#(a).#
Ethanoic acid is a weak acid and dissociates:
#CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^+" "color(red)((1))#
The expression for #K_a# is:
#K_a=([CH_3COO^-][H^+])/([CH_3COOH])" "color(red)((2))#
These refer to equilibrium concentrations.
When the #NaOH_((aq))# is added the following neutralisation takes place:
#CH_3COOH_((aq))+OH_((aq))^(-)rarrCH_3COO_((aq))^(-)+H_2O_((l))" "color(red)((3))#
This tells us that for every mole of #OH_((aq))^-# added, 1 mole of #CH_3COO_((aq))^-# will form and 1 mole of #CH_3COOH_((aq))# will be consumed.
The number of moles of #OH_((aq))^-# added is given by:
#n_(OH^-)=cxxv=0.2xx20/1000=4xx10^(-3)#
From #color(red)((3))# we can therefore say:
#n_(CH_3COO^-)=4xx10^-3#
The initial moles of #CH_3COOH_((aq))# is given by:
Initial moles #=cxxv=0.2xx50/1000=10xx10^(-3)#
So the number of moles of #CH_3COOH_((aq))# remaining is given by:
#n_(CH_3COOH)=(10xx10^(-3))-(4xx10^(-3))=6xx10^(-3)#
Now we can set up an ICE table:
#CH_3COOH_((aq))" "rightleftharpoons" "CH_3COO_((aq))^(-)" "+H_((aq))^+#
#color(red)("I")" "6xx10^(-3)" "4xx10^(-3)" "0#
#color(red)("C")" "-x" "+x" "+x#
#color(red)("E")" "(6xx10^(-3)-x)" "(4xx10^(-3)+x)" "x#
Because the dissociation is so small I am going to assume that #x# is much smaller than the initial moles of both #CH_3COOH# and #CH_3COO^-#. This means that:
#(6xx10^(-3)-x)rArr6xx10^(-3)#
and
#(4xx10^(-3)-x)rArr4xx10^(-3)#
So rearranging #color(red)((2))# gives:
#[H^+]=k_axx([CH_3COOH])/([CH_3COO^-])" "color(red)((4))#
Since the total volume is common we can write:
#[H^+]=1.8xx10^(-5)xx[(6xxcancel(10^(-3))]/(4xxcancel(10^(-3))]]#
#:.[H^+]=1.8xx10^(-5)xx3/2=2.7xx10^(-5)#
#pH=-log(2.7xx10^(-5))#
#color(red)(pH=4.56#
#(b).#
Now the pH is raised by adding extra alkali. We can use the new pH to get the ratio of acid to salt:
#pH=4.74#
#:.-log[H^+]=4.74#
#:.[H^+]=1.82xx10^(-5)"mol/l"#
From #color(red)((4))# we can write:
#1.82xxcancel(10^(-5))=1.8xxcancel(10^(-5))xx([CH_3COOH])/([CH_3COO^(-)])#
#:.([CH_3COOH])/([CH_3COO^(-)])=1.82/1.8=1.01" "color(red)((5))#
The number of moles of #OH^-# is given by:
#n_(OH^-)=0.2xxV_(OH^-)#
#:.n_(CH_3COO^-)=0.2xxV_(OH^-)#
and
#n_(CH_3COOH)=(10xx10^(-3))-0.2xxV_(OH^-)#
Substituting these into #color(red)((5))# gives:
#(0.01-0.2V_(OH^-))/(0.2V_(OH^-))=1.01#
#:.1.01xx0.2V_(OH^-)=0.01-0.2V_(OH^-)#
#:.0.202V_(OH^-)+0.2V_(OH^-)=0.01#
#:.0.402V_(OH^-)=0.01#
#:.V_(OH^-)=0.01/0.402=0.02487"L"#
#V_(OH^-)=24.87"ml"#
Since #20"ml"# was used initially, this represents an extra #4.87"ml"# of #0.2"M"" "NaOH_((aq))#
In practice this would be #4.85"ml"# given the usual accuracy of a graduated pipette.
