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