <--> NH2- + H + CH3CH2OH Ktotal = (10^-35 ) / (10^-15.9 ) = 7.943*10^-20

CH3CH2O- + NH3(aq) <--> NH2- + CH3CH2OH Ktotal = 7.943*10^-20

which is not likely to occur