**(A)**

Carbonic acid is a weak diprotic acid with 2 ionisations:

#H_2CO_3rightleftharpoonsHCO_3^(-)+H^+" "K_(a1)=4.27xx10^(-7)"mol/l"#

#HCO_3^(-)rightleftharpoonsCO_3^(2-)+H^(+)" "K_(a2)=4.68xx10^(-11)"mol/l"#

In the flask we have #20.00"ml"# of #0.10"M"# sodium carbonate. As the acid is run in from the burette the following reaction occurs as the carbonate ion becomes protonated:

#CO_3^(2-)+H^+rarrHCO_3^(-)#

The initial number of moles of #CO_3^(2-)# is given by:

#nCO_3^(2-)=20.00xx0.10/1000=2.00xx10^(-3)#

At #0.05"ml"# before the end-point we have added #19.05"ml"# of #0.10"M"color(white)xHCl"_((aq))#.

The number of moles of #HCl# added is given by:

#nHCl=19.05xx0.10/1000=1.905xx10^(-3)#

This will be equal to the number of moles of #HCO_3^-# **formed**.

So the number of moles of #CO_3^(2-)# **remaining** is given by:

#nCO_3^(2-)=(2.00xx10^(-3)-1.905xx10^(-3))=0.095xx10^(-3)#

To get the number of moles of #H^+# released we can use an **ICE** table based on #"mol/l"#

#color(white)(xxx)HCO_3^(-)color(white)(xxxxx)rightleftharpoonscolor(white)(xxxx)CO_3^(2-)color(white)(xxxxx)+color(white)(xxxxx)H^+#

#color(red)"I"color(white)(xxx)1.905xx10^(-3)color(white)(xxxxxx)0.095xx10^(-3)color(white)(xxxxxxxx)0#

#color(red)"C"color(white)(xxxx)-xcolor(white)(xxxxxxxxxxxxxx)+xcolor(white)(xxxxxxxxx)+x#

#color(red)"E"color(white)(xxx)(1.905xx10^(-3))-xcolor(white)(xxx)(0.0905xx10^(-3))+xcolor(white)(xx)x#

Since #x# is much smaller than #nHCO_3^-# and #nCO_3^(2-)# we can say that at equilibrium:

#:.nHCO_3^(-)=(1.905xx10^(-3))-xrArr(1.905xx10^(-3))#

and:

#nCO_3^(2-)=(0.095xx10^(-3))+xrArr(0.095xx10^(-3))#

I will make this approximation in the rest of the calculations.

The expression for #K_(a2)# is:

#K_(a2)=([CO_3^(2-)][H^+])/[[HCO_3^-]]#

Since the total volume is common to all species we can write:

#:.[H^+]=4.68xx10^(-11)xx(1.905xxcancel10^(-3))/(0.095xxcancel(10^(-3))#

#:.[H^(+)]=9.38xx10^(-10)#

#:.pH=-log(9.38xx10^(-10))#

#color(red)(pH=9.02#

This represents the pH just before the 1st end-point.

**(B)**

At the end-point the 1st protonation is complete:

#CO_3^(2-)+H^(+)rarrHCO_3^-#

So we now have #40.00"ml"# of a solution containing #2.00xx10^(-3)# moles of #HCO_3^-# ions.

As we continue to add acid the 2nd protonation starts to happen:

#HCO_3^(-)+H^+rarrH_2CO_3#

After the end point at #20.00"ml"# we add #0.05"ml"# of #0.1"M"color(white)(x)"HCl"#:

The number of moles of #H^+# added is given by:

#nH^+=0.05xx0.10/1000=5.00xx10^(-6)#

From the equation of the reaction for this stage of the titration we can say that:

#nH_2CO_3# formed #=5.00xx10^(-6)#

So the number of moles of #HCO_3^-# remaining is given by:

