Calculate The Ph Of A 2 M Solution Of Nh4Br

Use this premium calculator to estimate the acidity of ammonium bromide solution at 25 C. NH4Br is an acidic salt because NH4+ acts as a weak acid in water, while Br- is essentially neutral.

Quick chemistry rule: NH4Br comes from weak base NH3 and strong acid HBr, so its aqueous solution is acidic.

How to calculate the pH of a 2 m solution of NH4Br

If you need to calculate the pH of a 2 m solution of NH4Br, the key idea is that ammonium bromide is a salt made from a weak base and a strong acid. The weak base is ammonia, NH3, and the strong acid is hydrobromic acid, HBr. In water, bromide ion does not appreciably hydrolyze, but ammonium ion does. That means the pH is controlled by the acid behavior of NH4+.

Students often look at NH4Br and wonder whether it should be neutral because it is a salt. It is not neutral. Not all salts give pH 7. A salt formed from a strong acid and a weak base produces an acidic solution, and NH4Br is a classic example. The chemistry is driven by the equilibrium:

NH4+ + H2O ⇌ NH3 + H3O+
Ka = [NH3][H3O+] / [NH4+]

To find the acid dissociation constant of ammonium, you start from the base dissociation constant of ammonia. At 25 C, a common textbook value is Kb = 1.8 x 10^-5 for NH3. Since Ka x Kb = Kw and Kw = 1.0 x 10^-14 at 25 C, the conjugate acid constant is:

Ka(NH4+) = Kw / Kb(NH3)
Ka = (1.0 x 10^-14) / (1.8 x 10^-5)
Ka ≈ 5.56 x 10^-10

Once you know Ka, the pH calculation follows the same pattern used for any weak acid. If a problem states a 2 m NH4Br solution, the strict meaning is 2 mol of NH4Br per kilogram of solvent. However, many classroom exercises treat 2 m as approximately 2 M when density data are not supplied. A more refined approach converts molality to molarity using solution density and molar mass. This calculator lets you do either.

Short answer: if you approximate 2 m NH4Br as 2.0 M at 25 C, the pH comes out to about 4.48. If you convert 2 m to molarity using an assumed density of 1.00 g/mL, the concentration becomes about 1.668 M and the pH is about 4.52. The exact value depends on density and, at very high ionic strength, on activity effects.

Step by step method

1. Write the ions produced by dissolution: NH4Br → NH4+ + Br-.
2. Recognize that Br- is the conjugate base of a strong acid and is effectively neutral.
3. Use NH4+ as the weak acid responsible for hydronium production.
4. Calculate Ka for NH4+ from the known Kb of NH3.
5. Set up an ICE table using the initial concentration of NH4+.
6. Solve for x = [H3O+].
7. Compute pH = -log10[H3O+].

Exact setup for a 2.0 M approximation

If you treat the 2 m solution as approximately 2.0 M, then the equilibrium table is straightforward:

Initial: [NH4+] = 2.0, [NH3] = 0, [H3O+] = 0
Change: -x, +x, +x
Equilibrium: [NH4+] = 2.0 – x, [NH3] = x, [H3O+] = x

Substitute these values into the equilibrium expression:

Ka = x^2 / (2.0 – x)
5.56 x 10^-10 = x^2 / (2.0 – x)

Because Ka is small, x is much smaller than 2.0, so the common approximation is:

x ≈ √(KaC) = √[(5.56 x 10^-10)(2.0)]
x ≈ 3.33 x 10^-5 M

Now convert hydrogen ion concentration to pH:

pH = -log10(3.33 x 10^-5) ≈ 4.48

This is the answer most instructors expect when the problem gives only concentration and no density. The solution is acidic, but not strongly acidic, because NH4+ is a weak acid.

What changes when the problem says 2 m instead of 2 M?

The distinction matters if you want a more rigorous answer. Molality is based on the mass of solvent, while molarity depends on solution volume. To convert molality to molarity, you need density and molar mass. For NH4Br, the molar mass is about 97.94 g/mol. If the solution density is assumed to be 1.00 g/mL, then 2.0 m converts to:

M = (1000 x density x molality) / (1000 + molality x molar mass)
M = (1000 x 1.00 x 2.0) / (1000 + 2.0 x 97.94)
M = 2000 / 1195.88 ≈ 1.668 M

Using 1.668 M rather than 2.0 M gives a slightly lower hydronium concentration and therefore a slightly higher pH. That is why the calculator above reports about 4.52 using this particular density assumption. In a real concentrated electrolyte solution, activity corrections can also shift the effective pH somewhat from the simple ideal-solution estimate.

