Calculate the pH of a 0.150 M Solution of Na2HPO4
Use this premium calculator to estimate the pH of sodium hydrogen phosphate solution with exact equilibrium, amphiprotic approximation, or method comparison. The default chemistry data correspond to phosphoric acid at 25 degrees C.
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Expert Guide: How to Calculate the pH of a 0.150 M Solution of Na2HPO4
When you are asked to calculate the pH of a 0.150 M solution of Na2HPO4, you are working with the chemistry of an amphiprotic ion. Sodium hydrogen phosphate dissociates in water to give sodium ions, Na+, and hydrogen phosphate ions, HPO42-. The sodium ions are spectators, but HPO42- is chemically interesting because it can both accept a proton and donate a proton. That dual behavior is the reason this calculation is richer than a simple strong acid or strong base problem.
In classroom chemistry, the quickest answer is often based on the amphiprotic formula:
pH approximately equals 1/2(pKa2 + pKa3)
For phosphoric acid, common values are pKa2 = 7.21 and pKa3 = 12.32, giving:
pH approximately equals 1/2(7.21 + 12.32) = 9.765
So the standard quick answer is usually pH approximately 9.77. However, if you solve the full equilibrium problem for a 0.150 M Na2HPO4 solution, you obtain a value that is slightly different, typically around pH 9.68 to 9.77 depending on assumptions, rounding, ionic strength treatment, and the set of constants used. This calculator uses a full charge balance approach for the exact mode and also shows the amphiprotic approximation for comparison.
What Na2HPO4 Does in Water
Sodium hydrogen phosphate is the salt of a polyprotic acid. Once dissolved, it separates as:
- Na2HPO4 → 2 Na+ + HPO42-
- Na+ does not significantly affect pH
- HPO42- participates in acid-base equilibrium
The hydrogen phosphate ion can react in two directions:
- As a base: HPO42- + H2O ⇌ H2PO4– + OH–
- As an acid: HPO42- + H2O ⇌ PO43- + H3O+
Because HPO42- sits in the middle of the phosphoric acid sequence, it is not purely basic and not purely acidic. It is amphiprotic. That is why the pH lands in the mildly basic region, rather than becoming extremely basic.
Phosphoric Acid Equilibrium Data
To solve this problem well, you need the dissociation constants for phosphoric acid. At 25 degrees C, common textbook values are shown below.
| Step | Equilibrium | pKa | Ka |
|---|---|---|---|
| First dissociation | H3PO4 ⇌ H+ + H2PO4– | 2.15 | 7.08 × 10-3 |
| Second dissociation | H2PO4– ⇌ H+ + HPO42- | 7.21 | 6.17 × 10-8 |
| Third dissociation | HPO42- ⇌ H+ + PO43- | 12.32 | 4.79 × 10-13 |
These values are why the amphiprotic estimate works. The HPO42- ion is bracketed between pKa2 and pKa3, and its pH tends to fall near their average.
Fast Method: Amphiprotic Approximation
If your course allows the standard approximation for amphiprotic species, calculate the pH like this:
- Identify the amphiprotic ion: HPO42-
- Use the two adjacent pKa values, pKa2 and pKa3
- Apply pH = 1/2(pKa2 + pKa3)
- Insert values: pH = 1/2(7.21 + 12.32)
- Result: pH = 9.765, or about 9.77
This method is very popular because it is quick, elegant, and usually accurate enough for hand calculations. It also shows an important conceptual idea: the pH of an amphiprotic salt often depends more strongly on the relevant pKa values than on the concentration, provided the solution is not extremely dilute.
