(a).(a).
Ethanoic acid is a weak acid and dissociates:
CH_3COOH_((aq))rightleftharpoonsCH_3COO_((aq))^(-)+H_((aq))^+" "color(red)((1))CH3COOH(aq)⇌CH3COO−(aq)+H+(aq) (1)
The expression for K_aKa is:
K_a=([CH_3COO^-][H^+])/([CH_3COOH])" "color(red)((2))Ka=[CH3COO−][H+][CH3COOH] (2)
These refer to equilibrium concentrations.
When the NaOH_((aq))NaOH(aq) is added the following neutralisation takes place:
CH_3COOH_((aq))+OH_((aq))^(-)rarrCH_3COO_((aq))^(-)+H_2O_((l))" "color(red)((3))CH3COOH(aq)+OH−(aq)→CH3COO−(aq)+H2O(l) (3)
This tells us that for every mole of OH_((aq))^-OH−(aq) added, 1 mole of CH_3COO_((aq))^-CH3COO−(aq) will form and 1 mole of CH_3COOH_((aq))CH3COOH(aq) will be consumed.
The number of moles of OH_((aq))^-OH−(aq) added is given by:
n_(OH^-)=cxxv=0.2xx20/1000=4xx10^(-3)nOH−=c×v=0.2×201000=4×10−3
From color(red)((3))(3) we can therefore say:
n_(CH_3COO^-)=4xx10^-3nCH3COO−=4×10−3
The initial moles of CH_3COOH_((aq))CH3COOH(aq) is given by:
Initial moles =cxxv=0.2xx50/1000=10xx10^(-3)=c×v=0.2×501000=10×10−3
So the number of moles of CH_3COOH_((aq))CH3COOH(aq) remaining is given by:
n_(CH_3COOH)=(10xx10^(-3))-(4xx10^(-3))=6xx10^(-3)nCH3COOH=(10×10−3)−(4×10−3)=6×10−3
Now we can set up an ICE table:
CH_3COOH_((aq))" "rightleftharpoons" "CH_3COO_((aq))^(-)" "+H_((aq))^+CH3COOH(aq) ⇌ CH3COO−(aq) +H+(aq)
color(red)("I")" "6xx10^(-3)" "4xx10^(-3)" "0I 6×10−3 4×10−3 0
color(red)("C")" "-x" "+x" "+xC −x +x +x
color(red)("E")" "(6xx10^(-3)-x)" "(4xx10^(-3)+x)" "xE (6×10−3−x) (4×10−3+x) x
Because the dissociation is so small I am going to assume that xx is much smaller than the initial moles of both CH_3COOHCH3COOH and CH_3COO^-CH3COO−. This means that:
(6xx10^(-3)-x)rArr6xx10^(-3)(6×10−3−x)⇒6×10−3
and
(4xx10^(-3)-x)rArr4xx10^(-3)(4×10−3−x)⇒4×10−3
So rearranging color(red)((2))(2) gives:
[H^+]=k_axx([CH_3COOH])/([CH_3COO^-])" "color(red)((4))[H+]=ka×[CH3COOH][CH3COO−] (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.
