Calculate the pH of a 0.010 m AlCl3 Solution
Use the hydrated aluminum ion acid hydrolysis model to estimate pH for aluminum chloride solutions at 25 degrees Celsius.
Interactive AlCl3 pH Calculator
Aluminum chloride dissociates to give Al3+ and Cl-. Chloride is the conjugate base of a strong acid and is essentially neutral, while the hydrated aluminum ion behaves as a weak acid in water.
How to calculate the pH of a 0.010 m AlCl3 solution
To calculate the pH of a 0.010 m aluminum chloride solution, you do not treat AlCl3 as a simple neutral salt. Aluminum chloride is formed from a strong acid component, HCl, and a highly charged metal cation, Al3+, that strongly hydrates in water. Once dissolved, the chloride ions are essentially spectators for pH purposes, but the aluminum ion polarizes the O-H bonds in coordinated water molecules and promotes proton release. That makes the resulting solution acidic.
In introductory and intermediate chemistry, the accepted model is to consider the hydrated aluminum ion, usually written as [Al(H2O)6]3+, as a weak acid with a first hydrolysis constant near 1.4 x 10^-5 at 25 degrees Celsius. For a 0.010 m solution, which is dilute enough that molality and molarity are often taken as approximately equal in classroom calculations, the pH is found from the weak-acid equilibrium rather than from full strong-acid dissociation.
Step-by-step chemistry behind the calculation
1. Write the dissociation of aluminum chloride
When aluminum chloride dissolves, it separates into ions:
AlCl3(aq) to Al3+(aq) + 3Cl-(aq)
The chloride ion is the conjugate base of hydrochloric acid, a strong acid, so Cl- does not hydrolyze enough to matter for pH under ordinary conditions. The important species is Al3+, but in water that ion is really a hydrated complex.
2. Identify the acidic species in solution
The actual acidic particle is the hexaaqua aluminum ion:
[Al(H2O)6]3+ + H2O ⇌ [Al(H2O)5OH]2+ + H3O+
This is a weak-acid equilibrium. The corresponding acid dissociation constant is commonly taken to be about 1.4 x 10^-5 at 25 degrees Celsius, although textbook values may vary slightly. Because the cation has a +3 charge and a relatively small radius, it strongly attracts electron density from coordinated water and makes proton loss much easier than in ordinary liquid water.
3. Set up the ICE table
Let the initial aluminum concentration be C = 0.010. Assuming one acidic proton is considered in the first hydrolysis step, we set:
- Initial: [Al(H2O)6]3+ = 0.010, [H3O+] = 0
- Change: -x, +x
- Equilibrium: 0.010 – x, x
The equilibrium expression becomes:
Ka = x2 / (0.010 – x)
Using Ka = 1.4 x 10^-5, solve for x, where x is the hydronium ion concentration.
4. Solve exactly with the quadratic formula
Rearrange the expression:
x2 + Ka x – KaC = 0
Substitute the values:
x2 + (1.4 x 10^-5)x – (1.4 x 10^-7) = 0
Now apply the quadratic formula:
x = [-Ka + sqrt(Ka2 + 4KaC)] / 2
This gives:
x ≈ 3.67 x 10^-4 M
Then calculate pH:
pH = -log10(3.67 x 10^-4) ≈ 3.44
5. Check the approximation method
If x is small compared with 0.010, then:
x ≈ sqrt(KaC) = sqrt((1.4 x 10^-5)(0.010)) = sqrt(1.4 x 10^-7) ≈ 3.74 x 10^-4
That produces:
pH ≈ -log10(3.74 x 10^-4) ≈ 3.43
The approximation is very close to the exact answer. Since x / C is about 0.036 to 0.037, the approximation is acceptable for many teaching problems, though the exact quadratic is the better method.
Final answer for a 0.010 m AlCl3 solution
Using the hydrated aluminum ion hydrolysis constant Ka = 1.4 x 10^-5 at 25 degrees Celsius and treating 0.010 m as approximately 0.010 M in a dilute aqueous solution, the pH is approximately 3.44. If you use the square-root weak-acid approximation instead of the exact quadratic, you get about 3.43. Both values describe the same acidic behavior, and the small difference comes from rounding and mathematical method.
Why AlCl3 is acidic while NaCl is neutral
This question comes up often because students first learn that salts come from acids and bases, and they may expect all salts to be close to neutral. That is not true. Sodium chloride is neutral because Na+ is the cation of a strong base and Cl- is the anion of a strong acid. Neither ion appreciably hydrolyzes in water. Aluminum chloride is different because Al3+ is a small, highly charged cation with strong Lewis acidity. It binds water strongly and weakens the O-H bonds in the hydration shell. That leads to proton release and lower pH.
