Determine the final pH of a 0.200M triprotic phosphoric acid solution?

The second part of the solution #darr#

There is not enough information given in the question to determine the pH at the 3rd equivalence point so I will assume we are starting with 10.00 ml of #sf(0.2Mcolor(white)(x)H_3PO_(4(aq)))# in the conical flask and #sf(0.2Mcolor(white)(x)NaOH_((aq)))# in the burette.

The pH can be monitored with a pH meter.

The curve given in the question appears to be a generic one for a triprotic acid. The phosphoric acid curve differs slightly:

As you can see there is no rapid change in pH at the 3rd equivalence point.

This is due to the effect of the auto - ionisation of water. #sf(K_(a3))# is close to the ionic product of water (#sf(10^(-14))#) and at high pH values like this the phosphate (#sf(PO_4^(3-))#) ion competes with #sf(OH^-)# ions for the #sf(H^+)# ions from water.

Phosphoric acid is triprotic and has 3 dissociations:

#sf(H_3PO_4rightleftharpoonsH_2PO_4^(-)+H^+" "color(red)((1)))#

#sf(K_(a1)=([H_2PO_4^(-)][H^+])/([H_3PO_4])=7.5xx10^(-3))#

#sf(pK_(a1)=2.148)#

#sf(H_2PO_4^(-)rightleftharpoonsHPO_4^(2-)+H^(+)" "color(red)((2)))#

#sf(K_(a2)=([HPO_4^(2-)][H^+])/([H_2PO_4^(2-)])=6.23xx10^(-8))#

#sf(pK_(a2)=7.198)#

#sf(HPO_4^(2-)rightleftharpoonsPO_4^(3-)+H^(+)" "color(red)((3)))#

#sf(K_(a3)=([PO_4^(3-)][H^(+)])/([HPO_4^(2-)])=4.8xx10^(-13))#

#sf(pK_(a3)=12.319)#

You can estimate the #sf(pK_a)#value experimentally from the graph:

If you consider #color(red)((1))# then when the titration is halfway to the 1st equivalence point we can say that :

#sf([H_3PO_4]=[H_2PO_4^(-)])#

Rearranging the expression for #sf(K_(a1))# gives:

#sf([H^(+)]=K_(a1)xx(cancel([H_3PO_4]))/(cancel([H_2PO_4^(-)])))#

So #sf(pH=pK_(a1))#

This is marked on the graph . You can read this off at 0.5 equivalents which, in our case, would be when 10 ml of base are added. This gives #sf(pK_(a1)# to be just over 2.

To get the pH at the first equivalence point again, you can read this off the graph but a good way of doing this is to split the difference between #sf(pK_(a1))# and #sf(pK_(a2)rArr)#

#sf(pH=1/2(pK_(a1)+pK_(a2))=1/2(2.148+7.198)=4.67)#

The same reasoning can be applied to the 2nd equivalence point.

At the 3rd equivalence point all the protons have been lost and we have a solution of sodium phosphate:

#sf(H_3PO_4+3NaOHrarrNa_3PO_4+3H_2O)#

From the graph you can read off the pH when 3 equivalents of base has been added which, in our case will be 30 ml of 0.2 M NaOH.

This gives a pH of just over 12.

We can calculate this as follows:

The number of moles of #sf(NaOH)# added = #sf(cxxv=0.2xx30=6mmol)#.

From the equation the number of moles of #sf(PO_4^(3-)# formed must be 1/3 of this.

#:.##sf(n_(PO_4^(3-))=6/3=2mmol)#

The total volume of the solution in the flask after titration = #sf(10.00+30.00 = 40.00color(white)(x) ml)#.

#:.##sf([PO_4^(3-)]=n/v=(2xx10^(-3))/(0.04)=0.05color(white)(x)"mol/l")#

The phosphate ion is quite basic and is hydrolysed by water:

#sf(PO_4^(3-)+H_2OrightleftharpoonsHPO_4^(2-)+OH^(-))#

For which #sf(K_b=([HPO_4^(2-)][OH^(-)])/([PO_4^(3-)]))#

To find #sf([OH^-])# and hence the pH we need to find #sf(K_b)#.

We can do this using:

#sf(K_(a3)xxK_b=K_w=10^(-14)color(white)(x)"mol".^2"l"^(-2))#

#:.##sf(K_b=(10^(-14))/(4.8xx10^(-13))=0.0208color(white)(x)"mol/l")#

Now we can set up an ICE table based on concentrations in #sf("mol/l"" "rArr)#

#sf(" "PO_4^(3-)+H_2OrightleftharpoonsHPO_4^(2-)+OH^(-))#

#sf(color(red)(I)" "0.05" "0" "0)#

#sf(color(red)(C)" "-x" "+x" "+x)#

#sf(color(red)(E)" "(0.05-x)" "x" "x)#

#:.##sf(K_b=(x^2)/((0.05-x))=0.0208)#

Because of the large value of #sf(K_b)# we cannot make the assumption that #sf((0.05-x)rArr0.05)# so we need to multiply this out to get:

#sf(x^2+0.0208x-0.001=0)#

This can be solved using the quadratic formula, which I won't go into here. Ignoring the -ve root this gives:

#sf(x=[OH^-]=0.0231color(white)(x)"mol/l")#

So #sf(pOH=-log[OH^(-)]=-log(0.0231)=1.635)#

Since #sf(pH+pOH=14)#

Then #sf(pH=14-pOH=14-1.635=color(red)(12.36))#