Calculate the pH of 0.100 M Sodium Propanoate, NaC3H5O2
This premium calculator solves the hydrolysis equilibrium for sodium propanoate, the conjugate base of propanoic acid. Enter concentration, acid dissociation data, and calculation mode to get pH, pOH, Kb, hydroxide concentration, and a chart of equilibrium species.
Sodium Propanoate pH Calculator
Kb = Kw / Ka
Exact: x² / (C – x) = Kb, where x = [OH-]
pOH = -log10[OH-], pH = 14.00 – pOH (for Kw = 1.0 × 10^-14 at 25°C)
Using the common 25°C value Ka = 1.34 × 10^-5 and C = 0.100 M, the pH should be close to 8.94.
Equilibrium Visualization
The chart compares the equilibrium concentrations of propanoate ion, propanoic acid formed by hydrolysis, and hydroxide ion. For a weak conjugate base like propanoate, only a tiny fraction reacts, which is why the solution is mildly basic rather than strongly alkaline.
Expert Guide: How to Calculate the pH of 0.100 M Sodium Propanoate, NaC3H5O2
To calculate the pH of 0.100 M sodium propanoate, you need to recognize what kind of substance you are dealing with. Sodium propanoate, written as NaC3H5O2, is the sodium salt of propanoic acid. In water, it dissociates essentially completely into Na+ and C3H5O2–. The sodium ion does not significantly affect pH, but the propanoate ion does. Because propanoate is the conjugate base of a weak acid, it undergoes hydrolysis with water and generates a small amount of hydroxide ion. That is why the final solution is basic.
The central idea is simple: sodium propanoate is not a strong base, but its anion does accept protons from water to a limited extent. This means the pH will be greater than 7, but nowhere near the pH of a concentrated sodium hydroxide solution. For a standard textbook problem, the accepted Ka of propanoic acid at 25°C is commonly taken as about 1.34 × 10-5, which corresponds to a pKa near 4.87. With that data, a 0.100 M sodium propanoate solution has a pH of approximately 8.94.
Step 1: Write the Dissociation and Hydrolysis Equations
First, write how the salt behaves in water:
- NaC3H5O2 → Na+ + C3H5O2–
The relevant acid-base process is the hydrolysis of the propanoate ion:
- C3H5O2– + H2O ⇌ HC3H5O2 + OH–
This equation tells you that hydroxide ion is produced, so the solution becomes basic.
Step 2: Convert Ka to Kb
Most data tables list the acid dissociation constant for propanoic acid, not the base dissociation constant for propanoate. So the first actual calculation is:
- Use the relation Kb = Kw / Ka
- At 25°C, take Kw = 1.0 × 10-14
- Use Ka = 1.34 × 10-5
Then:
Kb = (1.0 × 10-14) / (1.34 × 10-5) = 7.46 × 10-10
This value is quite small, which confirms that propanoate is only a weak base.
Step 3: Set Up the ICE Table
For the hydrolysis equilibrium:
C3H5O2– + H2O ⇌ HC3H5O2 + OH–
If the initial concentration of sodium propanoate is 0.100 M, then the initial concentration of propanoate ion is also 0.100 M.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| C3H5O2– | 0.100 | -x | 0.100 – x |
| HC3H5O2 | 0 | +x | x |
| OH– | 0 | +x | x |
Now substitute into the equilibrium expression:
Kb = [HC3H5O2][OH–] / [C3H5O2–]
So:
7.46 × 10-10 = x2 / (0.100 – x)
Step 4: Solve for Hydroxide Concentration
There are two acceptable ways to solve this problem. The first is the weak-base approximation, where you assume x is much smaller than 0.100. The second is the exact quadratic solution. In this case, both approaches give essentially the same answer because Kb is very small relative to the initial concentration.
Approximate solution:
If x is small, then 0.100 – x ≈ 0.100
That gives:
x2 = (7.46 × 10-10)(0.100) = 7.46 × 10-11
x = 8.64 × 10-6 M
Exact solution:
Rearrange to quadratic form:
x2 + Kb x – KbC = 0
With Kb = 7.46 × 10-10 and C = 0.100, the positive root is essentially:
x = 8.64 × 10-6 M
That means:
- [OH–] = 8.64 × 10-6 M
- [HC3H5O2] = 8.64 × 10-6 M
- [C3H5O2–] ≈ 0.099991 M
Step 5: Convert to pOH and pH
Now calculate pOH:
pOH = -log(8.64 × 10-6) = 5.06
At 25°C:
pH = 14.00 – 5.06 = 8.94
This is the standard final answer:
The pH of 0.100 M sodium propanoate is approximately 8.94.