- NaCl: no significant hydrolysis, pH near 7 in pure water.
- NH4Cl: acidic because NH4+ is a weak acid.
- AlCl3: acidic because hydrated Al3+ undergoes hydrolysis.
- Na2CO3: basic because CO32- hydrolyzes to form OH-.
Comparison table: pH of AlCl3 at different concentrations
The table below uses Ka = 1.4 x 10^-5 and the exact weak-acid quadratic model for the first hydrolysis of hydrated Al3+ at 25 degrees Celsius. These values are useful for seeing the trend: more concentrated AlCl3 solutions are more acidic.
| AlCl3 concentration | Estimated [H+] | Estimated pH | Percent ionized |
|---|---|---|---|
| 0.0010 | 1.12 x 10^-4 | 3.95 | 11.2% |
| 0.0050 | 2.58 x 10^-4 | 3.59 | 5.16% |
| 0.0100 | 3.67 x 10^-4 | 3.44 | 3.67% |
| 0.0500 | 8.30 x 10^-4 | 3.08 | 1.66% |
| 0.1000 | 1.18 x 10^-3 | 2.93 | 1.18% |
Comparison table: exact vs approximation for 0.010 AlCl3
Students are often asked whether the square-root approximation is valid. The comparison below shows that for 0.010 AlCl3, the approximation is close, but the exact quadratic remains the more rigorous result.
| Method | Equation used | [H+] | pH | Difference from exact |
|---|---|---|---|---|
| Exact quadratic | x = [-Ka + sqrt(Ka2 + 4KaC)] / 2 | 3.67 x 10^-4 | 3.44 | 0.00 pH units |
| Approximation | x ≈ sqrt(KaC) | 3.74 x 10^-4 | 3.43 | About 0.01 pH units |
Common mistakes when solving this problem
- Assuming AlCl3 is neutral because it is a salt. This misses hydrolysis of the hydrated Al3+ ion.
- Treating AlCl3 like a strong acid. Aluminum chloride itself does not simply release three protons as HCl would. Its acidity comes from metal-ion hydrolysis.
- Using the chloride concentration in the Ka expression. Chloride is not the acidic species.
- Ignoring the distinction between molality and molarity. In careful physical chemistry, m and M are different. For a dilute aqueous teaching example like 0.010 m, they are usually close enough for this calculation.
- Forgetting that Ka values can vary somewhat by source. Different references may report slightly different hydrolysis constants due to ionic strength, temperature, and model assumptions.
How accurate is the answer?
The value pH about 3.44 is an educational estimate based on the first hydrolysis equilibrium of the hydrated aluminum ion. In real laboratory solutions, the measured pH can vary because of ionic strength, activity corrections, temperature, and secondary hydrolysis equilibria. Aluminum chemistry in water is also more complicated than the single-step weak-acid model suggests, especially at higher concentrations or over time. Still, for standard general chemistry and many analytical chemistry contexts, this is the correct and accepted method for a 0.010 m AlCl3 problem.
Factors that can shift the experimental pH
- Temperature changes alter equilibrium constants.
- Higher ionic strength changes activity coefficients.
- Hydrated aluminum can participate in multiple hydrolysis steps.
- Exposure to atmospheric carbon dioxide can slightly affect pH.
- Actual prepared concentration may be molal, molar, or weight-based depending on procedure.
Authority sources and further reading
If you want to deepen your understanding of pH, aqueous metal ion chemistry, and water acidity, these authoritative resources are helpful:
Practical interpretation of the result
A pH around 3.44 means a 0.010 m aluminum chloride solution is distinctly acidic, though much less acidic than a 0.010 M strong acid such as HCl. A 0.010 M HCl solution would have pH about 2.00 because it dissociates essentially completely. By contrast, the aluminum aqua ion is only partially hydrolyzed, so the hydronium concentration is much lower. This makes AlCl3 an excellent example of why salt solutions cannot always be classified by looking only at the formula. The properties of the ions in water are what matter.
Short answer summary
To calculate the pH of a 0.010 m AlCl3 solution, model the hydrated aluminum ion as a weak acid with Ka about 1.4 x 10^-5. Solve Ka = x2 / (0.010 – x) for x = [H+]. The exact solution gives x about 3.67 x 10^-4 and therefore pH about 3.44. This is the standard chemistry answer under dilute aqueous conditions at 25 degrees Celsius.