Question #c1d7c

1 Answer
Mar 30, 2016

Answer:

#(a). pH=4.56#

#(b)#. An extra #4.85"ml"# must be added.

Explanation:

#(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.