Calculate The Ph Of 0.10 M Ammonium Bromide Nh4Br Solution

Use this premium chemistry calculator to determine the pH of an ammonium bromide solution from first principles. The tool models NH4Br as a salt of a weak base and a strong acid, calculates the acid dissociation of NH4+, and visualizes how pH changes with concentration.

NH4Br pH Calculator

How to calculate the pH of 0.10 M ammonium bromide NH4Br solution

If you need to calculate the pH of 0.10 M ammonium bromide NH4Br solution, the key idea is that ammonium bromide is not a strong acid by itself, but it does produce an acidic solution in water. This happens because NH4Br is made from the weak base ammonia, NH3, and the strong acid hydrobromic acid, HBr. When NH4Br dissolves, it dissociates essentially completely into NH4+ and Br-. The bromide ion is the conjugate base of a strong acid, so Br- does not significantly react with water. The ammonium ion, however, is the conjugate acid of the weak base ammonia, so NH4+ donates protons weakly to water and generates hydronium.

That means the pH is controlled by the acid hydrolysis of NH4+, not by bromide. In practical classroom chemistry, this is a standard weak acid equilibrium problem. The salt concentration gives the initial ammonium ion concentration, then you use the acid dissociation constant of NH4+ to solve for the hydrogen ion concentration and finally convert that to pH. For a 0.10 M ammonium bromide solution at 25 C using a typical ammonia base constant of 1.8 × 10^-5, the pH comes out to about 5.13.

Step 1: Write the dissociation and hydrolysis reactions

The first reaction is the complete dissolution of the salt:

NH4Br(aq) → NH4+(aq) + Br-(aq)

After dissociation, the ammonium ion reacts with water as a weak acid:

NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

Because Br- is the conjugate base of HBr, a strong acid, it does not measurably affect pH under standard general chemistry conditions. Therefore, all pH work centers on the NH4+ equilibrium.

Step 2: Convert Kb of ammonia into Ka for ammonium

Textbooks usually give the base dissociation constant for ammonia rather than the acid dissociation constant for ammonium. The two are related by the water ion-product expression:

Ka(NH4+) = Kw / Kb(NH3)

At 25 C, if Kb for ammonia is 1.8 × 10^-5 and Kw is 1.0 × 10^-14, then:

Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10

This is a very small Ka, which tells you ammonium is a weak acid. Even so, in a 0.10 M solution there is enough NH4+ present to make the pH noticeably acidic.

Step 3: Set up an ICE table

For the hydrolysis equilibrium

NH4+ + H2O ⇌ NH3 + H3O+

start with an initial ammonium concentration of 0.10 M. Before hydrolysis, the concentrations are approximately:

• [NH4+] = 0.10 M
• [NH3] = 0
• [H3O+] = 0 from the salt itself, if you ignore the tiny contribution from pure water

Let x be the amount of NH4+ that reacts:

• [NH4+] at equilibrium = 0.10 – x
• [NH3] at equilibrium = x
• [H3O+] at equilibrium = x

Now apply the equilibrium expression:

Ka = [NH3][H3O+] / [NH4+] = x² / (0.10 – x)

Step 4: Solve for x and pH

Because Ka is small, many instructors use the weak acid approximation x << 0.10. Then the denominator becomes approximately 0.10, so:

x² / 0.10 = 5.56 × 10^-10

x² = 5.56 × 10^-11

x = 7.46 × 10^-6 M

Since x = [H3O+], the pH is:

pH = -log(7.46 × 10^-6) = 5.13

The exact quadratic method gives essentially the same result for this concentration because the approximation is excellent. In fact, x is much smaller than 5 percent of 0.10 M, so the assumption is chemically justified.

Why NH4Br is acidic but not strongly acidic

A common source of confusion is that NH4Br contains bromide, which comes from a strong acid, and students sometimes think that means the whole salt must behave like a strong acid. That is not correct. Once the salt is dissolved, you evaluate the acid-base behavior of each ion independently. Br- is a spectator with respect to hydrolysis because it is too weak a base to react significantly with water. NH4+ is the species that matters. It is a weak acid, so the resulting solution is acidic, but only mildly so.

This is why a 0.10 M NH4Br solution has a pH close to 5 instead of something very low like 1 or 2. In chemistry terms, the cation is hydrolyzing, but only to a small extent. That small extent still produces enough hydronium to lower the pH below neutral. When you compare NH4Br with salts like NaBr or KBr, the difference is dramatic. Sodium and potassium ions do not hydrolyze appreciably, so NaBr and KBr solutions are nearly neutral, whereas NH4Br is mildly acidic.

Quick result for the target problem

• Salt: ammonium bromide, NH4Br
• Concentration: 0.10 M
• Kb of NH3: 1.8 × 10^-5
• Calculated Ka of NH4+: 5.56 × 10^-10
• [H3O+]: 7.46 × 10^-6 M
• Final pH: approximately 5.13

Comparison table: pH of ammonium bromide at different concentrations

The following values use Ka = 5.56 × 10^-10 for NH4+ at 25 C and the weak acid model. These are useful benchmark numbers when you want to estimate how concentration changes acidity.

