Calculate The Ph Of A 0.142 M Phosphoric Acid Solution

Use this premium calculator to estimate pH from phosphoric acid molality, solution density, and accepted acid dissociation constants. The default setup evaluates a 0.142 m H3PO4 solution at 25 C.

Phosphoric Acid pH Calculator

The calculator assumes 25 C and pure aqueous solution conditions. With the default density of 1.000 g/mL, 0.142 m converts to about 0.140 M.

Expert Guide: How to Calculate the pH of a 0.142 m Phosphoric Acid Solution

Finding the pH of a 0.142 m phosphoric acid solution looks simple at first, but a careful answer requires some chemistry judgment. Phosphoric acid, H₃PO₄, is a triprotic acid. That means each molecule can donate up to three protons, and each proton comes off with a different dissociation constant. In practice, the first ionization dominates the pH for moderately concentrated solutions, while the second and third ionizations make much smaller contributions. The calculator above gives you a fast answer, but this guide shows the full logic behind the number.

The first important detail is the unit. The question uses 0.142 m, where lowercase m means molality, not molarity. Molality is moles of solute per kilogram of solvent. Most equilibrium expressions are written in terms of concentration per liter, or molarity, so you usually need to convert molality to molarity before solving the acid equilibrium. If the solution is dilute and close to the density of water, then 0.142 m and about 0.14 M are very close. That approximation is often good enough for homework and quick engineering estimates.

For the default settings in this calculator, 0.142 m H₃PO₄ with a density of 1.000 g/mL converts to about 0.140 M, and the resulting pH is about 1.55.

Why phosphoric acid needs special handling

Phosphoric acid is not a strong acid in the same sense as hydrochloric acid. It does not dissociate completely in its first step. Instead, it establishes an equilibrium:

H3PO4 ⇌ H+ + H2PO4-

Then the conjugate acid formed in the first step can dissociate again:

H2PO4- ⇌ H+ + HPO4^2-

And finally:

HPO4^2- ⇌ H+ + PO4^3-

At 25 C, accepted values for the dissociation constants are roughly Ka1 = 7.1 × 10^-3, Ka2 = 6.3 × 10^-8, and Ka3 = 4.2 × 10^-13. The huge drop from Ka1 to Ka2 tells you why the first dissociation controls the pH. Once the first proton comes off, removing the second proton is much harder, and removing the third is harder still.

Dissociation step	Reaction	Ka at 25 C	pKa	Importance for pH near 0.14 M
First	H3PO4 ⇌ H+ + H2PO4-	7.1 × 10^-3	2.15	Dominant
Second	H2PO4- ⇌ H+ + HPO4^2-	6.3 × 10^-8	7.20	Minor under strongly acidic conditions
Third	HPO4^2- ⇌ H+ + PO4^3-	4.2 × 10^-13	12.38	Negligible here

Step 1: Convert 0.142 m to molarity

Take 1.000 kg of water as the solvent basis. A 0.142 m solution contains:

n = 0.142 mol H3PO4 per 1.000 kg solvent

The molar mass of phosphoric acid is about 97.994 g/mol, so the solute mass is:

mass of solute = 0.142 × 97.994 = 13.9 g

The total solution mass is therefore about 1000.0 + 13.9 = 1013.9 g. If you assume the solution density is 1.000 g/mL, the total solution volume is about 1013.9 mL or 1.0139 L. Molarity becomes:

M = 0.142 mol / 1.0139 L = 0.140 M

This is why many classroom solutions simply treat 0.142 m as roughly 0.14 M. The difference is small for a dilute aqueous acid.

Step 2: Write the first dissociation equilibrium

For the first ionization, let x be the amount of H⁺ formed from H₃PO₄. Then:

• Initial [H₃PO₄] = 0.140 M
• Change = -x, +x, +x
• Equilibrium [H₃PO₄] = 0.140 – x
• Equilibrium [H⁺] = x
• Equilibrium [H₂PO₄^–] = x

Substitute into the Ka expression:

Ka1 = x^2 / (0.140 – x) = 7.1 × 10^-3

This leads to the quadratic equation:

x^2 + Ka1x – Ka1C = 0

With C = 0.140 M, the physically meaningful root gives x about 0.0282 M, so:

pH = -log10(0.0282) ≈ 1.55

That is the key result. It already includes the fact that phosphoric acid is only partially dissociated in the first step. Because the second and third dissociations are so much weaker, they do not shift the answer much under these acidic conditions.

