Calculate the pH of 2.8 M KC4H7O2
This premium calculator evaluates the pH of a potassium butyrate solution by modeling the hydrolysis of the conjugate base C4H7O2- in water. Use the default values for 2.8 M KC4H7O2 or customize the acid constant data for deeper analysis.
Calculator
KC4H7O2 fully dissociates in water into K+ and C4H7O2-. Because K+ is a spectator ion and C4H7O2- is the conjugate base of butyric acid, the solution is basic.
Visual Analysis
The chart compares the acid constant, derived base constant, hydroxide concentration, and resulting pH for the selected conditions.
KC4H7O2(aq) → K+(aq) + C4H7O2-(aq)
C4H7O2-(aq) + H2O(l) ⇌ HC4H7O2(aq) + OH-(aq)
Kb = Kw / Ka
If x = [OH-], then Kb = x² / (C – x)
- K+ does not significantly affect pH.
- C4H7O2- is the conjugate base of butyric acid.
- A concentrated salt of a weak acid produces a basic solution.
How to calculate the pH of 2.8 M KC4H7O2
To calculate the pH of 2.8 M KC4H7O2, you first identify what kind of salt you are dealing with. KC4H7O2 is potassium butyrate. It comes from a strong base, KOH, and a weak acid, butyric acid, HC4H7O2. That means the potassium ion is essentially neutral in water, while the butyrate ion, C4H7O2-, behaves as a weak base. Because weak bases generate hydroxide ions through hydrolysis, the solution will have a pH above 7.
The central chemistry is not the direct ionization of the salt itself, because the salt dissociates almost completely when dissolved. Instead, the important equilibrium is the reaction of the conjugate base with water. Once KC4H7O2 dissolves, it produces K+ and C4H7O2-. The C4H7O2- ion then reacts with water to form butyric acid and OH-. That hydroxide production is why the pH rises into the basic range.
Step 1: Write the dissociation and hydrolysis equations
The first equation is complete dissociation:
- KC4H7O2(aq) → K+(aq) + C4H7O2-(aq)
The second equation is the base hydrolysis step that controls pH:
- C4H7O2-(aq) + H2O(l) ⇌ HC4H7O2(aq) + OH-(aq)
This is the equilibrium you solve for pH. Since the concentration is 2.8 M, the initial concentration of C4H7O2- is also 2.8 M if the salt is fully dissolved.
Step 2: Convert the acid data to a base constant
Most textbooks and data tables list butyric acid using Ka or pKa, not Kb. So you convert the weak acid constant into the corresponding weak base constant for butyrate. At 25 degrees C, a commonly used pKa for butyric acid is about 4.82. Converting that to Ka:
- Ka = 10^-pKa = 10^-4.82 ≈ 1.51 × 10^-5
- Kb = Kw / Ka = 1.0 × 10^-14 / 1.51 × 10^-5 ≈ 6.62 × 10^-10
That Kb value tells you butyrate is a weak base, but because the concentration is quite high at 2.8 M, the hydroxide concentration is still enough to push the pH noticeably above neutral.
Step 3: Set up the ICE table
For the hydrolysis reaction, the ICE setup looks like this:
- Initial: [C4H7O2-] = 2.8, [HC4H7O2] = 0, [OH-] = 0
- Change: -x, +x, +x
- Equilibrium: 2.8 – x, x, x
Substitute into the Kb expression:
- Kb = x² / (2.8 – x)
With Kb ≈ 6.62 × 10^-10, you can solve either by approximation or by the quadratic equation.
Step 4: Solve for hydroxide concentration
Because Kb is very small compared with the initial concentration, x is tiny relative to 2.8. That makes the weak base approximation valid:
- x ≈ √(Kb × C)
Substitute the values:
- x ≈ √((6.62 × 10^-10)(2.8))
- x ≈ √(1.8536 × 10^-9)
- x ≈ 4.31 × 10^-5 M
This x value is the hydroxide ion concentration. From there:
- pOH = -log(4.31 × 10^-5) ≈ 4.37
- pH = 14.00 – 4.37 ≈ 9.63
Therefore, the pH of 2.8 M KC4H7O2 is approximately 9.63 at 25 degrees C using pKa = 4.82 for butyric acid.
Why the answer is not extremely basic
Students sometimes expect a 2.8 M salt solution to have a pH close to that of a strong base. That is not the case here because butyrate is only a weak base. Even though there is a lot of it in solution, only a very small fraction hydrolyzes to produce OH-. The high initial concentration increases the hydroxide level, but the small Kb value limits how far the equilibrium can shift.
This balance between concentration and base strength is central in acid-base chemistry. A concentrated weak base can have a pH that is clearly basic without ever approaching the pH of something like 2.8 M NaOH. For example, a strong base at 2.8 M would produce [OH-] close to 2.8 M and a pH well above 14 in an idealized introductory calculation, while potassium butyrate only produces hydroxide on the order of 10^-5 M by hydrolysis.
