Question #f7626

1 Answer
Apr 14, 2016

Answer:

#(a) " "1.44"g"#

#(b)" ""pH"" = " 4.1#

Explanation:

#(a)#

I will refer to propanoic acid as #HA#. It is a weak acid and dissociates as follows:

#HA_((aq))rightleftharpoonsH_((aq))^(+)+A_((aq))^(-)#

For which:

#K_a=([H_((aq))^(+)][A_((aq))^(-)])/([HA_((aq))])=1.35xx10^(-5)"mol/l"#

The concentrations are at equilibrium. Rearranging gives:

#[H_((aq))^(+)]=K_axx[[HA_((aq))]]/[[A_((aq))^(-)]]" "color(red)((1))#

We are told that #pH=7.4=-log[H_((aq))^(+)]#

From which #[H_((aq))^(+)]=1.78xx10^(-5)"mol/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.