Can you use the simple weak-acid approximation?

Sometimes students use the shortcut x = √(KaC). For this case:

x ≈ √(0.0071 × 0.140) ≈ 0.0315 M

That gives a pH around 1.50. This is close, but not as accurate as the quadratic result because x is not tiny compared with the starting concentration. The percent dissociation is roughly 20 percent, so the small-x approximation is not ideal. The quadratic is the better classroom method, and a full triprotic solver is the best numerical method.

Why the second and third dissociations contribute very little

Once the first proton dissociates, the solution already contains a fairly high concentration of H⁺, near 0.028 M. That high proton concentration pushes the later equilibria to the left by Le Chatelier’s principle. Since Ka2 is only 6.3 × 10^-8, the second dissociation is strongly suppressed at this pH. Ka3 is smaller still, so the amount of PO₄^3- is effectively negligible.

That is why the species distribution chart generated by the calculator usually shows mostly H₃PO₄ and H₂PO₄^–, with very small amounts of HPO₄^2- and essentially no PO₄^3-.

Practical calculation workflow

1. Identify whether the concentration is given as molality or molarity.
2. Convert molality to molarity if density data are available.
3. Use Ka1 for the main pH calculation.
4. Solve the quadratic rather than using the small-x shortcut when percent dissociation is significant.
5. Check whether Ka2 and Ka3 are likely to matter. For a pH near 1.5, they do not contribute much.

Reference values for pH versus phosphoric acid concentration

The table below gives approximate pH values for phosphoric acid at 25 C using the first-dissociation quadratic method. These numbers are useful for benchmarking your intuition and checking whether a calculated pH is reasonable.

Approximate concentration (M)	[H+] from quadratic (M)	Estimated pH	Interpretation
0.010	0.00558	2.25	Moderately acidic
0.050	0.0156	1.81	Clearly acidic
0.100	0.0233	1.63	Strongly acidic
0.140	0.0282	1.55	Close to the 0.142 m case
0.500	0.0561	1.25	Very acidic solution

Common mistakes when solving this problem

• Confusing molality with molarity. If you skip the unit check, your final pH may be slightly off.
• Treating phosphoric acid as a strong acid. It is not fully dissociated in the first step.
• Ignoring the quadratic when x is not small. For this concentration, the approximation introduces noticeable error.
• Overcomplicating the role of Ka2 and Ka3. They matter more near neutral and basic pH, not strongly acidic conditions like this.
• Using outdated or inconsistent Ka values. Small changes in constants slightly change the last decimal place of pH.

Interpreting the final answer

A pH of about 1.55 means the solution is strongly acidic, but not as acidic as a strong monoprotic acid of the same formal concentration. That distinction matters in lab safety, corrosion control, and chemical process design. Phosphoric acid is widely used in food processing, fertilizer production, metal treatment, and buffer preparation, so understanding its speciation is valuable in both academic and industrial settings.

If you are solving the problem in a general chemistry class, the standard answer is usually obtained with Ka1 and the quadratic. If you are modeling a real process or comparing with measured pH data, then density, ionic strength, and activity corrections can all matter. The calculator on this page gives you both the practical classroom answer and a fuller numerical equilibrium option.

Authoritative references for phosphoric acid data and pH concepts

For deeper study, review these credible sources:

• PubChem, National Institutes of Health: Phosphoric Acid
• NIST Chemistry WebBook: Phosphoric Acid Thermochemical and Chemical Data
• U.S. Environmental Protection Agency: What pH Means

Bottom line

To calculate the pH of a 0.142 m phosphoric acid solution, first convert molality to molarity if possible, then solve the first dissociation equilibrium with Ka1. With a density near 1.000 g/mL, the solution is about 0.140 M, the hydrogen ion concentration is about 0.028 M, and the pH is approximately 1.55. A full triprotic treatment changes the answer very little under these conditions, which is why the first-dissociation quadratic remains the standard method for this problem.