Calculate The Ph Of A 0.160 M Phosphoric Acid Solution

Use this interactive calculator to solve the pH of phosphoric acid from concentration and dissociation constants, then review a detailed expert guide explaining each chemistry step.

Expert Guide: How to Calculate the pH of a 0.160 M Phosphoric Acid Solution

Calculating the pH of a 0.160 M phosphoric acid solution is a classic acid-base equilibrium problem because phosphoric acid, H₃PO₄, is not a strong acid and it does not release all three acidic protons equally. Instead, it is a triprotic weak acid with three successive dissociation steps, each one weaker than the last. That matters because the pH is controlled mostly by the first dissociation under ordinary conditions, especially at a moderately high concentration like 0.160 M.

If you only remember one principle, remember this: for a 0.160 M phosphoric acid solution, the first ionization dominates the hydrogen ion concentration, while the second and third ionizations contribute very little to the final pH. That means we can usually calculate the pH accurately by solving the first equilibrium with the acid dissociation constant K_a1.

Phosphoric Acid Dissociation Steps

Phosphoric acid ionizes in three stages:

H3PO4 ⇌ H+ + H2PO4-
H2PO4- ⇌ H+ + HPO4^2-
HPO4^2- ⇌ H+ + PO4^3-

At 25°C, commonly used dissociation constants are:

Step	Equilibrium	Ka Value	pKa	Relative Strength
First	H3PO4 ⇌ H+ + H2PO4-	7.11 × 10^-3	2.15	Most important for pH at 0.160 M
Second	H2PO4- ⇌ H+ + HPO4^2-	6.32 × 10^-8	7.20	Very small contribution
Third	HPO4^2- ⇌ H+ + PO4^3-	4.50 × 10^-13	12.35	Negligible in acidic solution

The first K_a is much larger than the second and third, so the first proton is the one that matters most when calculating the pH of this acidic solution.

Set Up the Equilibrium Expression

Start with the first dissociation only:

H3PO4 ⇌ H+ + H2PO4-

Let the initial concentration of phosphoric acid be 0.160 M, and let x represent the amount dissociated.

• Initial: [H₃PO₄] = 0.160, [H⁺] = 0, [H₂PO₄^–] = 0
• Change: -x, +x, +x
• Equilibrium: [H₃PO₄] = 0.160 – x, [H⁺] = x, [H₂PO₄^–] = x

The equilibrium expression becomes:

Ka1 = [H+][H2PO4-] / [H3PO4] = x² / (0.160 – x)

Substitute K_a1 = 7.11 × 10^-3:

7.11 × 10^-3 = x² / (0.160 – x)

Solve Using the Quadratic Equation

Cross-multiplying gives:

x² = 0.00711(0.160 – x)

x² = 0.0011376 – 0.00711x

x² + 0.00711x – 0.0011376 = 0

Now use the quadratic formula:

x = [-b ± √(b² – 4ac)] / 2a

Here:

• a = 1
• b = 0.00711
• c = -0.0011376

The physically meaningful root is:

x = (-0.00711 + √(0.00711² + 4 × 0.0011376)) / 2

This gives:

x ≈ 0.0304 M

Since x is the hydrogen ion concentration from the first dissociation, we have:

[H+] ≈ 0.0304 M

Then calculate pH:

pH = -log10[H+]

pH = -log10(0.0304) ≈ 1.52

Final answer: the pH of a 0.160 M phosphoric acid solution is approximately 1.52 when calculated with the standard first-dissociation equilibrium method.

Why the Second and Third Dissociations Barely Change the Answer

After the first dissociation establishes a hydrogen ion concentration near 3.0 × 10^-2 M, the solution is already strongly acidic. That high hydrogen ion concentration suppresses additional dissociation of H₂PO₄^– and HPO₄^2- through the common ion effect. Because K_a2 is only 6.32 × 10^-8, its contribution to [H⁺] is tiny compared with the first dissociation. K_a3 is smaller still.

In practical classroom and lab calculations, treating phosphoric acid as a weak acid governed by K_a1 gives an excellent pH estimate at 0.160 M. If you were doing very high-precision speciation work, you would use a full equilibrium model, but your pH would still remain very close to 1.52.

Approximation Method vs Quadratic Method

Students often ask whether they can use the weak-acid approximation, where 0.160 – x is replaced by 0.160. If we do that, then:

x ≈ √(Ka × C) = √(0.00711 × 0.160) = √0.0011376 ≈ 0.0337 M

This approximation predicts:

pH ≈ -log10(0.0337) ≈ 1.47

That is somewhat close, but the approximation is not ideal because the dissociation is large enough that x is not negligible compared with 0.160. In fact, percent ionization is around 19%, which is well beyond the usual 5% guideline for using the simple approximation safely. So for this specific problem, the quadratic method is the better choice.