Why the Answer Is Not 7.00
Students often wonder why a salt solution can be basic. The reason is that not all salts are neutral. Salts formed from a strong acid and a strong base, such as sodium chloride, are generally neutral. Salts formed from a weak acid and a strong base, such as sodium propanoate, are basic because the anion can react with water. In contrast, a salt formed from a strong acid and a weak base would be acidic.
| Salt type | Example | Parent acid/base strength | Typical pH behavior |
|---|---|---|---|
| Strong acid + strong base | NaCl | HCl + NaOH | Approximately neutral |
| Weak acid + strong base | NaC3H5O2 | Propanoic acid + NaOH | Basic |
| Strong acid + weak base | NH4Cl | HCl + NH3 | Acidic |
| Weak acid + weak base | NH4CH3COO | Both weak | Depends on Ka vs Kb |
How Concentration Changes the pH
One useful way to understand the chemistry is to compare several sodium propanoate concentrations while keeping Ka fixed at 1.34 × 10-5 and using 25°C conditions. Because hydroxide concentration from hydrolysis depends on both Kb and the initial salt concentration, higher concentration produces a slightly higher pH.
| [NaC3H5O2] initial (M) | Kb used | Estimated [OH–] (M) | Estimated pH |
|---|---|---|---|
| 0.010 | 7.46 × 10-10 | 2.73 × 10-6 | 8.44 |
| 0.050 | 7.46 × 10-10 | 6.11 × 10-6 | 8.79 |
| 0.100 | 7.46 × 10-10 | 8.64 × 10-6 | 8.94 |
| 0.500 | 7.46 × 10-10 | 1.93 × 10-5 | 9.29 |
This trend is important. The solution becomes more basic as concentration increases, but the pH does not skyrocket because the base is weak. Even at 0.500 M, the pH remains in the mildly basic range.
Approximation Check and Why It Works
Whenever you use the weak-base shortcut, you should verify that it is valid. For the 0.100 M problem, x is 8.64 × 10-6 M. The percent ionization is:
(8.64 × 10-6 / 0.100) × 100 = 0.00864%
That is far below 5%, so the approximation is excellent. This is why chemistry instructors often accept the shortcut method for problems like this. Still, using the exact quadratic is a good habit if your calculator or software makes it easy.
Common Errors to Avoid
- Using Ka directly to calculate pH. Since sodium propanoate contains the conjugate base, you need Kb, not Ka.
- Calling the solution neutral. A weak-acid salt with a strong-base cation is basic.
- Forgetting the hydrolysis equation. The pH comes from OH– produced by the anion reacting with water.
- Assuming the sodium ion matters. Na+ is a spectator ion in this context.
- Using pH = -log[OH-]. That gives pOH, not pH.
Practical Relevance of Sodium Propanoate pH
Sodium propanoate and related carboxylate salts appear in food chemistry, preservation, analytical chemistry, and buffer systems. Their behavior matters because mild basicity can affect reaction rates, solubility, and microbial growth. In practical lab work, the exact pH can shift slightly due to ionic strength, temperature, and the source used for Ka. However, for general chemistry calculations, the result around pH 8.9 is the standard and chemically meaningful answer.
Authoritative References for Further Study
- NIST Chemistry WebBook (.gov)
- U.S. Environmental Protection Agency pH overview (.gov)
- Purdue University equilibrium and pH help (.edu)
Final Answer Summary
- Identify sodium propanoate as the salt of a weak acid and strong base.
- Write the hydrolysis reaction for propanoate.
- Convert Ka of propanoic acid to Kb using Kb = Kw/Ka.
- Solve for [OH–] using the equilibrium expression.
- Find pOH, then convert to pH.
For 0.100 M sodium propanoate at 25°C with Ka = 1.34 × 10-5, the result is:
pH ≈ 8.94
If you want a fast check, remember this conceptual shortcut: a moderate concentration of a weak conjugate base from a carboxylic acid usually lands in the mildly basic region, commonly between pH 8 and 9.5. Sodium propanoate fits that pattern perfectly.