Calculate The Ph Of A 0.150 M Na2Hpo4 Solution

Use this interactive chemistry calculator to estimate the pH of sodium hydrogen phosphate solution with either the amphiprotic approximation or a more rigorous equilibrium calculation. Default values are set for a 0.150 M Na2HPO4 aqueous solution at 25 degrees Celsius.

Expert guide: how to calculate the pH of a 0.150 M Na2HPO4 solution

Sodium hydrogen phosphate, written as Na2HPO4, is one of the most important phosphate salts used in chemistry, biochemistry, analytical labs, water treatment, and buffer preparation. When this salt dissolves in water, it forms sodium ions and the amphiprotic phosphate species HPO4 2-. The central question is straightforward: what is the pH of a 0.150 M Na2HPO4 solution? The answer is mildly basic, and under standard 25 degrees Celsius conditions the pH is typically around 9.7 to 9.8 when calculated from the amphiprotic behavior of hydrogen phosphate. An exact equilibrium treatment gives a very similar result.

To understand why, you need to see how HPO4 2- behaves in water. It is called amphiprotic because it can both accept a proton and donate a proton. In one direction, HPO4 2- can act as a base, reacting with water to make OH- and H2PO4-. In the other direction, it can act as a weak acid, donating a proton to form PO4 3-. Because these two tendencies are balanced by the second and third dissociation constants of phosphoric acid, the resulting pH sits between pKa2 and pKa3. That is why the classic shortcut for an amphiprotic salt is:

pH approximately equals 1/2(pKa2 + pKa3)

Using common phosphate constants at 25 degrees Celsius, pKa2 is about 7.21 and pKa3 is about 12.32. Substituting them gives:

pH approximately equals 1/2(7.21 + 12.32) = 9.765

That value is the standard textbook estimate for the pH of an aqueous Na2HPO4 solution, including the 0.150 M case. Because the concentration is moderate and the amphiprotic approximation is excellent for this system, the concentration has only a small effect on the answer over ordinary lab ranges. In practice, ionic strength and temperature can shift the measured pH somewhat, but 9.76 is the benchmark value you should expect from a theoretical calculation.

Why Na2HPO4 makes a basic solution

Na2HPO4 dissociates essentially completely in water:

Na2HPO4 → 2 Na+ + HPO4 2-

The sodium ion is a spectator ion because it comes from the strong base NaOH and does not significantly hydrolyze. The chemistry is controlled by HPO4 2-. Two competing equilibria matter:

• As a base: HPO4 2- + H2O ⇌ H2PO4- + OH-
• As an acid: HPO4 2- ⇌ PO4 3- + H+

The basic behavior is stronger than the acidic behavior in the pH region relevant to this solution, so the net result is an alkaline solution. However, it is not strongly basic like NaOH because phosphate hydrolysis is limited by relatively small equilibrium constants.

The amphiprotic shortcut method

If a solution contains an amphiprotic species such as HCO3-, H2PO4-, or HPO4 2-, a very useful approximation exists. For a species that lies between two acid dissociation steps, the pH can be estimated from the average of the two adjacent pKa values. For HPO4 2-, those adjacent steps are:

1. H2PO4- ⇌ H+ + HPO4 2- with pKa2
2. HPO4 2- ⇌ H+ + PO4 3- with pKa3

So for Na2HPO4:

1. Identify pKa2 and pKa3
2. Add them together
3. Divide by 2

Example:

• pKa2 = 7.21
• pKa3 = 12.32
• pH = 1/2(7.21 + 12.32) = 9.765

This is the fastest route if your instructor, textbook, or exam expects the amphiprotic treatment. It is also often the cleanest conceptual answer when concentration effects and activity corrections are neglected.

The exact equilibrium method

An advanced calculation uses full phosphate speciation rather than just the shortcut. In that approach, you write the acid dissociation constants for phosphoric acid, combine them with water autoionization, include the sodium concentration from the dissolved salt, and apply charge balance. For a formal phosphate concentration of 0.150 M, the total dissolved phosphate must be distributed among H3PO4, H2PO4-, HPO4 2-, and PO4 3-. At the pH of interest, H3PO4 is negligible, while H2PO4-, HPO4 2-, and a very small amount of PO4 3- account for almost all phosphate.

The charge balance for the solution is conceptually:

2C + [H+] = [OH-] + [H2PO4-] + 2[HPO4 2-] + 3[PO4 3-]

where C is the formal concentration of Na2HPO4. Solving this numerically gives a pH extremely close to the amphiprotic estimate. That agreement is the reason chemists are comfortable using the simpler formula for many practical cases.

