Calculate the pH of a 0.96 M Solution of K3PO4
Use this interactive calculator to estimate the pH of potassium phosphate, K3PO4, by treating PO43- as a weak base in water. The default inputs are set for a 0.96 M solution at 25 degrees Celsius using the third dissociation constant of phosphoric acid.
Interactive Calculator
Enter or confirm the values above, then click Calculate pH. Default values are configured for a 0.96 M solution of K3PO4.
Expert Guide: How to Calculate the pH of a 0.96 M Solution of K3PO4
If you need to calculate the pH of a 0.96 M solution of K3PO4, the key is to recognize what kind of compound potassium phosphate actually is when it dissolves in water. K3PO4, or tripotassium phosphate, dissociates essentially completely into three K+ ions and one PO43- ion. The potassium ions do not affect acid-base chemistry in any meaningful way because they are spectator ions derived from the strong base KOH. The phosphate ion, however, is the conjugate base of hydrogen phosphate, HPO42-, and it reacts with water to produce hydroxide ions. That hydrolysis step is what makes the solution basic.
In practical terms, a 0.96 M solution of K3PO4 is not just mildly alkaline. It is strongly basic. To quantify that basicity, you do not use the first or second dissociation constants of phosphoric acid. Instead, you use the third one, because PO43- is the conjugate base associated with the equilibrium HPO42- ⇌ H+ + PO43-. The third acid dissociation constant is commonly represented as Ka3, and the corresponding base constant for PO43- is Kb = Kw / Ka3.
Step 1: Write the Dissociation and Hydrolysis Reactions
Start by writing what happens when the salt dissolves:
K3PO4(aq) → 3K+(aq) + PO43-(aq)
Now write the base hydrolysis of phosphate:
PO43-(aq) + H2O(l) ⇌ HPO42-(aq) + OH–(aq)
This second equation is the one that determines the pH. Once K3PO4 dissolves, the initial concentration of PO43- is approximately the formal concentration of the salt, so for this problem we take [PO43-]0 = 0.96 M.
Step 2: Convert pKa3 to Ka3 and Then to Kb
A typical literature value for phosphoric acid has pKa3 near 12.32 at 25 degrees Celsius. Using that value:
- Ka3 = 10-12.32 ≈ 4.79 × 10-13
- Kw = 1.00 × 10-14
- Kb = Kw / Ka3 ≈ (1.00 × 10-14) / (4.79 × 10-13) ≈ 2.09 × 10-2
That Kb is fairly large for a weak base equilibrium, which already tells you that the hydrolysis of PO43- is substantial. In other words, you should not assume only a tiny fraction reacts.
Step 3: Set Up the ICE Table
Use an ICE table for the hydrolysis equilibrium:
- Initial: [PO43-] = 0.96, [HPO42-] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.96 – x, x, x
The base equilibrium expression is:
Kb = [HPO42-][OH–] / [PO43-] = x2 / (0.96 – x)
Substituting Kb ≈ 0.0209 gives:
0.0209 = x2 / (0.96 – x)
Step 4: Solve for Hydroxide Concentration
You can solve this exactly with the quadratic formula or estimate x using the square-root approximation. For a highly basic phosphate solution, the approximation is not terrible, but the exact quadratic is better and is what this calculator uses by default.
Quadratic setup:
x2 + 0.0209x – 0.0201 ≈ 0
Solving gives a positive root near:
x ≈ 0.131 M
So:
- [OH–] ≈ 0.131 M
- [HPO42-] ≈ 0.131 M
- [PO43-] ≈ 0.829 M
Step 5: Convert Hydroxide Concentration to pH
Once [OH–] is known:
- pOH = -log[OH–] = -log(0.131) ≈ 0.882
- pH = 14.00 – 0.882 ≈ 13.118
Rounded reasonably, the pH of a 0.96 M solution of K3PO4 is about 13.12.
Why the Third Dissociation Constant Matters
This is one of the most common places students lose points. Phosphoric acid is triprotic, so it has three dissociation constants:
- Ka1 for H3PO4 ⇌ H+ + H2PO4–
- Ka2 for H2PO4– ⇌ H+ + HPO42-
- Ka3 for HPO42- ⇌ H+ + PO43-
Since the dissolved salt provides PO43-, the conjugate acid pair involved is HPO42-/PO43-. That means Ka3 is the relevant acid constant, and Kb for PO43- is derived from Ka3, not Ka1 or Ka2. If you accidentally use the wrong equilibrium constant, your pH estimate can be dramatically wrong.
