# 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.

Sep 20, 2016

The pH of triprotic phosphoric acid 0.200 M will be 1.41

#### Explanation: The second part of the solution $\downarrow$ Sep 22, 2016

$\textsf{p H = 12.36}$

#### 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 $\textsf{0.2 M \textcolor{w h i t e}{x} {H}_{3} P {O}_{4 \left(a q\right)}}$ in the conical flask and $\textsf{0.2 M \textcolor{w h i t e}{x} N a O {H}_{\left(a q\right)}}$ 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. $\textsf{{K}_{a 3}}$ is close to the ionic product of water ($\textsf{{10}^{- 14}}$) and at high pH values like this the phosphate ($\textsf{P {O}_{4}^{3 -}}$) ion competes with $\textsf{O {H}^{-}}$ ions for the $\textsf{{H}^{+}}$ ions from water.

Phosphoric acid is triprotic and has 3 dissociations:

$\textsf{{H}_{3} P {O}_{4} r i g h t \le f t h a r p \infty n s {H}_{2} P {O}_{4}^{-} + {H}^{+} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{{K}_{a 1} = \frac{\left[{H}_{2} P {O}_{4}^{-}\right] \left[{H}^{+}\right]}{\left[{H}_{3} P {O}_{4}\right]} = 7.5 \times {10}^{- 3}}$

$\textsf{p {K}_{a 1} = 2.148}$

$\textsf{{H}_{2} P {O}_{4}^{-} r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}^{+} \text{ } \textcolor{red}{\left(2\right)}}$

$\textsf{{K}_{a 2} = \frac{\left[H P {O}_{4}^{2 -}\right] \left[{H}^{+}\right]}{\left[{H}_{2} P {O}_{4}^{2 -}\right]} = 6.23 \times {10}^{- 8}}$

$\textsf{p {K}_{a 2} = 7.198}$

$\textsf{H P {O}_{4}^{2 -} r i g h t \le f t h a r p \infty n s P {O}_{4}^{3 -} + {H}^{+} \text{ } \textcolor{red}{\left(3\right)}}$

$\textsf{{K}_{a 3} = \frac{\left[P {O}_{4}^{3 -}\right] \left[{H}^{+}\right]}{\left[H P {O}_{4}^{2 -}\right]} = 4.8 \times {10}^{- 13}}$

$\textsf{p {K}_{a 3} = 12.319}$

You can estimate the $\textsf{p {K}_{a}}$value experimentally from the graph:

If you consider $\textcolor{red}{\left(1\right)}$ then when the titration is halfway to the 1st equivalence point we can say that :

$\textsf{\left[{H}_{3} P {O}_{4}\right] = \left[{H}_{2} P {O}_{4}^{-}\right]}$

Rearranging the expression for $\textsf{{K}_{a 1}}$ gives:

$\textsf{\left[{H}^{+}\right] = {K}_{a 1} \times \frac{\cancel{\left[{H}_{3} P {O}_{4}\right]}}{\cancel{\left[{H}_{2} P {O}_{4}^{-}\right]}}}$

So $\textsf{p H = p {K}_{a 1}}$

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 $\textsf{p {K}_{a 1}}$ and $\textsf{p {K}_{a 2} \Rightarrow}$

$\textsf{p H = \frac{1}{2} \left(p {K}_{a 1} + p {K}_{a 2}\right) = \frac{1}{2} \left(2.148 + 7.198\right) = 4.67}$

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:

$\textsf{{H}_{3} P {O}_{4} + 3 N a O H \rightarrow N {a}_{3} P {O}_{4} + 3 {H}_{2} O}$

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 $\textsf{N a O H}$ added = $\textsf{c \times v = 0.2 \times 30 = 6 m m o l}$.

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

$\therefore$$\textsf{{n}_{P {O}_{4}^{3 -}} = \frac{6}{3} = 2 m m o l}$

The total volume of the solution in the flask after titration = $\textsf{10.00 + 30.00 = 40.00 \textcolor{w h i t e}{x} m l}$.

$\therefore$$\textsf{\left[P {O}_{4}^{3 -}\right] = \frac{n}{v} = \frac{2 \times {10}^{- 3}}{0.04} = 0.05 \textcolor{w h i t e}{x} \text{mol/l}}$

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

$\textsf{P {O}_{4}^{3 -} + {H}_{2} O r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + O {H}^{-}}$

For which $\textsf{{K}_{b} = \frac{\left[H P {O}_{4}^{2 -}\right] \left[O {H}^{-}\right]}{\left[P {O}_{4}^{3 -}\right]}}$

To find $\textsf{\left[O {H}^{-}\right]}$ and hence the pH we need to find $\textsf{{K}_{b}}$.

We can do this using:

$\textsf{{K}_{a 3} \times {K}_{b} = {K}_{w} = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol".^2"l}}^{- 2}}$

$\therefore$$\textsf{{K}_{b} = \frac{{10}^{- 14}}{4.8 \times {10}^{- 13}} = 0.0208 \textcolor{w h i t e}{x} \text{mol/l}}$

Now we can set up an ICE table based on concentrations in $\textsf{\text{mol/l"" } \Rightarrow}$

$\textsf{\text{ } P {O}_{4}^{3 -} + {H}_{2} O r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + O {H}^{-}}$

$\textsf{\textcolor{red}{I} \text{ "0.05" "0" } 0}$

$\textsf{\textcolor{red}{C} \text{ "-x" "+x" } + x}$

$\textsf{\textcolor{red}{E} \text{ "(0.05-x)" "x" } x}$

$\therefore$$\textsf{{K}_{b} = \frac{{x}^{2}}{\left(0.05 - x\right)} = 0.0208}$

Because of the large value of $\textsf{{K}_{b}}$ we cannot make the assumption that $\textsf{\left(0.05 - x\right) \Rightarrow 0.05}$ so we need to multiply this out to get:

$\textsf{{x}^{2} + 0.0208 x - 0.001 = 0}$

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

$\textsf{x = \left[O {H}^{-}\right] = 0.0231 \textcolor{w h i t e}{x} \text{mol/l}}$

So $\textsf{p O H = - \log \left[O {H}^{-}\right] = - \log \left(0.0231\right) = 1.635}$

Since $\textsf{p H + p O H = 14}$

Then $\textsf{p H = 14 - p O H = 14 - 1.635 = \textcolor{red}{12.36}}$