Calculate the pH of a 0.10 M Solution of H3PO4
Use this premium phosphoric acid calculator to estimate the pH of a 0.10 M H3PO4 solution with either a full triprotic equilibrium model or a first-dissociation approximation. The tool also visualizes phosphate species distribution so you can see how much acid remains as H3PO4 versus H2PO4-, HPO4-2, and PO4-3.
Expert Guide: How to Calculate the pH of a 0.10 M Solution of H3PO4
To calculate the pH of a 0.10 M solution of H3PO4, you need to remember one key chemical fact: phosphoric acid is a weak triprotic acid. That means each molecule can donate up to three protons, but it does so in stages rather than all at once. In practice, the first dissociation is by far the most important for a 0.10 M solution, while the second and third dissociations are much weaker and contribute only a small additional amount of hydrogen ion.
If you are solving this problem for homework, exam review, lab prep, or process chemistry, the most useful insight is this: a 0.10 M phosphoric acid solution does not have the same pH as a 0.10 M strong acid. A 0.10 M strong monoprotic acid such as HCl would have a pH of 1.00 because it fully dissociates. H3PO4, however, ionizes only partially in the first step and much less in later steps, so its pH is higher. A realistic result at 25 degrees C is about pH 1.63, depending on the exact constants and model used.
Step 1: Write the acid dissociation reactions
Phosphoric acid dissociates in three sequential steps:
- H3PO4 ⇌ H+ + H2PO4-
- H2PO4- ⇌ H+ + HPO4-2
- HPO4-2 ⇌ H+ + PO4-3
At 25 degrees C, common acid dissociation constants are approximately:
- Ka1 = 7.11 × 10-3
- Ka2 = 6.32 × 10-8
- Ka3 = 4.49 × 10-13
Because Ka1 is much larger than Ka2 and Ka3, the first equilibrium dominates the pH calculation for a moderately concentrated solution such as 0.10 M.
Step 2: Set up the first-dissociation equilibrium table
For the first ionization, let x be the amount of H3PO4 that dissociates:
H3PO4 ⇌ H+ + H2PO4-
- Initial: [H3PO4] = 0.10, [H+] = 0, [H2PO4-] = 0
- Change: -x, +x, +x
- Equilibrium: [H3PO4] = 0.10 – x, [H+] = x, [H2PO4-] = x
The equilibrium expression is:
Ka1 = [H+][H2PO4-] / [H3PO4] = x2 / (0.10 – x)
Substitute Ka1 = 7.11 × 10-3:
7.11 × 10-3 = x2 / (0.10 – x)
Step 3: Solve for x
You can solve this equation with the quadratic formula:
x2 + (7.11 × 10-3)x – (7.11 × 10-4) = 0
The physically meaningful root gives:
x ≈ 0.0233 M
Since x = [H+], the hydrogen ion concentration is approximately 0.0233 M. Therefore:
pH = -log10(0.0233) ≈ 1.63
Why the later dissociations matter so little
Students often ask why a triprotic acid does not simply produce three times as much hydrogen ion. The reason is that each successive proton is harder to remove. The first dissociation constant is on the order of 10-3, but the second is on the order of 10-8, and the third is on the order of 10-13. That is a massive drop in acidity strength from one step to the next.
Once the first dissociation has already generated about 2.3 × 10-2 M hydrogen ion, the equilibrium for the second dissociation is pushed strongly to the left by the common-ion effect. As a result, only a tiny extra amount of H+ is produced from H2PO4-. The third dissociation contributes even less.
