Calculate The Ph Of A 0.1 M K3Po4

This premium calculator estimates the pH of potassium phosphate, K3PO4, by treating PO4^3- as a basic ion that hydrolyzes in water. It uses standard 25 C equilibrium constants by default and shows the pH, pOH, hydroxide concentration, and percent hydrolysis.

K3PO4 pH Calculator

How to calculate the pH of a 0.1 M K3PO4 solution

To calculate the pH of a 0.1 M K3PO4 solution, you need to recognize what happens when potassium phosphate dissolves in water. K3PO4 is a soluble ionic compound that separates essentially completely into 3 K⁺ ions and 1 PO4^3- ion. The potassium ions come from the strong base KOH and are spectators in acid base chemistry. The phosphate ion is the important species because it is the conjugate base of hydrogen phosphate, HPO4^2-. That makes phosphate a base in water.

The key hydrolysis reaction is:

PO4^3- + H2O ⇌ HPO4^2- + OH^–

Because hydroxide is produced, the final solution is basic. For a first pass, this single equilibrium is enough to estimate the pH very well for a 0.1 M solution. The second hydrolysis of HPO4^2- is much weaker and contributes only a small additional amount of OH^–, so many textbook and exam solutions safely ignore it in the main calculation.

Step 1: Convert Ka3 to Kb for PO4^3-

Since PO4^3- is the conjugate base of HPO4^2-, its base dissociation constant is found from:

Kb = Kw / Ka3

Using standard 25 C values:

• Kw = 1.0 × 10^-14
• Ka3 ≈ 4.5 × 10^-13

So:

Kb = (1.0 × 10^-14) / (4.5 × 10^-13) ≈ 2.22 × 10^-2

Step 2: Set up the ICE table

Let the initial concentration of PO4^3- be 0.1 M. If x mol/L reacts, then:

• [PO4^3-] starts at 0.1 and becomes 0.1 – x
• [HPO4^2-] starts at 0 and becomes x
• [OH^–] starts near 0 and becomes x

Substitute into the equilibrium expression:

Kb = x² / (0.1 – x)

With Kb = 2.22 × 10^-2:

2.22 × 10^-2 = x² / (0.1 – x)

Step 3: Solve for hydroxide concentration

For this particular problem, the base is not weak enough for the tiny x approximation to be ideal, so solving the quadratic is better. Rearranging gives:

x² + Kb x – KbC = 0

where C is the initial concentration, 0.1 M. Solving gives x ≈ 0.037 M. This means:

• [OH^–] ≈ 0.037 M
• pOH = -log(0.037) ≈ 1.43
• pH = 14.00 – 1.43 ≈ 12.57

Final answer: the pH of a 0.1 M K3PO4 solution is approximately 12.57 at 25 C when the dominant first hydrolysis equilibrium is used.

Why K3PO4 is strongly basic

Students often ask why K3PO4, which is a salt, gives such a high pH. The answer lies in the parent acid and base. K3PO4 forms from phosphoric acid, H3PO4, and potassium hydroxide, KOH. KOH is a strong base. H3PO4 is a weak acid with three dissociation steps. Its third dissociation is extremely weak, which means the conjugate base PO4^3- has a comparatively large tendency to pull a proton from water. That creates OH^– and raises the pH.

In acid base salt analysis, a useful shortcut is this:

1. If the cation comes from a strong base, it is usually neutral.
2. If the anion comes from a weak acid, it behaves as a base.
3. The weaker the acid was, the stronger its conjugate base tends to be.

For phosphate chemistry, the third acid dissociation constant is very small. That is why the fully deprotonated phosphate ion is a significantly basic species in water.

Important equilibrium data for phosphoric acid

The following table summarizes common 25 C values used in classroom and laboratory calculations. These constants explain the acid base behavior of phosphate salts and buffers.

Equilibrium step	Reaction	Typical value at 25 C	Interpretation
Ka1	H3PO4 ⇌ H^+ + H2PO4^–	7.1 × 10^-3 (pKa1 ≈ 2.15)	First proton is moderately acidic
Ka2	H2PO4^– ⇌ H^+ + HPO4^2-	6.3 × 10^-8 (pKa2 ≈ 7.20)	Second proton is weakly acidic
Ka3	HPO4^2- ⇌ H^+ + PO4^3-	4.5 × 10^-13 (pKa3 ≈ 12.35)	Third proton is very weakly acidic, so PO4^3- is basic
Kb of PO4^3-	PO4^3- + H2O ⇌ HPO4^2- + OH^–	2.2 × 10^-2	Main constant used to estimate pH of K3PO4

Comparison of phosphate salts at 0.1 M

Not all phosphate salts have the same pH. Their behavior depends on which phosphate species is present. This comparison helps explain why K3PO4 is much more basic than KH2PO4 or K2HPO4.

