Question

Given the following:

CH3CH2OH + NH3 -- CH3CH2O- + NH4+

pka of Ethanol = 15.9.

A) Calculate the Keq for the acid-base reaction.

B) Explain different result between reactions of ammonia with acetic adis, and with ethanol.

Answer 1

a)

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only

If Keq > 1, this favours products, since this relates to a higher amount of C + D

If Keq < 1, this favours reactants, since this relates to a higher amount of A + B

If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios

Keq = [CH3CH2O-][NH4+]/([CH3CH2OH][NH3])

for ethanol

CH3CH2OH <--> CH3CH2O- + H+ pKa ethanol = 15.9

NH3 + H+ <--> NH4+ pKa = 9.25

Add all

NH3 + H+ + CH3CH2OH <--> CH3CH2O- + H+ + NH4+ pKa = 15.9 + 9.25 = 25.15

pKeq = 25.15

Keq = 10^-pKeq = 10^-25.15

b)

Note that ammonia is not likely to form NH4+ vs. CH3CH2O since ethanol is not likely to donate readily to ammonia

acetic acids --> R-COOH groups are likely to donate H+ readily, thereofre K vlaue will be favuring this

NH3 + H+ --> NH4+ only if COOH group is strong enough to be "acidic"