b calculate the ph of 0.380 m potassium propionate kc3h5o2
Use this premium chemistry calculator to determine the pH of a potassium propionate solution. The tool applies weak-base hydrolysis from the conjugate base of propionic acid and shows the full acid-base pathway, including Ka, Kb, pOH, and final pH.
How to calculate the pH of 0.380 M potassium propionate, KC3H5O2
When you see the prompt b calculate the ph of 0.380 m potassium propionate kc3h5o2, the chemistry behind it is a classic weak-acid-conjugate-base problem. Potassium propionate is a soluble ionic salt. In water, it dissociates essentially completely into potassium ions and propionate ions:
The potassium ion is a spectator ion because it comes from the strong base KOH and does not significantly affect pH. The important species is propionate, C3H5O2-, which is the conjugate base of propionic acid, HC3H5O2. Because propionic acid is a weak acid, its conjugate base can react with water to generate hydroxide:
That hydroxide formation is the reason the solution becomes basic. So the entire problem reduces to finding the base hydrolysis of propionate in a 0.380 M solution.
Step 1: Identify the acid-base pair
Potassium propionate is the salt of:
- Strong base: potassium hydroxide, KOH
- Weak acid: propionic acid, HC3H5O2
A salt from a strong base and weak acid always gives a basic solution, assuming no other acid-base effects dominate. That gives you a quick conceptual check before doing any math: the pH must end up above 7.
Step 2: Use the pKa or Ka for propionic acid
A commonly used value for the dissociation of propionic acid at 25 degrees C is pKa = 4.87. From this, we can calculate Ka:
Once Ka is known, the base dissociation constant for propionate is found from the water ion product:
This is a small Kb, which tells us propionate is a weak base. However, because the concentration is relatively high at 0.380 M, it still generates enough hydroxide to make the solution clearly basic.
Step 3: Set up the ICE table
For the hydrolysis reaction
the initial concentration of propionate is 0.380 M, while the initial concentrations of HC3H5O2 and OH- from hydrolysis are effectively zero. Let x represent the amount of propionate that reacts.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| C3H5O2- | 0.380 | -x | 0.380 – x |
| HC3H5O2 | 0 | +x | x |
| OH- | 0 | +x | x |
Now insert those values into the Kb expression:
Step 4: Use the weak-base approximation
Because Kb is much smaller than the initial concentration, x will be small relative to 0.380, so we can use the approximation:
That simplifies the equation to:
Now calculate pOH:
Finally, convert pOH to pH:
Answer: the pH of 0.380 M potassium propionate is approximately 9.23 at 25 degrees C, using pKa = 4.87 for propionic acid.
Step 5: Check whether the approximation is valid
The 5 percent rule is a fast way to test whether ignoring x in the denominator was justified. Compare x with the original concentration:
That is far below 5 percent, so the approximation is excellent. In other words, the simplified solution and the quadratic solution will agree to many significant figures in this case.
Why this salt is basic
Students sometimes memorize rules about salts but do not connect them to equilibrium chemistry. Potassium propionate is basic because the anion is willing to accept a proton from water. The stronger the parent acid, the weaker its conjugate base. Propionic acid is weak, not strong, so propionate has noticeable basicity. Still, it is only a weak base, which is why the pH lands around 9.2 rather than 12 or 13.
You can compare this to other common salts:
- KCl comes from a strong acid and strong base, so its solution is close to neutral.
- NH4Cl comes from a weak base and strong acid, so its solution is acidic.
- KC3H5O2 comes from a weak acid and strong base, so its solution is basic.