Real Chemistry Context: Why This Solution Is So Basic
A calculated pH above 13 may seem surprisingly high for a salt solution, but it makes chemical sense. K3PO4 is the salt of a strong base and a weak acid. Salts of strong bases and weak acids hydrolyze to form hydroxide ions in water. Because phosphate carries a 3- charge, it is especially basic relative to many other common anions. At high concentration, that effect becomes very noticeable.
In laboratory and industrial settings, tripotassium phosphate is used in cleaning formulations, buffering systems, food processing, and specialty chemical applications. However, concentrated phosphate solutions can deviate from ideal behavior, especially near molar concentrations. Activities, ionic strength effects, temperature shifts, and additional proton transfer equilibria can slightly alter the exact pH measured in the lab. The value around 13.12 is therefore the standard textbook result under idealized conditions.
Comparison Table: Dissociation Constants of Phosphoric Acid
| Equilibrium | Approximate pKa at 25 degrees C | Approximate Ka | Why It Matters |
|---|---|---|---|
| H3PO4 ⇌ H+ + H2PO4– | 2.15 | 7.08 × 10-3 | Used when the acidic form is H3PO4 |
| H2PO4– ⇌ H+ + HPO42- | 7.20 | 6.31 × 10-8 | Important in biological buffering near neutral pH |
| HPO42- ⇌ H+ + PO43- | 12.32 | 4.79 × 10-13 | Required for K3PO4 pH calculations |
Comparison Table: Estimated pH of K3PO4 at Different Concentrations
| K3PO4 Concentration | Estimated [OH–] | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.10 M | 0.036 M | 1.45 | 12.55 |
| 0.25 M | 0.064 M | 1.19 | 12.81 |
| 0.50 M | 0.097 M | 1.01 | 12.99 |
| 0.96 M | 0.131 M | 0.88 | 13.12 |
| 1.50 M | 0.163 M | 0.79 | 13.21 |
Approximation Versus Exact Quadratic Method
The square-root approximation assumes x is small relative to the initial concentration, leading to x ≈ √(KbC). For this example:
x ≈ √(0.0209 × 0.96) ≈ 0.142 M
That gives pOH ≈ 0.85 and pH ≈ 13.15. This is close, but a bit higher than the more accurate quadratic result of about 13.12. Because x is not extremely small compared with 0.96, the approximation has noticeable error. For homework or exam work where precision matters, the quadratic method is preferred.
Important Assumptions Behind the Answer
- The solution behaves ideally and concentration is treated as activity.
- The value of Kw is taken as 1.00 × 10-14 at 25 degrees Celsius.
- The hydrolysis of PO43- is treated as the dominant source of OH–.
- Secondary equilibria and ionic strength corrections are not explicitly included.
- The notation 0.96 m is treated here as approximately 0.96 M for a standard instructional pH calculation.
How to Avoid Common Mistakes
- Do not treat K3PO4 as a neutral salt. It is strongly basic in water.
- Do not use Ka1 or Ka2. The correct acid constant is Ka3.
- Do not forget to convert pKa to Ka before finding Kb.
- Do not confuse pOH and pH at the end of the calculation.
- Check whether approximation assumptions are valid before using them.
Authoritative Chemistry References
For reliable background on acid-base equilibria, phosphate chemistry, and water ionization, consult these authoritative resources:
- LibreTexts Chemistry educational resources
- U.S. Environmental Protection Agency water chemistry resources
- NIST Chemistry WebBook
Bottom Line
To calculate the pH of a 0.96 M solution of K3PO4, treat phosphate as a weak base, derive Kb from the third dissociation constant of phosphoric acid, solve the hydrolysis equilibrium, and then convert hydroxide concentration to pH. Using standard values at 25 degrees Celsius, the result is about pH = 13.12. That answer is chemically reasonable, mathematically defensible, and consistent with the expectation that tripotassium phosphate forms a strongly alkaline aqueous solution.