Question

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

Socratic, Let's work together to find the answer to this question. There is no specific answer and it is open to intepretation.

Answer 1

The pH of triprotic phosphoric acid 0.200 M will be 1.41

Explanation:

The second part of the solution `darr`

Answer 2

`sf(pH=12.36)`

Explanation:

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))`