Calculate the pH of a 1.51 m Solution of CH3NH3Br
This premium calculator solves the acid hydrolysis problem for methylammonium bromide, CH3NH3Br. Enter the concentration, choose how you want to treat the concentration unit, and the tool will calculate Ka, hydrogen ion concentration, pH, and pOH using a full weak-acid equilibrium approach.
Expert Guide: How to Calculate the pH of a 1.51 m Solution of CH3NH3Br
If you need to calculate the pH of a 1.51 m solution of CH3NH3Br, the key idea is that this salt is not neutral in water. Methylammonium bromide dissociates into CH3NH3+ and Br–. The bromide ion is the conjugate base of a strong acid, HBr, so Br– does not significantly affect pH. The methylammonium ion, however, is the conjugate acid of the weak base methylamine, CH3NH2, so it does react with water and produces hydronium ions. That makes the solution acidic.
In other words, CH3NH3Br behaves as an acidic salt. Once it dissolves, the important equilibrium is:
Because hydronium ion is produced, the pH falls below 7. For a relatively concentrated solution such as 1.51, the acid is still weak, so the pH is not extremely low. It typically ends up around the low 5 range when standard 25°C constants are used.
Step 1: Identify the Acidic Species
CH3NH3Br is composed of:
- CH3NH3+, the conjugate acid of methylamine
- Br–, the conjugate base of hydrobromic acid
Since HBr is a strong acid, Br– is negligibly basic in water. Therefore, the pH depends almost entirely on CH3NH3+.
Step 2: Convert Kb of Methylamine into Ka of Methylammonium
Most reference tables list the base dissociation constant of methylamine rather than the acid dissociation constant of methylammonium. The relationship is:
At 25°C, a commonly used value is:
- Kb(CH3NH2) = 4.4 × 10-4
- Kw = 1.0 × 10-14
So:
This is a very small Ka, which confirms that CH3NH3+ is a weak acid. Even so, in a 1.51 concentration solution, it still generates enough H+ to make the pH clearly acidic.
Step 3: Set Up the ICE Table
Assume the 1.51 m value is treated numerically as the concentration used in the equilibrium setup, which is the standard simplification in many classroom problems when density is not supplied. Let x represent the amount of CH3NH3+ that ionizes:
Plug this into the Ka expression:
Since Ka is very small, x is much smaller than 1.51, so the denominator can be approximated as 1.51:
Thus:
- [H3O+] ≈ 5.86 × 10-6 M
- pH = -log(5.86 × 10-6) ≈ 5.23
Why This Salt Is Acidic Instead of Neutral
Students often remember that salts come from acids and bases, but the parent acid and parent base matter. A salt formed from a strong acid and weak base gives an acidic solution. CH3NH3Br fits that pattern exactly. HBr is a strong acid, and CH3NH2 is a weak base. Because the cation comes from a weak base, its conjugate acid can donate protons to water. That hydrolysis process lowers pH.
| Salt Type | Parent Acid | Parent Base | Expected pH | Example |
|---|---|---|---|---|
| Strong acid + strong base | Strong | Strong | About 7 | NaCl |
| Strong acid + weak base | Strong | Weak | Below 7 | CH3NH3Br |
| Weak acid + strong base | Weak | Strong | Above 7 | CH3COONa |
| Weak acid + weak base | Weak | Weak | Depends on Ka vs Kb | NH4CH3COO |
Important Note About Molality Versus Molarity
The problem statement uses 1.51 m, which formally means molality, or moles of solute per kilogram of solvent. Strictly speaking, equilibrium expressions are written in terms of concentration or activity, not molality alone. In many general chemistry exercises, however, if density is not provided, the numerical value is used directly as an approximation for molarity, especially when the purpose is to test acid-base equilibrium concepts rather than solution thermodynamics.
If you were doing a more rigorous physical chemistry treatment, you would need the density of the solution, and possibly activity coefficients, to convert properly between molality and molarity and to refine the result. Still, for most instructional contexts, the standard answer remains close to pH 5.23.
