# Phosphoric Acid Problem

What is the $\ce{[HPO4^2-]}$ of a solution labeled "$\pu{0.10 M}$ Phosphoric Acid"? Given the $K_\mathrm{a}$ values of $7.1 \times 10^{-3}$, $6.3 \times 10^{-8}$ and $4.2 \times 10^{-13}$ for the first, second and third ionizations, respectively.

Some sources have shown that the answer is $6.3 \times 10^{-8}$, but following through the long calculation through each ionization to get the equilibrium concentration of $\ce{HPO4^{2-}}$ I get an answer of $3.8 \times 10^{-5}$. Hoping someone can enlighten me on this matter.

Attempt at solving:

So first what I did was to individually find the concentrations using the given $K_\mathrm{a}$ values. Working from the top, I calculated for the concentration of $\ce{H2PO4-}$ from the given concentration of $\ce{H3PO4}$. Had to go quadratic with this one since the first ionization constant is a relatively big number. After getting the concentration of $\ce{H2PO4-}$, went on to get the concentration of $\ce{HPO4^2-}$.

After this one, used the final ionization constant to get the concentration of $\ce{PO4^3-}$ produced, which I found to be so small that the $\ce{HPO4^2-}$ from earlier isn't all that much affected. And after all that I get the answer of $\pu{3.8e-5 M}$ for $\ce{HPO4^2-}$.

• It would be very beneficial if you could include your whole approach. That way we can better follow your train of thought and may point out where you went wrong. – Martin - マーチン Jun 28 '17 at 7:19

Let's list what's going on here. Orthophosphoric acid dissociates stepwise:

\begin{align} \ce{H3PO4 + H2O & <=>[$K_\mathrm{a1}$] H3O+ + H2PO4–} \tag{R1}\\ \ce{H2PO4- + H2O & <=>[$K_\mathrm{a2}$] H3O+ + HPO4^2–} \tag{R2}\\ \ce{HPO4^2- + H2O & <=>[$K_\mathrm{a3}$] H3O+ + PO4^3–} \tag{R3} \end{align}

Mass balances for protons and phosporous, respectively:

\begin{align} C_\ce{H} & = [\ce{H3O+}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + [\ce{H3PO4}] = [\ce{H3O+}] + \sum_{n=1}^3 [\ce{H_$n$PO4^$(3-n)-$}] \tag{1}\\ C_\ce{P} & = [\ce{PO4^3-}] + [\ce{HPO4^2–}] + [\ce{H2PO4–}] + \ce{H3PO4} \tag{2} \end{align}

Now we compare the equilibrium constants:

\begin{align} K_\mathrm{a1} & = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.1 \times 10^{−3} \tag{3}\\ K_\mathrm{a2} & = \frac{[\ce{H3O+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.3 \times 10^{−8} \tag{4}\\ K_\mathrm{a3} & = \frac{[\ce{H3O+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 4.2 \times 10^{−13} \tag{5} \end{align}

Notice that $K_\mathrm{a1} >> K_\mathrm{a2}$ (more that $10^5$), therefore by the first dissociation step phosphoric acid behaves as an acid of medium strength, producing equal amounts of $\ce{H3O+}$ and $\ce{H2PO4-}$ (considering charge balance):

$$[\ce{H3O+}] \approx [\ce{H2PO4-}] \tag{6}$$

which directly leads us to the answer. Using expression for $K_\mathrm{a2}$:

$$[\ce{HPO4^{2-}}] = K_\mathrm{a2} \times \frac{[\ce{H2PO4-}]}{[\ce{H3O+}]} = K_\mathrm{a2} \tag{7}$$

That's it, $[\ce{HPO4^{2-}}] = \pu{6.3 \times 10^{−8} (mol/L)}$

Now for the fun part, if you really want to solve quadratic equation, we can also find solution pH and $[\ce{PO4^{3-}}]$.

Based on the values of equilibrium constants, it is safe to assume that pretty much only first dissociation step defines solution's pH. We can rewrite equation for $K_\mathrm{a1}$ as follows:

$$K_\mathrm{a1} = \frac{[\ce{H3O+}]^2}{C_0 - [\ce{H3O+}]},$$

where $C_0$ is the concentration of the solution ($\pu{0.10 M}$). Using the given numbers we have

$$7.1 \times 10^{-3} = \frac{[\ce{H3O+}]^2}{0.10 - [\ce{H3O+}]}$$ $$[\ce{H3O+}]^2 + 7.1 \times 10^{-3} [\ce{H3O+}] - 7.1 \times 10^{-4} = 0$$ $$[\ce{H3O+}] = \frac{-7.1 \times 10^{-3} + \sqrt{(7.1 \times 10^{-3})^2 + 4 \times 7.1 \times 10^{-4}}}{2} = \pu{2.33 \times 10^{-2} (mol/L)}$$

$$\mathrm{pH} = - \log{[\ce{H3O+}]} = 1.63$$

WolframAlpha thinks the same.

Now we know $[\ce{H2PO4-}] = \pu{2.33 \times 10^{-2} (mol/L)}$ and we can finally find $[\ce{PO4^3-}]$ from the expression for $K_\mathrm{a3}$:

$$[\ce{PO4^{3-}}] = K_\mathrm{a3} \times \frac{[\ce{HPO4^2-}]}{[\ce{H3O+}]} = 4.2 \times 10^{−13} \times \frac{6.3 \times 10^{−8}}{2.33 \times 10^{-2}} = \pu{1.1 \times 10^{-18} (mol/L)}$$

This demonstrates that in pretty concentrated solution of phosphoric acid anions $\ce{PO4^3-}$ are practically non-existing.