More Detailed Method: Full Equilibrium and Charge Balance
If you want the most rigorous answer, you treat the total phosphate concentration as 0.150 M and write the charge balance with phosphate distribution fractions. In this setup:
- Total phosphate concentration, CT = 0.150 M
- Sodium concentration, [Na+] = 2 × 0.150 = 0.300 M
- All phosphate species are linked through Ka1, Ka2, and Ka3
The charge balance is:
[H+] + [Na+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-]
When this equation is solved numerically, the result is close to the amphiprotic estimate but often slightly lower. That difference comes from the fact that the amphiprotic formula is an approximation, while the charge balance accounts for all species at the chosen concentration.
| Method | Formula or approach | Typical result for 0.150 M Na2HPO4 | Comment |
|---|---|---|---|
| Amphiprotic approximation | 1/2(pKa2 + pKa3) | 9.77 | Fast and standard for hand work |
| Base only approximation | Kb = Kw/Ka2, then solve for OH– | About 10.19 | Usually overestimates pH because it ignores acid behavior of HPO42- |
| Exact equilibrium solve | Mass balance plus charge balance | About 9.68 to 9.77 | Best for precise modeling |
Why the Base Only Method Is Not the Best Choice
Some students try to treat HPO42- only as a base. That gives:
Kb = Kw / Ka2
Using Ka2 = 6.17 × 10-8, we get:
Kb approximately 1.62 × 10-7
Then, for C = 0.150 M:
[OH-] approximately sqrt(Kb × C) = sqrt(1.62 × 10-7 × 0.150)
[OH-] approximately 1.56 × 10-4
This leads to pOH approximately 3.81 and pH approximately 10.19.
That value is significantly higher than the accepted amphiprotic estimate. The reason is simple: the base only method ignores the fact that HPO42- can also donate a proton to form PO43-. Once both directions are considered, the final pH drops.
Species Distribution Near the Final pH
At a pH near 9.7, phosphate is distributed mainly between H2PO4–, HPO42-, and a smaller amount of PO43-. H3PO4 is negligible. In practical terms:
- HPO42- is the dominant species
- H2PO4– remains significant
- PO43- is present but still a minority species
- The mixture behaves as a basic phosphate buffer region
This is one reason phosphate solutions are so important in chemistry and biochemistry. The phosphate system covers multiple buffer regions and appears everywhere from laboratory reagents to biological fluids.
Step by Step Summary for Exams and Homework
- Recognize that Na2HPO4 produces HPO42- in water.
- Recognize that HPO42- is amphiprotic.
- Use the neighboring dissociation constants, pKa2 and pKa3.
- Apply the amphiprotic formula: pH = 1/2(pKa2 + pKa3).
- Insert pKa2 = 7.21 and pKa3 = 12.32.
- Obtain pH = 9.765.
- Report the final answer as pH approximately 9.77.
If your instructor expects more rigor, mention that a full equilibrium calculation produces a close value, often a bit lower. Including that note shows strong chemical understanding.
Common Mistakes to Avoid
- Using pKa1 and pKa2 instead of pKa2 and pKa3. For HPO42-, the relevant adjacent values are the ones on either side of that species.
- Treating HPO42- as only a base. This leads to an overestimated pH.
- Forgetting the species identity. Na2HPO4 is not the same as NaH2PO4. They give different pH values because they contain different phosphate ions.
- Ignoring conditions. pKa values shift slightly with temperature and ionic strength, which can matter in high precision work.
Why Concentration Often Has a Smaller Effect Here
Students are often surprised that changing the concentration of an amphiprotic salt does not drastically alter the simple pH estimate. That happens because the amphiprotic formula comes from the balance of acid and base tendencies within the same species. For moderate concentrations, the pH is governed mainly by the neighboring pKa values. At very low concentrations, water autoionization becomes more important. At higher ionic strengths, activity corrections may shift the answer somewhat. But for many standard textbook problems, the pKa average gives a very reliable estimate.
Authoritative Chemistry References
If you want to verify constants or review acid-base equilibrium theory, these resources are useful:
- PubChem, phosphoric acid data, U.S. National Library of Medicine
- NIST Chemistry WebBook, phosphoric acid reference data
- MIT OpenCourseWare, chemical equilibrium and acid-base concepts
Final Answer
For the classic textbook question, calculate the pH of a 0.150 M solution of Na2HPO4, the accepted amphiprotic answer is:
pH approximately 9.77
If you apply a full equilibrium model with charge balance, you will obtain a value in the same neighborhood, commonly a little lower depending on the constants and assumptions used. For most general chemistry and analytical chemistry contexts, reporting 9.77 is correct and expected.