Problem
The term sodium phosphate can refer to several phosphate salts. In this problem, sodium phosphate is taken to mean trisodium phosphate, \(\mathrm{Na_3PO_4}\). A \(0.0500\,\mathrm{M}\) \(\mathrm{Na_3PO_4}\) solution is prepared at \(25^\circ\mathrm{C}\). Given \(K_{a3}(\mathrm{H_3PO_4})=4.2\times 10^{-13}\) and \(K_w=1.0\times 10^{-14}\), determine the pH.
Key chemistry: \(\mathrm{Na_3PO_4}\) is a salt of a weak, polyprotic acid (phosphoric acid). The anion \(\mathrm{PO_4^{3-}}\) is a base and raises pH by hydrolysis in water.
Solution
1) Dissociation of sodium phosphate and the hydrolyzing species
In water, \(\mathrm{Na_3PO_4}\) dissociates essentially completely:
Sodium ions are spectators for acid–base behavior, so the pH is controlled by \(\mathrm{PO_4^{3-}}\). The relevant hydrolysis step is:
2) Convert \(K_{a3}\) to \(K_b\) for \(\mathrm{PO_4^{3-}}\)
The third dissociation of phosphoric acid is:
The conjugate-base relationship gives:
3) ICE table for hydrolysis and equilibrium equation
From the salt dissociation, the initial phosphate concentration is \([\mathrm{PO_4^{3-}}]_0 = 0.0500\,\mathrm{M}\). Let \(x\) be the amount hydrolyzed (in \(\mathrm{M}\)).
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| \(\mathrm{PO_4^{3-}}\) | \(0.0500\) | \(-x\) | \(0.0500 - x\) |
| \(\mathrm{HPO_4^{2-}}\) | \(0\) | \(+x\) | \(x\) |
| \(\mathrm{OH^-}\) | \(\approx 0\) | \(+x\) | \(x\) |
The base constant expression is:
4) Solve for \([\mathrm{OH^-}]=x\)
Set \(K_b=2.38\times 10^{-2}\) and solve:
Rearrange to a quadratic:
Use the positive root:
Thus \([\mathrm{OH^-}] \approx 2.46\times 10^{-2}\,\mathrm{M}\). The value is below \(0.0500\,\mathrm{M}\), so the equilibrium concentration \(0.0500-x\) remains positive as required.
5) Convert to pOH and pH
Why one hydrolysis step is sufficient: the next step, \(\mathrm{HPO_4^{2-}}+\mathrm{H_2O}\rightleftharpoons \mathrm{H_2PO_4^-}+\mathrm{OH^-}\), has \(K_b=K_w/K_{a2}\) and is typically far smaller than \(2.38\times 10^{-2}\), so its additional \([\mathrm{OH^-}]\) is negligible compared with \(2.46\times 10^{-2}\,\mathrm{M}\).
Visualization
Final answer
For a \(0.0500\,\mathrm{M}\) sodium phosphate solution interpreted as \(\mathrm{Na_3PO_4}\) at \(25^\circ\mathrm{C}\), \(K_b=K_w/K_{a3}=2.38\times 10^{-2}\) gives \([\mathrm{OH^-}]\approx 2.46\times 10^{-2}\,\mathrm{M}\), \(\mathrm{pOH}\approx 1.61\), and \(\mathrm{pH}\approx 12.39\).