(a)(a)
I will refer to propanoic acid as HAHA. It is a weak acid and dissociates as follows:
HA_((aq))rightleftharpoonsH_((aq))^(+)+A_((aq))^(-)HA(aq)⇌H+(aq)+A−(aq)
For which:
K_a=([H_((aq))^(+)][A_((aq))^(-)])/([HA_((aq))])=1.35xx10^(-5)"mol/l"Ka=[H+(aq)][A−(aq)][HA(aq)]=1.35×10−5mol/l
The concentrations are at equilibrium. Rearranging gives:
[H_((aq))^(+)]=K_axx[[HA_((aq))]]/[[A_((aq))^(-)]]" "color(red)((1))[H+(aq)]=Ka×[HA(aq)][A−(aq)] (1)
We are told that pH=7.4=-log[H_((aq))^(+)]pH=7.4=−log[H+(aq)]
From which [H_((aq))^(+)]=1.78xx10^(-5)"mol/l"[H+(aq)]=1.78×10−5mol/l
:.1.78xx10^(-5)=1.34xx10^(-5)xx([HA_((aq))])/([A_((aq))^(-)])
:.([HA_((aq))])/([A_((aq))^(-)]]=(1.78xxcancel(10^(-5)))/(1.34xxcancel(10^(-5)))=1.33
The total volume is common to both HA and A^(-) so we can write moles n instead of concentration.
:.(nHA)/(nA^(-))=1.33
I will assume that we have 0.02 "mol " of HA at equilibrium as the amount that dissociates is very small in comparison.
:.nA^(-)=0.02/1.33=0.015
The M_r of sodium propanoate =96.06
So the mass needed is given by:
m=0.015xx96.06=1.44"g"
Again, I have neglected the tiny amount of A^(-) formed from the dissociation of the acid.
There may be a volume change on mixing but again, this is common to acid and co-base so will cancel.
(b)
The buffer works because there is a large reserve of the co-base A^(-) which can absorb the addition of small amounts of H^(+) ions:
H_((aq))^(+)+A_((aq))^(-)rightleftharpoonsHA_((aq))
The position of equilibrium lies well to the right.
Since they react in a 1:1 molar ratio we can see that adding 0.01"mol" of H^(+) will consume 0.01"mol" of A^(-) and form 0.01"mol" of HA.
:.nA^(-) remaining =0.015-0.01=0.005
The no. moles HA will increase by 0.01 so:
nHA=0.02+0.01=0.03
Using the usual assumptions already described we can insert these values into color(red)((1))rArr
[H_((aq))^(+)]=1.35xx10^(-5)xx0.03/0.005
[H_((aq))^(+)]=8.1xx10^(-5)"mol/l"
:.pH=-log[8.1xx10^(-5)]
color(red)"pH=4.1"
We are asked to compare this with the pH of 0.01"M"" " "HCl" solution.
pH=-log[H_((aq))^(+)]
The "HCl" solution can be assumed to be 100% dissociated so:
pH=-log(0.01)
color(red)"pH=2"
This is, therefore, considerably more acidic when not made up in a buffer.