Why NH4Br is acidic and NaBr is not

A useful comparison is NH4Br versus NaBr. Both are bromide salts, but sodium ion is the conjugate acid of a strong base, NaOH, so Na+ does not acidify water. Ammonium ion is the conjugate acid of weak base NH3, so it does. That difference completely changes the pH behavior.

• NH4Br: acidic because NH4+ donates protons weakly to water.
• NaBr: essentially neutral because both Na+ and Br- come from strong species.
• NH4Cl: also acidic for the same reason as NH4Br, since Cl- is likewise neutral.

Key constants and properties at 25 C

Property	Value	Why it matters
Molar mass of NH4Br	97.94 g/mol	Needed for converting molality to molarity.
Kb of NH3	1.8 x 10^-5	Starting point for finding Ka of NH4+.
pKb of NH3	4.74	Log form of base strength.
Ka of NH4+	5.56 x 10^-10	Controls hydronium formation in NH4Br solution.
pKa of NH4+	9.25	Shows NH4+ is a weak acid.
Kw of water	1.0 x 10^-14	Links Ka and Kb at 25 C.

Comparison table: estimated pH of NH4Br at different concentrations

The following values use the standard weak acid approximation with Ka = 5.56 x 10^-10 and assume concentration in molarity for comparison. These estimates are very close to the exact quadratic results because NH4+ is weak and x remains much smaller than the formal concentration.

NH4Br concentration	Estimated [H3O+]	Estimated pH	Interpretation
0.01 M	2.36 x 10^-6 M	5.63	Mildly acidic
0.10 M	7.46 x 10^-6 M	5.13	More acidic as concentration rises
1.00 M	2.36 x 10^-5 M	4.63	Clearly acidic
2.00 M	3.33 x 10^-5 M	4.48	Typical textbook answer for 2 m approximated as 2 M

Common mistakes when solving this problem

• Assuming all salts are neutral. Salts can be acidic, basic, or neutral depending on the parent acid and base.
• Using Kb directly for NH4+. NH4+ is an acid, so you need Ka, not Kb.
• Forgetting to convert from molality to molarity. If the problem specifically emphasizes 2 m and gives density, do the conversion.
• Treating Br- as basic. Bromide is the conjugate base of a strong acid and contributes negligibly to pH.
• Ignoring activity in concentrated solutions. At high ionic strength, ideal formulas are still useful estimates, but not perfect experimental predictions.

Exam shortcut for NH4Br pH problems

In many introductory chemistry settings, your instructor is looking for a fast but chemically justified answer. Here is the shortcut:

1. Identify NH4+ as a weak acid.
2. Find Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10.
3. Use [H+] ≈ √(KaC).
4. With C = 2.0, get [H+] ≈ 3.33 x 10^-5.
5. Report pH ≈ 4.48.

That approach is fast, defendable, and usually sufficient. If the problem is from analytical chemistry, physical chemistry, or chemical engineering, your instructor may expect you to discuss concentration units, density, and activity coefficients as well.

How the calculator on this page helps

The calculator above handles both the classic textbook approximation and a more careful molality conversion. You can enter the concentration as molality or molarity, choose an exact quadratic or approximate weak acid method, and adjust the solution density if your source provides it. The chart gives a visual summary of how much NH4+ is present initially versus how little actually dissociates to generate hydronium. That visual is useful because it reinforces an important weak acid idea: even though the solution is acidic, only a tiny fraction of NH4+ ionizes.

Reliable chemistry references

For broader background on pH and water chemistry, see the official resources from USGS and EPA. For compound specific reference information on ammonium bromide, see PubChem.

Final answer

To calculate the pH of a 2 m solution of NH4Br, treat NH4+ as a weak acid with Ka ≈ 5.56 x 10^-10. If you approximate the solution concentration as 2.0 M, the hydrogen ion concentration is about 3.33 x 10^-5 M and the pH is approximately 4.48. If you strictly convert 2 m to molarity using an assumed density of 1.00 g/mL, the solution is about 1.668 M and the pH is about 4.52. Either way, the important chemical conclusion is the same: NH4Br gives an acidic aqueous solution.