Calculate the pH of a 0.152 M C5H5NHCl Solution
Use this interactive tool to compute the pH of pyridinium chloride from its concentration and the base dissociation constant of pyridine. The calculator applies weak acid equilibrium chemistry using Ka = Kw / Kb.
- Find Ka from the conjugate base
- Solve equilibrium for [H3O+]
- Display pH, pOH, and ion concentrations
Calculator
How to calculate the pH of a 0.152 M C5H5NHCl solution
To calculate the pH of a 0.152 M C5H5NHCl solution, you first identify what species actually controls acidity. C5H5NHCl is pyridinium chloride. In water, the chloride ion is essentially a spectator ion, while the pyridinium ion, C5H5NH+, acts as a weak acid because it is the conjugate acid of pyridine, C5H5N. That means the problem is not handled like a strong acid dissociation. Instead, you use weak acid equilibrium chemistry.
The key relationship is:
Ka = Kw / Kb
If you know the base dissociation constant for pyridine, usually taken near 1.7 x 10^-9 at 25 C, then the acid dissociation constant for pyridinium is:
Ka = (1.0 x 10^-14) / (1.7 x 10^-9) = 5.88 x 10^-6
Once Ka is known, set up the weak acid equilibrium:
C5H5NH+ + H2O ⇌ C5H5N + H3O+
Start with an initial concentration of 0.152 M pyridinium ion. Let x be the amount that dissociates. At equilibrium:
- [C5H5NH+] = 0.152 – x
- [C5H5N] = x
- [H3O+] = x
The equilibrium expression becomes:
Ka = x^2 / (0.152 – x)
Solving exactly with the quadratic formula gives x close to 9.42 x 10^-4 M, so:
pH = -log(9.42 x 10^-4) ≈ 3.03
Why C5H5NHCl is acidic
Students often wonder why a salt can produce an acidic solution. The answer depends on the parent acid and parent base. Pyridinium chloride comes from the reaction of pyridine, a weak base, with hydrochloric acid, a strong acid. The chloride ion does not hydrolyze to any meaningful extent, but the pyridinium ion can donate a proton to water. Because of that proton transfer, hydronium is formed and the solution becomes acidic.
This is a standard pattern in aqueous chemistry:
- Strong acid + weak base salt gives an acidic solution.
- Weak acid + strong base salt gives a basic solution.
- Strong acid + strong base salt gives a nearly neutral solution.
Step by step setup for this exact problem
If your assignment specifically asks you to calculate the pH of a 0.152 M C5H5NHCl solution, follow this workflow:
- Write the acidic species as C5H5NH+.
- Use the known Kb of pyridine to find Ka for pyridinium.
- Build an ICE table for the acid dissociation in water.
- Substitute equilibrium concentrations into the Ka expression.
- Solve for x, where x = [H3O+].
- Convert [H3O+] into pH using pH = -log[H3O+].
This is a powerful method because it works not only for this one concentration, but also for any other concentration of pyridinium chloride as long as temperature and equilibrium constants are known.
Exact solution versus approximation
For weak acids, chemists often approximate 0.152 – x as simply 0.152 if x is small relative to the initial concentration. That leads to:
x ≈ sqrt(KaC)
Plugging in values:
x ≈ sqrt((5.88 x 10^-6)(0.152)) ≈ 9.45 x 10^-4 M
Then:
pH ≈ 3.02
The approximation is extremely close to the exact answer because the percent ionization is well below 5 percent. This is why many general chemistry courses accept either the exact value or the approximation, provided you show the logic clearly.
| Quantity | Typical value at 25 C | Use in calculation |
|---|---|---|
| Initial salt concentration | 0.152 M | Initial [C5H5NH+] |
| Kb of pyridine | 1.7 x 10^-9 | Needed to find Ka |
| Kw of water | 1.0 x 10^-14 | Ka = Kw / Kb |
| Ka of pyridinium | 5.88 x 10^-6 | Weak acid equilibrium |
| Calculated [H3O+] | 9.42 x 10^-4 M | Determines pH |
| Final pH | 3.03 | Reported answer |
Comparison of pH at different concentrations of pyridinium chloride
It is useful to compare how concentration affects pH. As the initial concentration of C5H5NHCl rises, the hydronium concentration also rises and pH decreases. Because this is a weak acid system, the relationship is not perfectly linear, but the trend is very clear.
