Calculate the pH of a 29 m Solution of KC7H5O2
Use this interactive potassium benzoate pH calculator to estimate hydroxide concentration, pOH, pH, and percent hydrolysis using standard weak-base salt equilibria at 25 degrees Celsius.
Potassium Benzoate pH Calculator
Click Calculate pH to solve for the pH of a 29 m solution of KC7H5O2.
Expert Guide: How to Calculate the pH of a 29 m Solution of KC7H5O2
To calculate the pH of a 29 m solution of KC7H5O2, you need to recognize what this compound is and how it behaves in water. KC7H5O2 is potassium benzoate, a salt made from potassium ion, K+, and benzoate ion, C7H5O2–. Potassium comes from a strong base, potassium hydroxide, while benzoate is the conjugate base of benzoic acid, which is a weak acid. Because the cation from a strong base is essentially pH neutral in aqueous solution, the acid-base behavior is controlled almost entirely by the benzoate ion. That means the solution is basic, not acidic or neutral.
The central chemistry idea is hydrolysis. When potassium benzoate dissolves, it separates into ions. The benzoate ion then reacts with water to generate a small amount of hydroxide ion:
C7H5O2- + H2O ⇌ HC7H5O2 + OH-
That hydroxide production raises the pH above 7. At a standard classroom level, the calculation uses the acid dissociation constant of benzoic acid, converts it into the base dissociation constant of benzoate, and then solves the weak-base equilibrium. For benzoic acid at 25 degrees Celsius, a common value is Ka = 6.3 × 10-5. With Kw = 1.0 × 10-14, the benzoate ion has:
Kb = Kw / Ka = (1.0 × 10^-14) / (6.3 × 10^-5) ≈ 1.59 × 10^-10
Step 1: Identify the active acid-base species
This is the first checkpoint. Students often try to classify the whole salt as if it were an acid or base directly, but salts must be analyzed by their ions:
- K+ comes from the strong base KOH and is effectively neutral.
- C7H5O2– is the conjugate base of benzoic acid, so it is weakly basic.
- Therefore, the pH depends on benzoate hydrolysis.
This immediately tells you the pH must be greater than 7. Once you know that, the rest of the problem becomes a weak-base equilibrium problem.
Step 2: Convert Ka to Kb
Because benzoate is the conjugate base of benzoic acid, you generally do not look up Kb directly. Instead, you use the conjugate relation:
Ka × Kb = Kw
So:
- Take the acid constant of benzoic acid.
- Divide Kw by Ka.
- Use the resulting Kb in the hydrolysis calculation.
Using Ka = 6.3 × 10-5 and Kw = 1.0 × 10-14 gives Kb ≈ 1.59 × 10-10. That is a small value, which means benzoate is only a weak base. Even so, because the concentration is high, enough hydroxide forms to make the pH distinctly basic.
Step 3: Set up the equilibrium expression for a 29 m solution
The problem statement uses 29 m, which typically means 29 molal, or 29 moles of solute per kilogram of solvent. In rigorous physical chemistry, molality and molarity are not interchangeable, and for a highly concentrated solution like 29 m, real activity effects become important. However, in many educational pH problems, the concentration is approximated as if the listed molality were the effective analytical concentration used in the hydrolysis expression. That is the assumption built into this calculator.
Let the starting concentration or effective concentration of benzoate be C = 29. Then if x is the hydroxide concentration generated by hydrolysis:
- Initial: [C7H5O2–] = 29, [HC7H5O2] = 0, [OH–] = 0
- Change: [C7H5O2–] decreases by x, [HC7H5O2] increases by x, [OH–] increases by x
- Equilibrium: [C7H5O2–] = 29 – x, [HC7H5O2] = x, [OH–] = x
Now apply the base equilibrium expression:
Kb = x² / (29 – x)
Step 4: Solve for hydroxide concentration
There are two standard ways to solve the equation. The first is the approximation commonly taught in general chemistry. If x is much smaller than 29, then 29 – x is approximately 29:
x ≈ √(Kb × C)
Substituting values:
x ≈ √(1.59 × 10^-10 × 29) ≈ √(4.61 × 10^-9) ≈ 6.79 × 10^-5
So the hydroxide concentration is approximately 6.79 × 10-5. You can verify the approximation is valid because x is tiny compared with 29. The percent hydrolysis is extremely small even though the final pH is noticeably basic.
The more exact method solves the quadratic equation:
x = (-Kb + √(Kb² + 4KbC)) / 2
For this problem, the exact and approximate answers are essentially the same to practical reporting precision. That is why the approximation is usually accepted for classroom work.