When the Approximation Is Reasonable
- The textbook or assignment gives only molality and no density information
- The course level is general chemistry or introductory analytical chemistry
- The intent is to practice conjugate acid hydrolysis
- The weak-acid dissociation is small compared with the formal concentration
When You Would Need More Detail
- Very concentrated solutions where nonideal behavior matters
- Precise laboratory pH prediction
- Thermodynamic calculations using activities instead of simple concentrations
- Temperature conditions far from 25°C
Exact Quadratic Solution Versus Approximation
For weak acids, the square-root approximation is usually enough:
Here, with Ka = 2.27 × 10-11 and C = 1.51, that gives [H3O+] ≈ 5.86 × 10-6. If you solve the full quadratic equation, the result is essentially the same to normal reporting precision. This tells you the approximation is very safe in this problem.
| Method | Ka Used | Formal Concentration | [H3O+] | Calculated pH |
|---|---|---|---|---|
| Square-root approximation | 2.27 × 10-11 | 1.51 | 5.86 × 10-6 | 5.23 |
| Full quadratic solution | 2.27 × 10-11 | 1.51 | 5.86 × 10-6 | 5.23 |
Reference Data You Should Know
Several real numerical constants support this calculation. At 25°C, pure water has Kw = 1.0 × 10-14, which means the neutral pH point is 7.00 under idealized conditions. Methylamine is a weak base with Kb commonly listed near 4.4 × 10-4. From that, the conjugate acid methylammonium has a Ka on the order of 10-11. This is why CH3NH3+ is acidic, but only weakly acidic.
- Write the dissolution: CH3NH3Br → CH3NH3+ + Br–
- Ignore Br– as a pH contributor because it is the conjugate base of a strong acid
- Use Ka = Kw / Kb
- Set up the weak-acid equilibrium for CH3NH3+
- Solve for [H3O+]
- Take the negative log to get pH
Common Mistakes to Avoid
- Treating CH3NH3Br as a neutral salt. It is acidic because CH3NH3+ hydrolyzes.
- Using Kb directly in the acid expression. You must first convert Kb of CH3NH2 into Ka of CH3NH3+.
- Letting bromide affect the pH. Br– is effectively neutral in this context.
- Confusing m with M. The problem may require a classroom approximation unless density is given.
- Forgetting the logarithm sign. pH is negative log base 10 of hydronium concentration.
How Concentration Affects the pH
For a weak acid like CH3NH3+, increasing concentration lowers pH, but not in a directly proportional way. The hydronium concentration tends to scale with the square root of the formal acid concentration when the approximation is valid. That means a tenfold increase in concentration changes pH by about 0.5 units rather than a full 1 unit for a weak acid. This is why a fairly concentrated salt solution can still have a pH that is only moderately acidic.
Quick Interpretation of the 1.51 Result
A pH near 5.23 means the solution is acidic but nowhere close to the strength of a strong acid. For comparison, pure water at 25°C sits at pH 7.00, while common strong acid laboratory solutions can have pH values below 1. The CH3NH3Br solution is clearly acidic enough to matter in equilibrium and buffer calculations, but it remains governed by weak-acid chemistry.
Authoritative Chemistry Resources
For learners who want to verify the theory behind salt hydrolysis, acid-base equilibria, and pH calculations, these authoritative resources are useful:
- LibreTexts Chemistry educational resources (.edu affiliated hosting and institutional use)
- National Institute of Standards and Technology, NIST (.gov)
- NIST Chemistry WebBook for chemical reference data (.gov)
Bottom Line
To calculate the pH of a 1.51 m solution of CH3NH3Br, recognize that the methylammonium ion is a weak acid. Use the known Kb of methylamine to find Ka for CH3NH3+, set up the acid equilibrium, and solve for hydronium concentration. With Kb = 4.4 × 10-4 and Kw = 1.0 × 10-14 at 25°C, you obtain Ka = 2.27 × 10-11, [H3O+] ≈ 5.86 × 10-6, and pH ≈ 5.23. That is the standard answer expected in most chemistry coursework.