| [C5H5NHCl] (M) | Ka used | Approximate [H3O+] (M) | Approximate pH |
|---|---|---|---|
| 0.010 | 5.88 x 10^-6 | 2.43 x 10^-4 | 3.61 |
| 0.050 | 5.88 x 10^-6 | 5.42 x 10^-4 | 3.27 |
| 0.100 | 5.88 x 10^-6 | 7.67 x 10^-4 | 3.12 |
| 0.152 | 5.88 x 10^-6 | 9.45 x 10^-4 | 3.02 |
| 0.200 | 5.88 x 10^-6 | 1.08 x 10^-3 | 2.97 |
Common mistakes to avoid
- Treating the salt as a strong acid. C5H5NHCl is not the same as HCl. Only the pyridinium cation contributes to acidity, and it does so weakly.
- Using Kb directly in the pH equation. You need the acid constant of the conjugate acid, so first convert Kb to Ka.
- Including chloride in the equilibrium expression. Cl- is a spectator ion here and does not enter the acid hydrolysis expression.
- Forgetting temperature effects on Kw. If temperature changes, Kw changes, and Ka derived from Kw / Kb changes too.
- Using the weak acid approximation without checking it. The 5 percent rule is a good quick validation.
How the ICE table works here
The ICE method, meaning Initial, Change, Equilibrium, is one of the most reliable ways to organize weak acid and weak base calculations. For this solution:
- Initial: [C5H5NH+] = 0.152, [C5H5N] = 0, [H3O+] = 0
- Change: -x, +x, +x
- Equilibrium: 0.152 – x, x, x
Inserting those terms into the Ka expression gives the full equation. If you solve with the quadratic formula, your answer is mathematically rigorous. If you use the square root approximation, you gain speed with almost no loss in accuracy for this concentration range.
Interpretation of the answer
A pH near 3.03 means the solution is clearly acidic, but not nearly as acidic as a 0.152 M strong acid solution. For comparison, a 0.152 M strong acid would have [H3O+] around 0.152 M and pH about 0.82. That huge difference highlights why equilibrium matters. Weak acid systems release only a small fraction of the available protons into solution.
The percent ionization for this case is roughly:
(9.42 x 10^-4 / 0.152) x 100 ≈ 0.62%
Since this is much less than 5 percent, the approximation is justified and the acid is indeed weak in water.
Authority sources and reference chemistry data
For readers who want to confirm acid-base constants, aqueous equilibrium concepts, or the molecular identity of pyridine related systems, these references are useful:
- NIST Chemistry WebBook for compound data and physical chemistry references.
- University of California Davis educational chemistry material on acid-base properties of salts.
- Michigan State University chemistry notes covering acidity, basicity, and conjugate acid-base relationships.
Practical exam tip
If this problem appears on a quiz or exam, most instructors are looking for correct species identification and proper equilibrium setup. A clean solution typically includes four things: identify C5H5NH+ as a weak acid, compute Ka from Kb, write the equilibrium expression, and calculate pH from the resulting hydronium concentration. Even if your numerical answer differs slightly because of a different accepted Kb value, the method is what earns most of the credit.
Bottom line
To calculate the pH of a 0.152 M C5H5NHCl solution, recognize that pyridinium chloride behaves as a weak acid salt. Use the Kb of pyridine to determine Ka for pyridinium, then solve the weak acid equilibrium. With Kb = 1.7 x 10^-9 at 25 C, the pH comes out to approximately 3.03. This result makes sense chemically because the conjugate acid of a weak base produces an acidic aqueous solution, but only partial ionization occurs.