#nHCO_3^(-)=(2.00xx10^(-3))-(5.00xx10^(-6))=0.001995#

#K_(a1)=([HCO_3^-][H^+])/([H_2CO_3])#

#:.[H^+]=K_(a1)xx([H_2CO_3])/([HCO_3^-])#

#:.[H^+]=4.27xx10^(-7)xx(5.00xx10^(-6))/(0.001995)=1.0675xx10^(-9)#

#:.pH=-log(1.0675xx10^(-9))#

#color(red)(pH=8.97)#

Phenolphthalein is an indicator that changes colour over the pH range 8.3 - 10 so is a suitable indicator for this end-point.

**(C)**

Now the titration continues to #0.05"ml"# short of the 2nd end-point which occurs at #40.00"ml"# of #"HCl"# added.

Since the 1st end-point we have added #19.05"ml"# of #0.10"M"color(white)(x)" HCl"#.

#:.nH^+# added #=19.05xx0.10/1000=1.905xx10^(-3)#

#:.nH_2CO_3# formed #=1.905xx10^(-3)#

#:.nHCO_3^-# remaining = #(2.00xx10^(-3))-(1.905xx10^(-3))=0.095xx10^(-3)#

#K_(a1)=([H^+][HCO_3^-])/([H_2CO_3])#

#:.[H^+]=K_(a1)xx[[H_2CO_3]]/[[HCO_3^(-)]]#

#:.[H^+]=4.27xx10^(-7)xx(1.905xxcancel(10^(-3)))/(0.095xxcancel(10^(-3)#

#[H^+]=8.561xx10^(-6)#

#pH=-log(8.561xx10^(-6))#

#color(red)(pH=5.06)#

**(D)**

When we get to the end point this reaction has reached completion:

#Na_2CO_(3(aq))+2HCl_((aq))rarrH_2CO_(3(aq))+2NaCl_((aq))#

So we have # 60"ml"# of a solution that contains #2xx10^(-3)# moles of #H_2CO_3#.

We can find the pH of this solution. I will ignore the 2nd ionisation because #K_(a2)# is much smaller than #K_(a1)#.

#K_(a1)=([HCO_3^-][H^+])/([H_2CO_3])#

Since #[H^+]=[HCO_3^-]# we can write:

#[H^+]^(2)=K_(a1)xx[H_2CO_3]#

We can find #[H_2CO_3]# using #c=n/v:#

#[H_2CO_3]=(2.00xx10^(-3))/(60.00/1000)=0.033"mol/l"#

#:.[H^+]^2=4.27xx10^(-7)xx0.033#

#[H^+]^2=1.41xx10^(-8)#

#:.[H^+]=1.187xx10^(-4)"mol/l"#

#pH=-log(1.187xx10^(-4)) #

#pH=3.92#

Now we add an extra #0.05"ml"# of # 0.1"M"color(white)(x)" HCl"#. The extra number of moles of #H^+# added is given by:

#0.10xx0.05/1000=5.00xx10^(-6)#

The number of moles of #H^+# which are already there due to the dissociation of #H_2CO_3# is given by:

#n=cxxv=1.187xx10^(-4)xx60.00/1000=7.122xx10^(-6)#

So we add this to the extra #H^+# added to get the total moles of #H^(+)# present:

#nH^(+)=(7.122+5)xx10^(-6)#

#nH^(+)=1.2122xx10^(-5)#

The new total volume is #60.05"ml".#

#:.[H^+]=(1.2122xx10^(-5))/(60.05/1000)=2.00xx10^(-4)"mol/l"#

#pH=-log(2xx10^(-4))#

#color(red)(pH=3.7)#

So you can see the pH has dropped slightly.

I have ignored any #H^+# from the ionisation of water.

The pH curve looks like this:

(In our example A is at 20.00ml, B is at 40.00ml)

Methyl Orange is an indicator which changes colour over the range pH 3.1 - 4.4 so is suitable for this end-point.

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