Calculate The Ph Of 0.10 M Na3Po4

Use this premium phosphate solution calculator to estimate the pH of trisodium phosphate in water. The tool applies either an exact charge-balance model or a fast hydrolysis approximation using standard 25 C phosphoric acid constants.

Core chemistry

PO₄^3- + H₂O ⇌ HPO₄^2- + OH^–

K_b1 = K_w / K_a3

Charge balance: [Na⁺] + [H⁺] = [OH^–] + [H₂PO₄^–] + 2[HPO₄^2-] + 3[PO₄^3-]

Default example: 0.10 M Na₃PO₄. For accurate teaching and lab review, use the exact model. The approximation is helpful for quick homework checks.

How to calculate the pH of 0.10 M Na₃PO₄

To calculate the pH of 0.10 M sodium phosphate, written as Na₃PO₄, you need to recognize that this salt is not neutral in water. It comes from a strong base, NaOH, and a weak acid, phosphoric acid, H₃PO₄. That means the sodium ions do not affect pH very much, but the phosphate ion, PO₄^3-, does. It behaves as a base by pulling a proton from water and generating hydroxide ions. Because hydroxide raises alkalinity, the solution becomes strongly basic.

The most important first step is understanding the role of phosphate as the conjugate base of HPO₄^2-. In water, the dominant hydrolysis reaction is:

PO₄^3- + H₂O ⇌ HPO₄^2- + OH^–

This reaction is governed by a base dissociation constant, K_b, which is linked to the third acid dissociation constant of phosphoric acid:

K_b1 = K_w / K_a3

At 25 C, phosphoric acid is a classic triprotic acid. Its three acid constants are widely reported and extremely useful for pH calculations involving phosphate buffers and phosphate salts. Since Na₃PO₄ contains the fully deprotonated phosphate ion, the third dissociation constant is the one that matters most in a quick hand calculation.

Phosphoric acid step	Equilibrium	Typical 25 C value	pK value
First dissociation	H3PO4 ⇌ H^+ + H2PO4^–	Ka1 = 7.11 × 10^-3	pKa1 ≈ 2.15
Second dissociation	H2PO4^– ⇌ H^+ + HPO4^2-	Ka2 = 6.32 × 10^-8	pKa2 ≈ 7.20
Third dissociation	HPO4^2- ⇌ H^+ + PO4^3-	Ka3 = 4.49 × 10^-13	pKa3 ≈ 12.35

Quick approximation method for 0.10 M Na₃PO₄

If you are solving a typical general chemistry or analytical chemistry problem, you can usually begin with the first hydrolysis step only. For phosphate, that gives:

1. Compute the base constant using the third acid constant.
2. Write an ICE table for PO₄^3- hydrolysis.
3. Use the equilibrium expression to solve for [OH^–].
4. Convert to pOH, then to pH.

First, calculate the base constant:

K_b1 = 1.0 × 10^-14 / 4.49 × 10^-13 ≈ 2.23 × 10^-2

Now write the equilibrium setup for a 0.10 M solution:

• Initial [PO₄^3-] = 0.10 M
• Initial [HPO₄^2-] = 0
• Initial [OH^–] = 0, ignoring water autoionization in the setup

Let x be the amount of phosphate that hydrolyzes. Then:

• [PO₄^3-] = 0.10 – x
• [HPO₄^2-] = x
• [OH^–] = x

The expression becomes:

K_b1 = x² / (0.10 – x)

Substitute 2.23 × 10^-2 for K_b1:

2.23 × 10^-2 = x² / (0.10 – x)

Because K_b is not tiny, the small x shortcut is not ideal here. Solving the quadratic gives x near 0.037 M. That means:

• [OH^–] ≈ 0.037 M
• pOH ≈ 1.43
• pH ≈ 12.57

This is the standard textbook-level answer and is already very good. It explains why sodium phosphate is strongly basic in aqueous solution.

Why the exact answer can differ slightly

A more rigorous treatment does not stop at the first hydrolysis step. Once some HPO₄^2- forms, that ion can also react with water:

HPO₄^2- + H₂O ⇌ H₂PO₄^– + OH^–

That second basic reaction is much weaker than the first, but not completely zero. A full charge-balance model includes all phosphate species simultaneously:

• H₃PO₄
• H₂PO₄^–
• HPO₄^2-
• PO₄^3-

It also includes:

• Total phosphate concentration
• Charge balance with the three sodium ions per formula unit
• Water autoionization, K_w

Using that more advanced approach, the pH of 0.10 M Na₃PO₄ at 25 C is about 12.58. In other words, the exact and approximate methods agree very closely for this concentration, which is useful news for students and working chemists alike.

Bottom line: For most classroom and many laboratory purposes, the pH of 0.10 M Na₃PO₄ is approximately 12.6. A more exact calculation gives a value very close to that, usually around 12.58 depending on the constants chosen.

