Calculate the pH of 5 M Solution of AlBr3
Use this interactive aluminum bromide pH calculator to estimate the acidity of a concentrated AlBr3 solution. The calculation treats Al3+ as a weak acidic metal ion in water and solves the hydrolysis equilibrium using either an exact quadratic method or a quick approximation.
Results
Expert Guide: How to Calculate the pH of a 5 M Solution of AlBr3
If you need to calculate the pH of a 5 M solution of AlBr3, the most important idea is that aluminum bromide is not handled like a simple neutral salt. In water, AlBr3 dissociates into Al3+ and Br-. The bromide ion is the conjugate base of hydrobromic acid, a strong acid, so Br- contributes essentially no basicity. The acidic behavior comes from the highly charged aluminum ion. Once hydrated in water, Al3+ strongly polarizes nearby O-H bonds in coordinated water molecules, making proton release easier. That hydrolysis step is what lowers the pH.
In introductory chemistry, this is often summarized by saying that the hexaaqua aluminum ion acts as a weak acid: [Al(H2O)6]3+ ⇌ [Al(H2O)5OH]2+ + H+. The corresponding acid dissociation constant is commonly reported in the neighborhood of 1 x 10^-5 to 1.4 x 10^-5 at room temperature, depending on the source and the exact convention used. For a practical calculator, taking Ka = 1.4 x 10^-5 is a very standard classroom assumption. With an AlBr3 concentration of 5.0 M, the initial aluminum ion concentration is also 5.0 M because one formula unit provides one Al3+ ion.
Step 1: Write the dissociation and hydrolysis reactions
Start with the salt dissociation:
- AlBr3(aq) → Al3+(aq) + 3Br-(aq)
Then identify which ion affects pH. Bromide is negligible as a base because it comes from the strong acid HBr. Aluminum is the reactive species for acid-base chemistry in water:
- Al3+ + H2O ⇌ AlOH2+ + H+
In a more chemically realistic notation, the reacting species is a hydrated aluminum complex. However, for pH work, the shorthand hydrolysis equation above is usually enough and gives the same usable equilibrium setup.
Step 2: Set up the ICE table
Let the initial Al3+ concentration be 5.0 M. If x is the amount that hydrolyzes, then:
- Initial: [Al3+] = 5.0, [H+] = 0
- Change: [Al3+] decreases by x, [H+] increases by x
- Equilibrium: [Al3+] = 5.0 – x, [H+] = x
The acid dissociation expression is:
Ka = [AlOH2+][H+] / [Al3+] = x^2 / (5.0 – x)
If Ka = 1.4 x 10^-5, then:
1.4 x 10^-5 = x^2 / (5.0 – x)
Step 3: Solve for hydrogen ion concentration
There are two common approaches. The quick approach assumes x is very small compared with 5.0:
x^2 / 5.0 ≈ 1.4 x 10^-5
x^2 ≈ 7.0 x 10^-5
x ≈ 8.37 x 10^-3 M
Since x = [H+], we get:
pH = -log(8.37 x 10^-3) ≈ 2.08
The exact quadratic method gives nearly the same answer:
x^2 + Kax – KaC = 0
x = (-Ka + √(Ka^2 + 4KaC)) / 2
Substituting Ka = 1.4 x 10^-5 and C = 5.0 gives x ≈ 8.36 x 10^-3 M, again leading to pH ≈ 2.08. This confirms that the approximation is excellent here.
Why AlBr3 solution is acidic
The acidity of AlBr3 solutions surprises many students because bromide itself is not acidic. The reason is the very high charge density on Al3+. Small, highly charged metal ions attract electron density strongly from water ligands, weakening O-H bonds and making proton donation easier. This is a classic trend in inorganic chemistry: as cation charge increases and ionic radius decreases, hydrated cations often become more acidic.
By contrast, a salt such as NaBr forms Na+ and Br-. Sodium does not hydrolyze enough to matter, and bromide is neutral in water, so NaBr solutions stay close to pH 7. Aluminum bromide behaves very differently because the aluminum ion is strongly hydrolyzing.