Approximate species present in the solution
At the calculated pH near 1.63, phosphoric acid exists mostly as undissociated H3PO4 and the conjugate base H2PO4-. The higher deprotonated forms are almost negligible. That chemical picture is consistent with the acid constants and helps validate the answer.
| Phosphate species | Typical fraction at pH about 1.63 | Chemical meaning |
|---|---|---|
| H3PO4 | About 76.7% | Most molecules remain in the fully protonated acid form. |
| H2PO4- | About 23.3% | Produced mainly by the first dissociation step. |
| HPO4-2 | About 0.00006% | Essentially negligible in strongly acidic solution. |
| PO4-3 | Far below 0.000000001% | Negligible under these conditions. |
Comparison with strong acids and common acidic liquids
Another helpful way to understand the result is to compare this solution with other acidic systems. A 0.10 M strong monoprotic acid would have [H+] = 0.10 M and pH = 1.00. Phosphoric acid at the same formal concentration is less acidic because it is weak in the equilibrium sense, even though it has more than one ionizable proton.
| Solution or liquid | Typical pH | Notes |
|---|---|---|
| 0.10 M HCl | 1.00 | Essentially complete dissociation for a strong acid. |
| 0.10 M H3PO4 | About 1.63 | Weak triprotic acid with first dissociation dominant. |
| Lemon juice | About 2.0 to 2.6 | Acidic due mainly to citric acid. |
| Cola soft drink | About 2.3 to 2.8 | Often acidified with phosphoric acid and carbonic acid. |
| Black coffee | About 4.8 to 5.1 | Mildly acidic compared with mineral acid solutions. |
| Pure water at 25 degrees C | 7.00 | Neutral reference point. |
Can you use the small-x approximation?
Sometimes chemistry students are taught to simplify weak-acid problems by assuming x is small compared with the initial concentration. For H3PO4 at 0.10 M, that approximation is not ideal for the first dissociation because the computed x is around 0.0233 M, which is more than 20% of the starting concentration. That is far above the common 5% rule. So in this particular problem, the quadratic method is the better classroom approach.
If you tried the small-x assumption anyway, you would get:
x ≈ √(Ka1 × C) = √(7.11 × 10-3 × 0.10) ≈ 0.0267 M
That would give a pH near 1.57, which is noticeably off from the more accurate value near 1.63. This is a good reminder that weak-acid approximations are useful only when the dissociation is truly small relative to the starting concentration.
Full equilibrium vs classroom approximation
There are two defensible ways to approach the problem:
- Classroom method: solve only the first dissociation exactly with a quadratic equation and ignore Ka2 and Ka3.
- More rigorous method: use a charge-balance and mass-balance model for the full triprotic acid system.
For 0.10 M H3PO4, both methods produce essentially the same pH to two decimal places. The full method is more chemically complete, but the first-dissociation treatment is typically what instructors expect unless the problem specifically asks for a full equilibrium analysis.
Practical significance of phosphoric acid pH calculations
Phosphoric acid is used in fertilizers, metal treatment, food processing, beverages, buffer preparation, and laboratory reagent systems. Knowing the pH of a given concentration matters because pH affects corrosion, reaction rates, biological compatibility, and buffer performance. In food and beverage contexts, phosphoric acid contributes tartness and microbial control. In industrial cleaning and metal finishing, acidity influences oxide removal and surface preparation. In environmental chemistry, phosphate speciation changes with pH and determines how phosphorus behaves in natural waters and treatment processes.
That is why understanding a simple problem like the pH of a 0.10 M H3PO4 solution is more than an academic exercise. It introduces the bigger concept of speciation: the idea that a compound can exist in several chemically distinct forms whose abundance depends on pH.
Common mistakes to avoid
- Assuming complete dissociation: H3PO4 is weak, so you cannot set [H+] = 0.10 M.
- Multiplying by three: the acid is triprotic, but the later proton losses are much weaker.
- Using the small-x approximation automatically: here it introduces meaningful error.
- Ignoring the correct Ka order: Ka1 is the primary driver, not Ka2 or Ka3.
- Mixing up molarity and molality: most introductory pH calculations use molarity, written as M.
Final answer
If you are asked to calculate the pH of a 0.10 M solution of H3PO4 at 25 degrees C, the accepted result is:
pH ≈ 1.63
This value comes from solving the first dissociation equilibrium exactly. A full triprotic equilibrium model gives nearly the same answer and confirms that the solution is dominated by H3PO4 and H2PO4-.