What makes Na₃PO₄ more basic than many other salts

Many students first learn that salts can be acidic, basic, or neutral depending on the parent acid and base. Sodium chloride is neutral because it comes from a strong acid and strong base. Ammonium chloride is acidic because ammonium is the conjugate acid of a weak base. Sodium phosphate is strongly basic because phosphate is the conjugate base of a very weak acid step, namely the third deprotonation of phosphoric acid.

That is why Na₃PO₄ often produces significantly higher pH values than salts such as NaH₂PO₄ or Na₂HPO₄. The phosphate family spans a wide range of acid-base behavior depending on how many hydrogens remain attached to the phosphate group.

Na3PO4 concentration	Approximate pH at 25 C	Interpretation
0.001 M	11.70 to 11.75	Basic, but much less alkaline than 0.10 M
0.010 M	12.20 to 12.25	Strongly basic, typical teaching problem range
0.100 M	About 12.58	Standard reference case for this calculator
1.00 M	12.85 to 12.90	Very strongly basic, non-ideal effects may matter in real systems

Step by step exact method using charge balance

If you want the more advanced derivation, start by defining the total phosphate concentration C_T as 0.10 M. Since each unit of Na₃PO₄ contributes three sodium ions, the total sodium concentration is 0.30 M. Then write the fractional composition of all phosphate species as functions of [H⁺]. For phosphoric acid, the denominator is:

D = [H⁺]³ + K_a1[H⁺]² + K_a1K_a2[H⁺] + K_a1K_a2K_a3

Then each phosphate species is:

• [H₃PO₄] = C_T[H⁺]³ / D
• [H₂PO₄^–] = C_TK_a1[H⁺]² / D
• [HPO₄^2-] = C_TK_a1K_a2[H⁺] / D
• [PO₄^3-] = C_TK_a1K_a2K_a3 / D

Next apply charge balance:

0.30 + [H⁺] = [OH^–] + [H₂PO₄^–] + 2[HPO₄^2-] + 3[PO₄^3-]

With [OH^–] = K_w / [H⁺], you solve numerically for [H⁺]. This is the exact strategy used in the calculator above. It handles the coupled equilibria correctly and is the preferred method when precision matters.

Species distribution near the final pH

At a pH around 12.58, the phosphate is not 100 percent PO₄^3-. A significant fraction has already been converted into HPO₄^2-. That is an important insight because it explains why the solution does not become even more alkaline. As hydroxide forms, the equilibrium shifts and partially protonates the phosphate. This buffers the system somewhat, although the solution remains clearly basic.

In practice, near pH 12.6 you can expect:

• PO₄^3- to remain the major species
• HPO₄^2- to be the next most important species
• H₂PO₄^– and H₃PO₄ to be negligible

That composition is consistent with the pK_a3 value near 12.35. When pH is slightly above pK_a3, the more deprotonated form, PO₄^3-, is favored, but not completely overwhelming.

Common mistakes when solving this problem

1. Treating Na₃PO₄ as a neutral salt. It is not neutral. Phosphate hydrolyzes strongly enough to create a basic solution.
2. Using the wrong acid constant. For PO₄^3- acting as a base, you need K_a3, not K_a1 or K_a2.
3. Assuming x is negligible without checking. Because K_b1 is on the order of 10^-2, the small x shortcut is not very reliable here.
4. Forgetting to convert pOH to pH. Once you know [OH^–], you calculate pOH first, then pH = 14.00 – pOH at 25 C.
5. Ignoring concentration units. A solution entered in mM must be converted to M before equilibrium calculations are performed.

Why this calculation matters in real chemistry

Trisodium phosphate has practical significance in cleaning formulations, water treatment discussions, materials processing, and educational chemistry labs. Its strongly basic behavior affects corrosion, surface treatment, precipitation reactions, and phosphate speciation. In biological or environmental contexts, phosphate chemistry also matters because pH controls the distribution among phosphate species, which in turn influences solubility and reactivity.

When concentration becomes high, real solutions can deviate from ideal behavior because activity effects become more important. For routine educational work, concentration-based calculations are usually acceptable. For advanced analytical chemistry, however, especially at ionic strengths approaching 0.1 M or higher, activity corrections can refine the answer further.

Authoritative sources for phosphate and pH chemistry

If you want to verify constants or learn more from high-quality references, these resources are excellent starting points:

• PubChem, Trisodium Phosphate
• PubChem, Phosphoric Acid
• MIT OpenCourseWare, Acid-Base Equilibria

Final answer

For a 0.10 M Na₃PO₄ solution at 25 C, the pH is approximately 12.58. If you use a quick first-hydrolysis equilibrium approach, you will usually get a value very close to 12.57 to 12.58. That is the correct order of magnitude and the accepted chemistry result for this system.