Comparison table: hydrolyzing metal ions and acidity strength
| Hydrated metal ion | Typical charge | Approximate pKa of first hydrolysis | Acidic tendency in water |
|---|---|---|---|
| [Na(H2O)n]+ | +1 | Very large pKa, effectively nonacidic | Negligible effect on pH |
| [Mg(H2O)6]2+ | +2 | About 11.4 | Very weakly acidic |
| [Zn(H2O)6]2+ | +2 | About 9.0 | Weakly acidic |
| [Fe(H2O)6]3+ | +3 | About 2.2 | Strongly hydrolyzing |
| [Al(H2O)6]3+ | +3 | About 5.0 | Moderately acidic, important for pH calculations |
This table shows why aluminum salts often produce acidic solutions. Their acidity is nowhere near that of a strong acid like HCl, but it is still substantial enough to push the pH well below neutral, especially at high concentration.
Computed pH values for different AlBr3 concentrations
One useful way to understand the chemistry is to compare several concentrations. Using Ka = 1.4 x 10^-5 and solving the equilibrium with the weak acid model for Al3+, we obtain the following approximate pH values:
| AlBr3 concentration (M) | [Al3+] initial (M) | Estimated [H+] at equilibrium (M) | Estimated pH |
|---|---|---|---|
| 0.01 | 0.01 | 3.67 x 10^-4 | 3.44 |
| 0.10 | 0.10 | 1.18 x 10^-3 | 2.93 |
| 0.50 | 0.50 | 2.64 x 10^-3 | 2.58 |
| 1.00 | 1.00 | 3.73 x 10^-3 | 2.43 |
| 5.00 | 5.00 | 8.36 x 10^-3 | 2.08 |
Common mistakes when calculating the pH of AlBr3
- Assuming the solution is neutral. This is wrong because Al3+ hydrolyzes water and produces H+.
- Treating AlBr3 like a strong acid directly. The acidity does not come from AlBr3 dissociating into H+. It comes from the weak acid behavior of hydrated Al3+.
- Using bromide in the equilibrium expression. Br- is usually ignored for pH because it is the conjugate base of a strong acid.
- Forgetting stoichiometry. A 5 M AlBr3 solution gives 5 M Al3+, not 15 M Al3+. The 3 applies to bromide only.
- Ignoring concentration effects and real solution behavior. At very high ionic strength, such as 5 M, real activity effects can matter. Classroom calculations usually still use concentration, but advanced treatments may differ.
Does 5 M concentration change the interpretation?
Yes, at least conceptually. A 5 M solution is highly concentrated. In real solutions this concentrated, ion pairing, nonideal activity coefficients, and altered water structure can affect measured pH relative to a simple textbook model. However, most educational and exam settings expect the standard weak acid hydrolysis calculation. Under that assumption, the best answer remains about pH 2.08. If you were doing research-grade solution chemistry, you would likely use activities rather than raw molar concentrations and might need literature data for ionic strength corrections.
Practical interpretation of the answer
A pH near 2.08 means the solution is clearly acidic. It is less acidic than a 0.01 M strong acid, which has pH 2 exactly, but it is still corrosive enough to demand proper handling and eye protection in any lab environment. This also helps explain why many aluminum salts can affect water chemistry, hydrolysis equilibria, and buffering behavior in environmental and analytical settings.
If your instructor asks for the “pH of a 5 M AlBr3 solution,” the shortest complete answer is usually: “AlBr3 dissociates to Al3+ and Br-. Br- is neutral, while Al3+ hydrolyzes as a weak acid with Ka about 1.4 x 10^-5. Solving x^2/(5.0 – x) = 1.4 x 10^-5 gives [H+] ≈ 8.36 x 10^-3 M, so pH ≈ 2.08.”
Recommended authoritative references
- LibreTexts Chemistry educational resources
- U.S. Environmental Protection Agency
- NIST Chemistry WebBook
- University of Wisconsin Chemistry resources
Final answer summary
To calculate the pH of a 5 M solution of AlBr3, assume complete dissociation to give 5 M Al3+ and 15 M Br-. Only Al3+ affects pH significantly because it hydrolyzes as a weak acid. Using Ka = 1.4 x 10^-5 for hydrated aluminum:
- Ka = x^2 / (5.0 – x)
- x = [H+] ≈ 8.36 x 10^-3 M
- pH = -log[H+] ≈ 2.08
Therefore, the pH of a 5 M AlBr3 solution is approximately 2.08 under standard textbook assumptions.