# 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. Commented Jun 28, 2017 at 7:19
• do you think the current answer is sufficient to accept?
– uhoh
Commented May 17, 2022 at 17:36

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.