Reference values commonly used in this problem
The exact pH can vary slightly depending on the textbook value selected for the acid dissociation constant of propionic acid. Different sources may round pKa to 4.86, 4.87, or 4.88. Those tiny differences shift the final pH only slightly.
| Parameter | Common value | Meaning for calculation |
|---|---|---|
| Propionic acid pKa | 4.87 | Used to compute Ka = 1.35 × 10^-5 |
| Ka of propionic acid | 1.35 × 10^-5 | Needed to find the conjugate base strength |
| Kw at 25 degrees C | 1.0 × 10^-14 | Used in Kb = Kw / Ka |
| Kb of propionate | 7.41 × 10^-10 | Base hydrolysis constant |
| [OH-] at 0.380 M salt | 1.68 × 10^-5 M | Predicted hydroxide concentration |
| Final pH | 9.23 | Basic solution |
How concentration changes the pH
Because weak-base hydrolysis depends on both Kb and concentration, more concentrated potassium propionate solutions produce slightly more hydroxide and therefore somewhat higher pH values. The relationship is not linear, because pH is logarithmic. Doubling the concentration does not double the pH number. Instead, it changes hydroxide concentration according to the square-root approximation.
| KC3H5O2 concentration (M) | Approx. [OH-] (M) | Approx. pOH | Approx. pH |
|---|---|---|---|
| 0.050 | 6.09 × 10^-6 | 5.22 | 8.78 |
| 0.100 | 8.61 × 10^-6 | 5.07 | 8.93 |
| 0.380 | 1.68 × 10^-5 | 4.77 | 9.23 |
| 0.500 | 1.93 × 10^-5 | 4.71 | 9.29 |
| 1.000 | 2.72 × 10^-5 | 4.57 | 9.43 |
Common mistakes when solving this problem
- Treating potassium propionate as a strong base. It is not. The salt dissociates completely, but the propionate ion only partially reacts with water.
- Using Ka directly instead of Kb. Once the salt is dissolved, you are analyzing the behavior of the conjugate base. So you must convert Ka to Kb.
- Forgetting that potassium is a spectator ion. K+ does not control the pH here.
- Reporting pOH as pH. After finding [OH-], you must calculate pOH and then convert to pH.
- Using the wrong acid. The corresponding weak acid is propionic acid, not acetic acid or another carboxylic acid.
When should you use the quadratic formula?
For this specific case, the square-root method is more than sufficient because x is tiny compared with 0.380 M. However, in very dilute weak-acid or weak-base problems, or when K is not much smaller than the concentration, the approximation can break down. In those cases, the exact quadratic method is preferable. This calculator includes both methods so you can verify that the answers match for this problem.
Practical interpretation of pH 9.23
A pH of 9.23 means the solution is mildly basic, not extremely caustic. In laboratory settings, potassium propionate may appear in discussions of buffers, food chemistry, and weak-acid salt systems. The number itself reflects a balance: propionate is basic enough to pull some protons from water, but not strong enough to generate a very large hydroxide concentration.
This also illustrates a broader principle in acid-base chemistry: the pH of a salt solution depends on the acid and base from which the salt was formed. Understanding conjugate pairs is often more important than memorizing isolated examples.
Fast exam shortcut for this exact question
If you are solving this under timed conditions, you can use the following compressed approach:
- Recognize KC3H5O2 as the salt of a weak acid and strong base.
- Conclude the solution is basic.
- Use pKa = 4.87 to compute Ka = 1.35 × 10^-5.
- Compute Kb = 1.0 × 10^-14 / 1.35 × 10^-5 = 7.41 × 10^-10.
- Use [OH-] = √(KbC) = √[(7.41 × 10^-10)(0.380)] = 1.68 × 10^-5.
- Find pOH = 4.77 and pH = 9.23.
That sequence is usually enough for full credit when the instructor expects the weak-base hydrolysis method.
Authoritative references for acid-base data
For students who want to verify acid dissociation constants, water ion product values, and foundational acid-base theory, these sources are useful:
- NIST Chemistry WebBook (.gov)
- University-level explanation of salt hydrolysis chemistry (.edu-hosted course content reference)
- University of Wisconsin acid-base equilibria resource (.edu)
Final answer
If the question is simply b calculate the ph of 0.380 m potassium propionate kc3h5o2, the concise final result is:
That answer comes from treating the propionate ion as a weak base, calculating its Kb from the Ka of propionic acid, and solving for hydroxide concentration in a 0.380 M solution.