Phosphoric acid equilibrium data	Typical 25 degrees Celsius value	Why it matters for Na2HPO4
pKa1	2.15	Mostly irrelevant at pH near 9.8 because H3PO4 is negligible.
pKa2	7.21	Controls the H2PO4- / HPO4 2- balance.
pKa3	12.32	Controls the HPO4 2- / PO4 3- balance.
Predicted pH of Na2HPO4	About 9.76	From the amphiprotic average 1/2(pKa2 + pKa3).

Does the 0.150 M concentration matter?

Yes, but less than many students expect. For amphiprotic salts, the first estimate of pH does not depend explicitly on concentration. That surprises people because weak acid and weak base problems often do depend on concentration. The reason is that amphiprotic species sit between two linked equilibria, and their pH is governed mainly by the two neighboring pKa values. At moderate concentrations such as 0.150 M, the approximation remains very good.

Still, in a real laboratory environment, concentration can influence the measured value through activity effects. A pH meter measures hydrogen ion activity, not idealized concentration. As ionic strength rises, activity coefficients drift away from 1, and the measured pH can differ slightly from the ideal textbook result. This is especially relevant in buffer preparation, environmental chemistry, and high precision analytical work.

Formal Na2HPO4 concentration	Ideal amphiprotic estimate	Expected practical observation
0.010 M	9.76	Usually close to the ideal estimate in dilute laboratory solutions.
0.150 M	9.76	Common teaching example; measured pH may shift slightly with ionic strength.
0.500 M	9.76	Greater non-ideal effects can appear, especially without activity correction.

Species distribution at the calculated pH

One of the best ways to check your reasoning is to examine which phosphate species dominate around pH 9.76. Since pH is above pKa2, HPO4 2- should exceed H2PO4-. Since pH is still well below pKa3, PO4 3- should remain a minor component. That is exactly what the equilibrium calculation shows. Most of the phosphate remains as HPO4 2-, a smaller fraction appears as H2PO4-, and only a trace becomes PO4 3-.

This distribution matters in buffer chemistry. The phosphate buffer system is widely used in biology and chemistry because it offers useful buffering in the neighborhood of pKa2, around neutral pH. A pure Na2HPO4 solution alone is not an optimized phosphate buffer at pH 7.2 because it contains too much of the basic form. To build a physiological phosphate buffer, chemists typically combine NaH2PO4 and Na2HPO4 in the proper ratio.

Common mistakes when solving this problem

• Treating HPO4 2- as only a weak base. If you use only Kb = Kw/Ka2 and ignore the acid side, you often predict a pH that is too high.
• Using the wrong pKa pair. For HPO4 2-, you must average pKa2 and pKa3, not pKa1 and pKa2.
• Forgetting sodium dissociation. Na2HPO4 gives two sodium ions per formula unit, which matters in the exact charge balance setup.
• Confusing molarity with molality. The problem states 0.150 M, meaning moles per liter of solution.
• Ignoring temperature. pKa values change with temperature, so the final pH can shift slightly if conditions move away from 25 degrees Celsius.

Step by step solution for students

1. Recognize that Na2HPO4 dissociates into 2 Na+ and HPO4 2-.
2. Identify HPO4 2- as an amphiprotic ion.
3. Look up the neighboring phosphoric acid constants: pKa2 and pKa3.
4. Apply the amphiprotic formula: pH = 1/2(pKa2 + pKa3).
5. Insert the values 7.21 and 12.32.
6. Calculate pH = 9.765.
7. Round according to your course convention, often to 9.77.

If your course requires a more rigorous treatment, use the exact equilibrium method implemented in the calculator above. It numerically solves the phosphate system and returns species fractions, pOH, and hydroxide concentration in addition to pH.

How this connects to real laboratory work

Na2HPO4 is used in biochemical buffers, cleaning formulations, water treatment chemistry, food processing, and pharmaceutical preparations. In many of those settings, knowing the pH of the stock salt solution matters because it affects enzyme stability, solubility, ionic interactions, and downstream titrations. In biochemistry labs, sodium phosphate salts are often mixed to make buffers with tightly controlled pH. In environmental systems, phosphate speciation influences nutrient chemistry and precipitation equilibria with calcium and magnesium.

Because phosphate chemistry is so central, it is worth consulting authoritative references when you need exact constants or analytical methods. Good starting points include the U.S. Environmental Protection Agency, the NIST Chemistry WebBook, and university resources such as LibreTexts Chemistry. These sources help verify equilibrium data, acid-base theory, and phosphate applications.

Final answer

For a 0.150 M Na2HPO4 solution at 25 degrees Celsius, the best standard theoretical result is:

pH approximately 9.77

That result comes from the amphiprotic formula using phosphate constants pKa2 and pKa3. A full charge balance and speciation calculation gives a closely matching value and confirms that the dominant phosphate species in solution is HPO4 2-.