Method	[H+]	Predicted pH	Difference from Quadratic	Use Case
Quadratic first-dissociation	0.0304 M	1.52	Reference value	Recommended for 0.160 M H3PO4
Weak-acid approximation	0.0337 M	1.47	About 0.05 pH units low	Quick estimate only
Strong-acid assumption	0.160 M	0.80	Far too acidic	Incorrect for phosphoric acid

Percent Ionization of 0.160 M Phosphoric Acid

Percent ionization tells you how much of the original acid concentration actually releases H⁺ in the first equilibrium step. Use:

Percent ionization = (x / C) × 100

Substitute the values:

Percent ionization = (0.0304 / 0.160) × 100 ≈ 19.0%

That result confirms why the shortcut approximation is less reliable here. Nearly one-fifth of the acid dissociates in the first step, so x is not tiny relative to the starting concentration.

How 0.160 M Phosphoric Acid Compares with Other Acids

Looking at comparison data helps build intuition. A 0.160 M strong monoprotic acid such as HCl would fully dissociate and give [H⁺] = 0.160 M, corresponding to a pH of about 0.80. A weak acid like acetic acid at the same concentration would have a much higher pH because its K_a is only around 1.8 × 10^-5. Phosphoric acid sits in between: clearly acidic, but nowhere near as strong as HCl.

Acid	Concentration	Key Ka or Behavior	Approximate pH	Interpretation
Hydrochloric acid, HCl	0.160 M	Essentially complete dissociation	0.80	Much stronger than phosphoric acid
Phosphoric acid, H3PO4	0.160 M	Ka1 = 7.11 × 10^-3	1.52	Moderately strong weak acid in first step
Acetic acid, CH3COOH	0.160 M	Ka ≈ 1.8 × 10^-5	2.77	Substantially weaker than phosphoric acid

Common Mistakes When Solving This Problem

1. Treating phosphoric acid as strong. It is not fully dissociated in water, so using pH = -log(0.160) is incorrect.
2. Adding all three protons directly. The second and third protons are far less acidic and should not be counted as if they all dissociate completely.
3. Using the approximation without checking x. At 0.160 M, the first dissociation is too significant for the 5% rule.
4. Forgetting that K_a values depend on temperature. Standard values are usually given for 25°C.
5. Dropping units too early. Keep concentrations in mol/L throughout the calculation.

Step-by-Step Summary You Can Reuse

1. Write the first dissociation equilibrium for H₃PO₄.
2. Set up an ICE table using initial concentration 0.160 M.
3. Write the K_a1 expression: x² / (0.160 – x) = 7.11 × 10^-3.
4. Solve the quadratic equation for x.
5. Take pH = -log₁₀(x).
6. Report pH ≈ 1.52 and note that later dissociations are negligible.

Why This Matters in Real Chemistry

Phosphoric acid appears in analytical chemistry, beverage chemistry, fertilizer production, metal treatment, and biological buffer systems involving phosphate species. Understanding its pH behavior is important because pH influences reaction rates, corrosion, solubility, and equilibrium speciation. In laboratory settings, the difference between pH 1.52 and pH 0.80 is enormous in terms of chemical behavior, which is why it is important not to mistake phosphoric acid for a strong acid.

Phosphate equilibria are also foundational in environmental chemistry and biochemistry. Even though this specific example is a straightforward acid calculation, it introduces the broader idea that polyprotic acids must be analyzed stepwise, not as if all protons are equivalent. That principle applies to sulfurous acid, carbonic acid, citric acid, and many other multiprotic systems.

Authoritative Chemistry References

• NIST Chemistry WebBook for chemical data and thermodynamic reference material.
• LibreTexts Chemistry for acid-base equilibrium tutorials from academic contributors.
• U.S. Environmental Protection Agency for broader environmental chemistry context involving phosphate systems.

Bottom Line

To calculate the pH of a 0.160 M phosphoric acid solution, use the first dissociation constant and solve the equilibrium expression with the quadratic formula. The resulting hydrogen ion concentration is about 0.0304 M, giving a pH of approximately 1.52. This is the chemically correct result for a standard aqueous solution at room temperature, and it is much more accurate than either a strong-acid assumption or an unchecked weak-acid approximation.