# Solutions of Polyprotic Acid/Base Systems, Problem D

4.  Calculate the pH of a solution prepared by adding 55 ml of 0.098 M sodium phosphate to 65 ml of 0.136 M phosphoric acid.

\begin{align} \ce{&Phosphoric\: acid &&– H3PO4\\ &Sodium\: phosphate &&– Na3PO4 \:(which\: dissociates\: to\: PO4^3- )} \end{align}

Both of these species are in the phosphoric acid system shown below.  The two species in appreciable quantities are shown in boldface.

\begin{align} \ce{&\mathbf{H3PO4} + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\\ \ce{& H2PO4- + H2O \leftrightarrow HPO4^2- + H3O+} &&\mathrm{K_{a2}}\\ \ce{& HPO4^2- + H2O \leftrightarrow \mathbf{PO4^{3-}} + H3O+} &&\mathrm{K_{a3}} \end{align}

What we need to realize here is that H3PO4 is an acid, and $$\ce{PO4^3-}$$ is a base.  As such they will react with each other according to the following neutralization reaction.

$\ce{H3PO4 + PO4^3- \leftrightarrow H2PO4- + HPO4^2-}$

If we use the Ka for H3PO4 (7.11×10-3) and the Kb for $$\ce{PO4^3-}$$ (2.4×10-2) and solve for Kn using our established equation (Kn = (Ka×Kb)/Kw) we get a value of 1.7×1010, a very large number.  This neutralization will go to completion.

We can also take this a step further.  In almost all cases for a multistep equilibria system such as this, we can anticipate that only one reaction of the series will be important in determining the pH.  This does not happen in every situation (a little later in the course we will see an example where this does not rigorously work), but it does most of the time.  One expectation is that eventually we will wind up with appreciable amounts of both members of a conjugate pair, which constitutes a buffer.  The most straight forward way of working with this problem is to consider the moles of different chemicals that are present.

Moles of phosphate:                (0.098 mol/L)(0.055 L)  =  0.00539 moles

Moles of phosphoric acid:       (0.136 mol/L)(0.065 L)  =  0.00884 moles

\begin{align} & &&\ce{H3PO4}\hspace{25px} + &&\ce{PO4^3-} \hspace{25px}\leftrightarrow &&\ce{H2PO4-}\hspace{25px} + &&\ce{HPO4^2-}\\ &\ce{Initial} &&0.00884 &&0.00539 &&0 &&0\\ &\ce{Neutralization} &&0.00345 &&0 &&0.00539 &&0.00539 \end{align}

As seen by the data, all of the phosphate ion ($$\ce{PO4^3-}$$) has been used up in the solution.  At this point, which is still an intermediate one, it can be helpful to put in boldface the three species that are present in appreciable quantities in the solution.

\begin{align} &\mathbf{H_3PO_4} + \ce{H2O} \leftrightarrow \mathbf{\ce{H2PO4-}} + \ce{H3O+} &&\mathrm{K_{a1}}\\ &\mathbf{\ce{H2PO4-}} + \ce{H2O} \leftrightarrow \mathbf{\ce{HPO4^2-}} + \ce{H3O+} &&\mathrm{K_{a2}}\\ &\ce{HPO4^2-} + \ce{H2O} \leftrightarrow \ce{PO4^3-} + \ce{H3O+} &&\mathrm{K_{a3}} \end{align}

What we now need to examine is the possibility for neutralization between H3PO4 and $$\ce{HPO4^2-}$$.  Note that the product consists of two equivalents of the species $$\ce{H2PO4-}$$.

$\ce{H3PO4 + HPO4^2- \leftrightarrow 2H2PO4-}$

Evaluation of the Kn for this reaction involves the Ka value of H3PO4 (7.11×10-3) and the Kb value for $$\ce{HPO4^2-}$$ (Kb of Ka2 = 1.578×10-7).  Using our established equation for evaluating Kn, we get a value of 1.12×105, which is a large number.  This neutralization will essentially go to completion.

We can now assess the moles of material produced in the reaction.

\begin{align} & &&\ce{H3PO4}\hspace{25px} + &&\ce{HPO4^2-} \hspace{25px}\leftrightarrow &&\ce{2H2PO4-}\\ &\ce{Initial} &&0.00345 &&0.00539 &&0.00539 \\ &\ce{Neutralization} &&0 &&0.00194 &&0.01229 \end{align}

Be careful when calculating the amount of $$\ce{H2PO4-}$$ in the final solution.  There is an amount already present (0.00539 moles) and we produce two equivalents in the reaction above (2×0.00345 moles), hence the total of 0.01229 moles of $$\ce{H2PO4-}$$.

Once again, it is helpful to boldface the species in the series of reactions that are present in appreciable quantities.

\begin{align} \ce{&H3PO4 + H2O \leftrightarrow H2PO4- + H3O+} &&\mathrm{K_{a1}}\\ &\mathbf{\ce{H2PO4-}} + \ce{H2O \leftrightarrow \mathbf{HPO4^{2-}} + H3O+} &&\mathrm{K_{a2}}\\ \ce{&HPO4^2- + H2O \leftrightarrow PO4^3- + H3O+} &&\mathrm{K_{a3}} \end{align}

Note that we now have appreciable quantities of a conjugate pair.  Since the distribution of the species in a conjugate pair will not change (these two cannot neutralize each other since they will simply reform each other), we can now calculate the pH using the appropriate Henderson-Hasselbalch equation.

$\mathrm{pH=pK_a+ \log\left( \dfrac{[HPO_4^{2-}]}{[H_2PO_4^-]}\right )=7.198+\log\left( \dfrac{0.00194}{0.01229}\right )=6.4}$

It might be useful to check and make sure that ignoring the other two reactions was a reasonable thing to do.

We can calculate the concentration of H3PO4 using the expression for Kn.  We first need to convert the moles of the different species to molarity for the subsequent calculations.

$\mathrm{[H_2PO_4^-] =\dfrac{0.01229\: mol}{0.120\: L} = 0.102\: M}$

$\mathrm{[HPO_4^{2-}] =\dfrac{0.00194\: mol}{0.120\: L} = 0.016\: M}$

$\mathrm{K_n= \dfrac{[H_2PO_4^-]^2}{[H_3PO_4][HPO_4^{2-}]}=\dfrac{(0.102)^2}{(x)(0.016)}=1.12\times10^5}$

$\mathrm{x = [H_3PO_4] = 5.8\times 10^{-6}}$

This is a very small quantity of H3PO4, so ignoring the Ka1 reaction was justified.  We can calculate the amount of $$\ce{PO4^3-}$$ using Ka3

$\mathrm{K_{a3}= \dfrac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]} = \dfrac{(x)(3.98\times10^{-7})}{(0.016)} = 4.17\times10^{-13}}$

$\mathrm{x = [PO_4^{3-}] = 1.68\times 10^{-8}}$

Once again, we see that this is an exceptionally small quantity that can be ignored.  As has been the pattern so far, in problems involving polyprotic acids or bases, we see that only one reaction is significant in determining the pH of the solution.  This situation will occur in almost all cases for these reactions.  The approach to these problems is to identify the important reaction and solve for the pH using only that one, then use the values that are obtained to check that we could ignore the other ones.  Of course, if the pKa values for the reactions are appreciably different (value of 10 or greater), we would know ahead of time that only one reaction will be important.