Page 65 - A Working Method Approach For Introductory Physical Chemistry Calculations

Equilibrium 11 49
concentration of CH3C02, since each reaction acts as a source of
CH3CO$ ions.
4. (a) CH3C02Na(aq) Na+(aq) + CH3CO,(aq)
Initial 0.6 0 0
Final 0 0.6 0.6
At the end of the reaction, nothing remains of the salt, since
100% dissociation of CH3C02Na has occurred! Therefore, from
the dissociation of sodium ethanoate, 0.6 mol of ethanoate
anion, CH3C0,, is generated.
(b) CH3C02H + H20 + CH3C0, + H30+

Initial 0.8 0 0
Change -x +X +X
Final 0.8-x X X
But the dissociation of the salt produces 0.6 mol of CH3C0,.
Hence the actual equilibrium concentration of CH3COT is not x,
but (0.6 + x)!
i.e. CH3C02H + H20 + CH~COT + H3O+
Final 0.8-x 0.6 + x X

where Ka = { [CH3C0,][H30']}/[CH3C02H], since the activity
of H20, a = 1.
5. Hence, K, = ((0.6 + x)(x)}/(0.8-x) = 1.8 x lo-'.
6. Assume that x << 0.8. Hence, Ka becomes (0.6x)/(0.8) = 1.8 x
lov5 + x = 0.000024.
7. (0.000024/0.6)% = 0.004% and (0.000024/0.8) = 0.003%. Since
both are <5%, the assumption made was valid, and hence a
quadratic equation need not be solved in this particular case.
Therefore [CH3C02]- = 0.600024 M; [H3O+] = 0.000024 M
and [CH3CO*H] = 0.799976 M. Hence, % acid dissociated =
(x/0.8)% = 0.003%.
If the common ion effect had not occurred i.e. if there was no
salt, CH3C02Na present in solution, the equilibrium expression
would be the following:
CH3C02H + H20 + CH3CO: + H30+
Initial 0.8 0 0
Change -x +X +X
Final 0.8-x X X
Again assuming that x << 0.8, K, = x2/0.8 = 1.8 x i.e. x