Salt	Main acid base character	Dominant species in water	Approximate pH at 0.1 M
KH2PO4	Acidic	H2PO4^–	About 4.7
K2HPO4	Amphiprotic, mildly basic	HPO4^2-	About 9.7
K3PO4	Basic	PO4^3-	About 12.6

Can you use the square root approximation?

Many introductory problems teach the weak base shortcut:

[OH^–] ≈ √(KbC)

For K3PO4 at 0.1 M, this gives:

[OH^–] ≈ √((2.22 × 10^-2)(0.1)) ≈ 0.047 M

That estimate is noticeably larger than the quadratic result because x is not tiny compared with the initial concentration. In fact, 0.047 M is nearly half of the starting 0.1 M, so the small x assumption breaks down. This is why a careful calculation gives a more reliable pH near 12.57 instead of a rougher value near 12.67.

Rule of thumb

• If x is less than 5 percent of the initial concentration, the approximation is usually fine.
• If x is larger, solve the quadratic.
• For 0.1 M K3PO4, use the quadratic if you want a sound answer.

How concentration affects the pH

If you dilute K3PO4, the pH decreases, but it still remains basic over a broad range. The chart in the calculator displays this relationship. At lower concentrations, there is less phosphate available to generate hydroxide, so pOH rises and pH falls. At higher concentrations, the pH increases, but not linearly, because equilibrium chemistry is logarithmic and constrained by hydrolysis.

Here are useful benchmark values using the same first hydrolysis model and standard constants:

• 0.001 M K3PO4 gives a pH around 11.54
• 0.01 M K3PO4 gives a pH around 12.08
• 0.1 M K3PO4 gives a pH around 12.57
• 1.0 M K3PO4 gives a pH around 13.00

These values are strong evidence that phosphate salts can span a surprisingly wide pH range depending on which protonation state is present and how concentrated the solution is.

Common mistakes when solving this problem

1. Treating K3PO4 as a neutral salt. It is not neutral because PO4^3- is the conjugate base of a weak acid species.
2. Using Ka2 instead of Ka3. The hydrolysis of PO4^3- reverses the third dissociation step of phosphoric acid, so Ka3 is the one that matters first.
3. Ignoring the need for Kb. You must convert Ka3 to Kb with Kb = Kw/Ka3.
4. Using the square root shortcut blindly. For 0.1 M K3PO4, the approximation is not ideal.
5. Forgetting that pH + pOH = 14 at 25 C. Once you get [OH^–], the fastest route is often pOH first, then pH.

Exam style summary for a fast solution

If you need the shortest valid pathway on a quiz or homework set, use this sequence:

1. Write hydrolysis: PO4^3- + H2O ⇌ HPO4^2- + OH^–
2. Compute Kb = Kw/Ka3 = 1.0 × 10^-14 / 4.5 × 10^-13 = 2.22 × 10^-2
3. Set up Kb = x² / (0.1 – x)
4. Solve the quadratic to get x ≈ 0.037 M
5. Find pOH = -log(0.037) ≈ 1.43
6. Find pH = 14.00 – 1.43 = 12.57

Advanced note on exact treatment

An even more rigorous calculation would account for all phosphate equilibria simultaneously, include charge balance, and include the second hydrolysis of HPO4^2-. For many educational settings, that level of detail is unnecessary because the first hydrolysis dominates the result and the final pH remains very close to the value obtained here. The calculator on this page is designed for a practical, instruction friendly answer that matches standard chemistry teaching approaches.

Authoritative references for phosphate and pH chemistry

• PubChem, U.S. National Library of Medicine: Phosphoric Acid
• U.S. Environmental Protection Agency: pH Overview
• University of Wisconsin: Polyprotic Acid Chemistry

Bottom line

If you are asked to calculate the pH of a 0.1 M K3PO4 solution, the correct chemistry is to treat phosphate as a base. Convert Ka3 to Kb, solve the hydrolysis equilibrium, and then convert hydroxide concentration into pOH and pH. With standard constants at 25 C, the expected answer is approximately pH = 12.57. That is why K3PO4 is classified as a strongly basic salt in water, even though it is not itself a strong base like NaOH or KOH.