NH4Br concentration (M)	Estimated [H3O+] (M)	Estimated pH	Acidity description
0.001	7.45 × 10^-7	6.13	Slightly acidic
0.010	2.36 × 10^-6	5.63	Mildly acidic
0.050	5.27 × 10^-6	5.28	Mildly acidic
0.10	7.46 × 10^-6	5.13	Mildly acidic
0.50	1.67 × 10^-5	4.78	More acidic
1.00	2.36 × 10^-5	4.63	Moderately acidic for a salt solution

Comparison table: NH4Br versus related salts

This table helps show what controls pH. The identity of the cation is more important here than the identity of the anion, as long as the anion comes from a strong acid.

Salt at 0.10 M	Acid-base source	Main hydrolyzing ion	Typical pH behavior
NH4Br	Weak base + strong acid	NH4+	Acidic, about pH 5.13
NH4Cl	Weak base + strong acid	NH4+	Acidic, nearly same as NH4Br
NaBr	Strong base + strong acid	None significant	Approximately neutral, near pH 7
NH4CH3COO	Weak base + weak acid	Both ions can matter	Depends on Ka and Kb balance

When to use the exact quadratic method

For 0.10 M NH4Br, the weak acid approximation is entirely acceptable. Still, the exact quadratic solution is the most robust way to solve equilibrium problems, especially when concentrations are very low or when a problem asks for formal rigor. Starting from

Ka = x² / (C – x)

you rearrange to

x² + Ka x – Ka C = 0

Then solve with the quadratic formula:

x = [-Ka + √(Ka² + 4KaC)] / 2

Here, x is the hydronium concentration. For NH4Br at 0.10 M, this exact expression gives virtually the same value as the square-root shortcut. The calculator above can show either method, but by default it uses the exact form because it avoids approximation errors automatically.

Common mistakes to avoid

1. Assuming NH4Br is neutral just because it is a salt. Not all salts are neutral.
2. Using HBr to set the pH directly. HBr is not present as free strong acid after the salt is formed and dissolved.
3. Forgetting to convert Kb of NH3 into Ka of NH4+.
4. Treating Br- as basic. It is the conjugate base of a strong acid and is effectively neutral in this context.
5. Rounding too early. Keep enough significant digits until the final pH value.

What the chemistry means in plain language

Imagine dissolving ammonium bromide in water. The crystal lattice breaks apart into charged particles. The bromide ion mostly floats in solution without interacting with water in an acid-base way. The ammonium ion is different. It can donate a proton to water, forming a little ammonia and hydronium. That hydronium makes the solution acidic. Because the ammonium ion is only a weak acid, it does not donate many protons, which is why the pH does not drop dramatically. This is the same fundamental logic used across weak acid, weak base, and salt hydrolysis problems in introductory chemistry.

How concentration affects the answer

The pH of ammonium bromide depends on the concentration because the amount of NH4+ available for hydrolysis changes. If the solution is more dilute, the hydronium concentration generated by the weak acid equilibrium is lower, and the pH moves closer to neutral. If the solution is more concentrated, more hydronium is produced, so the pH decreases. The trend is not linear because pH is logarithmic and because weak acid equilibria depend on the square root of concentration in the approximation regime.

For example, increasing NH4Br from 0.010 M to 0.10 M lowers the pH from roughly 5.63 to 5.13. Increasing it further to 1.00 M lowers the pH to around 4.63. Those are meaningful shifts, but they still keep the system in the weakly acidic range rather than the strongly acidic range.

Useful authoritative references

For supporting acid-base concepts, equilibrium constants, and chemical data, these references are credible starting points:

• LibreTexts Chemistry educational materials
• National Institute of Standards and Technology (NIST.gov)
• United States Environmental Protection Agency (EPA.gov)
• University of California Berkeley Chemistry (.edu)

Final takeaway

To calculate the pH of 0.10 M ammonium bromide NH4Br solution, treat NH4Br as a fully dissociated salt whose acidity comes from NH4+. Convert the ammonia base constant into the acid constant of NH4+, set up the weak acid equilibrium, solve for hydronium, and convert to pH. Using Kb = 1.8 × 10^-5 for NH3 and Kw = 1.0 × 10^-14 at 25 C, the resulting pH is approximately 5.13. That makes ammonium bromide a mildly acidic salt solution, which is exactly what you should expect from a salt formed from a weak base and a strong acid.

Educational note: real laboratory pH can differ slightly from the ideal calculation because concentrated solutions can show activity effects, temperature shifts, and instrument calibration differences. For general chemistry coursework, the equilibrium treatment shown here is the standard accepted method.