Comparison table: weak acid data and conjugate base behavior
| Acid | Approximate pKa at 25 degrees C | Approximate Ka | Conjugate Base | Approximate Kb of Conjugate Base |
|---|---|---|---|---|
| Acetic acid | 4.76 | 1.74 × 10^-5 | Acetate | 5.75 × 10^-10 |
| Butyric acid | 4.82 | 1.51 × 10^-5 | Butyrate | 6.62 × 10^-10 |
| Benzoic acid | 4.20 | 6.31 × 10^-5 | Benzoate | 1.58 × 10^-10 |
| Formic acid | 3.75 | 1.78 × 10^-4 | Formate | 5.62 × 10^-11 |
The table shows why butyrate behaves similarly to acetate as a weak base. Its conjugate acid is a carboxylic acid with a pKa in the upper 4 range, making the conjugate base only mildly basic. That is why the calculated pH lands around 9 to 10 rather than 11 to 13.
Exact solution versus approximation
In many chemistry courses, the square root approximation is used because it is fast and normally accurate for weak acids and weak bases when x is much smaller than the starting concentration. For KC4H7O2, the approximation works very well because x is about 4.3 × 10^-5 while the initial concentration is 2.8. The percent change is tiny:
- (4.31 × 10^-5 / 2.8) × 100 ≈ 0.0015%
Since that is far less than 5%, the approximation is excellent. An exact quadratic solution gives nearly the same value to the displayed precision.
Common mistakes when solving this type of problem
- Treating KC4H7O2 as neutral. It is not neutral because it contains the conjugate base of a weak acid.
- Using Ka directly instead of converting to Kb. You need Kb for the hydrolysis of C4H7O2-.
- Forgetting that K+ is a spectator ion. Potassium does not significantly hydrolyze in water.
- Subtracting pH from 7 instead of 14. Once you calculate pOH, use pH = 14 – pOH at 25 degrees C.
- Assuming strong base behavior. Weak base hydrolysis creates much less OH- than complete ionization would.
How concentration changes the pH
Since the hydroxide concentration from a weak base scales approximately with the square root of concentration, the pH does rise as the salt concentration increases, but not linearly. Doubling the concentration does not double the pH shift. This is an important conceptual point because many students expect concentration and pH to move together in a simple one-to-one fashion. In weak electrolyte systems, equilibrium moderates that relationship.
| [KC4H7O2] (M) | Estimated [OH-] using Kb = 6.62 × 10^-10 | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.10 | 8.14 × 10^-6 M | 5.09 | 8.91 |
| 0.50 | 1.82 × 10^-5 M | 4.74 | 9.26 |
| 1.00 | 2.57 × 10^-5 M | 4.59 | 9.41 |
| 2.80 | 4.31 × 10^-5 M | 4.37 | 9.63 |
This table highlights a practical trend: even fairly concentrated weak base salts produce pH values in a moderate basic range rather than an extreme one. For potassium butyrate, moving from 0.10 M to 2.80 M raises the pH by less than one unit, which reflects the logarithmic nature of pH and the limited hydrolysis of the conjugate base.
What assumptions are built into the calculation
Like most introductory chemistry calculations, this result relies on a standard set of assumptions. First, it assumes 25 degrees C so that Kw = 1.0 × 10^-14. Second, it treats activities as if they were equal to concentrations. At 2.8 M, that ideal assumption becomes less accurate because ionic strength is high and activity effects can matter. Third, it assumes the salt dissociates completely and that only the butyrate hydrolysis equilibrium needs to be considered. These assumptions are fine for a classroom or homework answer, but a research-grade treatment could produce a slightly different value if activity coefficients were included.
When to use an exact equilibrium treatment
If your chemistry course is more advanced, or if your instructor expects careful equilibrium modeling, you may be asked to solve the quadratic equation instead of using the square root approximation. For this problem, the exact relation is:
- x² + Kb x – Kb C = 0
Then:
- x = (-Kb + √(Kb² + 4KbC)) / 2
Using C = 2.8 and Kb ≈ 6.62 × 10^-10 gives essentially the same hydroxide concentration as the approximation. This is why the calculator above lets you compare approximate and exact methods. For dilute solutions or stronger weak bases, the exact method can matter more.
Practical chemistry interpretation
Potassium butyrate is relevant in acid-base chemistry because it behaves as the salt of a weak acid and a strong base. That makes it an excellent teaching example for identifying whether a salt solution is acidic, basic, or neutral. It also demonstrates how the identity of the parent acid determines the pH outcome. If the acid is weak, the conjugate base can hydrolyze. If the acid is strong, the conjugate base would be negligibly basic. This type of reasoning helps students classify salts quickly before doing any math.
In the case of KC4H7O2, the expected qualitative answer is basic, and the quantitative answer is around pH 9.6. Those two pieces should agree. If you ever get a pH below 7 for this problem, that is a sign that a setup mistake was made.
Authoritative references for acid-base constants and water chemistry
For deeper study, consult these authoritative resources:
- National Institute of Standards and Technology (NIST)
- Chemistry LibreTexts
- United States Environmental Protection Agency (EPA)
- Michigan State University acid-base reference
- USGS pH and water science overview
Final answer
If you use a standard butyric acid pKa of 4.82 at 25 degrees C, the pH of 2.8 M KC4H7O2 is approximately 9.63. The solution is basic because C4H7O2- is the conjugate base of a weak acid and undergoes hydrolysis to produce OH-.