Step 5: Convert hydroxide concentration to pOH and pH
Once you know [OH–], the rest is straightforward:
pOH = -log10([OH-])
pH = 14 – pOH
Using [OH–] = 6.79 × 10-5:
- pOH ≈ 4.17
- pH ≈ 9.83
That is the standard estimated answer for the pH of a 29 m potassium benzoate solution under idealized 25 degree Celsius assumptions. The key conclusion is that the solution is basic because benzoate pulls a proton from water and forms hydroxide.
| Quantity | Value Used | Why It Matters |
|---|---|---|
| Solute | KC7H5O2 | Potassium benzoate dissociates into K+ and benzoate. |
| Acid constant of benzoic acid, Ka | 6.3 × 10-5 | Needed to determine the basic strength of benzoate. |
| Ion-product of water, Kw | 1.0 × 10-14 | Connects Ka and Kb through KaKb = Kw. |
| Base constant of benzoate, Kb | 1.59 × 10-10 | Controls the hydrolysis equilibrium producing OH–. |
| Effective concentration used in calculation | 29 | Treated as the analytical concentration for the standard educational estimate. |
| Estimated [OH–] | 6.79 × 10-5 | Directly gives pOH and pH. |
| Estimated pH | 9.83 | Final practical answer under idealized assumptions. |
Why the answer is not extremely basic
A common mistake is assuming that a concentrated salt of a weak acid should produce a pH near that of a strong base. That is not how conjugate-base hydrolysis works. Benzoate is only a weak base because benzoic acid is not extremely weak. The Kb value is on the order of 10-10, so only a tiny fraction of the benzoate reacts with water. The concentration is high, which raises [OH–], but the intrinsic equilibrium constant still limits hydrolysis strongly.
Another subtle point is that pH depends logarithmically on hydroxide concentration. A small amount of OH– can move pH noticeably above 7, but that does not mean a large fraction of the dissolved salt hydrolyzed. In this case, the hydrolyzed percentage is only a minute fraction of the starting benzoate.
Comparison table: pH versus concentration for potassium benzoate
The trend below uses the same Ka and Kw assumptions as the calculator. These values help show how pH changes with concentration and why 29 m gives a pH around 9.8 rather than 11 or 12.
| Effective concentration | Estimated [OH–] | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.010 | 1.26 × 10-6 | 5.90 | 8.10 |
| 0.100 | 3.99 × 10-6 | 5.40 | 8.60 |
| 1.00 | 1.26 × 10-5 | 4.90 | 9.10 |
| 10.0 | 3.99 × 10-5 | 4.40 | 9.60 |
| 29.0 | 6.79 × 10-5 | 4.17 | 9.83 |
Important accuracy note for a 29 m solution
From a strict solution chemistry perspective, 29 m is an extremely concentrated solution. At such high ionic strength, ideal behavior breaks down, and a more rigorous treatment would use activity coefficients rather than raw concentrations. In professional physical chemistry, pH in highly concentrated electrolyte systems is more complicated than the general chemistry model suggests. Even so, the standard textbook approach remains valuable because it teaches the right equilibrium logic:
- Classify the salt by its ions.
- Identify the conjugate acid-base pair.
- Convert Ka to Kb if needed.
- Set up the hydrolysis equilibrium.
- Solve for [OH–], then pOH and pH.
If your teacher, textbook, or exam is a typical general chemistry setting, this is almost certainly the intended method. If you are working in analytical chemistry, process chemistry, or advanced thermodynamics, then you would want a non-ideal treatment and possibly measured density and activity data.
Common mistakes when solving this problem
- Treating potassium benzoate as neutral. It is not neutral because benzoate is the conjugate base of a weak acid.
- Using Ka directly in the equilibrium expression. The reacting species is benzoate, so you need Kb.
- Assuming K+ affects pH. Potassium ion is a spectator for acid-base purposes.
- Forgetting to convert from pOH to pH. After finding hydroxide concentration, you need both steps.
- Ignoring the concentration basis. The classroom estimate treats the listed 29 m as the effective concentration, but advanced work may require activities.
Final answer summary
Under the standard weak-base hydrolysis model at 25 degrees Celsius, a 29 m solution of KC7H5O2 has an estimated pH of about 9.83. The logic is:
- KC7H5O2 dissociates to K+ and C7H5O2–.
- Benzoate is the conjugate base of benzoic acid.
- Use Ka of benzoic acid to calculate Kb of benzoate.
- Solve the hydrolysis equilibrium for [OH–].
- Convert to pOH and then pH.
If you want the fastest exam-ready shortcut, use:
pH ≈ 14 – [ -log10( √( (Kw / Ka) × C ) ) ]
With Ka = 6.3 × 10-5, Kw = 1.0 × 10-14, and C = 29, the result is approximately 9.83.
Authoritative references for acid-base data
For deeper study, consult these authoritative resources:
- NIST Chemistry WebBook for reliable thermodynamic and compound data.
- Chemistry LibreTexts for university-level explanations of weak acid and weak base equilibria.
- U.S. Environmental Protection Agency for broader water chemistry and pH background in environmental systems.