Calculate The Ph Of A 0.20 M Solution Of Nh4Br

Use this interactive chemistry calculator to find the pH of ammonium bromide solutions by applying weak acid hydrolysis, equilibrium relationships, and either an exact quadratic method or the common square root approximation.

NH4Br pH Calculator

How to Calculate the pH of a 0.20 m Solution of NH4Br

To calculate the pH of a 0.20 m solution of NH4Br, you need to recognize the acid base behavior of the ions produced when ammonium bromide dissolves in water. NH4Br is a salt made from the weak base ammonia, NH3, and the strong acid hydrobromic acid, HBr. Because HBr is a strong acid, Br- is essentially a spectator ion and does not significantly affect pH. The important species is NH4+, the conjugate acid of NH3. This means the solution is acidic, not neutral.

In a typical general chemistry or analytical chemistry setting, the phrase “0.20 m” is often used in a way that allows a dilute solution assumption, so the molality is treated approximately like molarity for equilibrium calculations. That gives an initial ammonium concentration of about 0.20. Once NH4+ is in water, it undergoes hydrolysis according to the equilibrium:

NH4+ + H2O ⇌ NH3 + H3O+

The equilibrium constant for this process is the acid dissociation constant of NH4+, written as Ka. Since NH4+ is the conjugate acid of NH3, you can find Ka from the relationship:

Ka = Kw / Kb

At 25°C, a commonly used value for the base dissociation constant of ammonia is 1.8 × 10^-5. The ionic product of water is 1.0 × 10^-14. Therefore:

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

Now let the hydronium ion concentration produced by hydrolysis be x. The equilibrium table becomes:

• Initial: [NH4+] = 0.20, [NH3] = 0, [H3O+] = 0
• Change: [NH4+] = -x, [NH3] = +x, [H3O+] = +x
• Equilibrium: [NH4+] = 0.20 – x, [NH3] = x, [H3O+] = x

Substitute into the Ka expression:

Ka = x² / (0.20 – x)

Because Ka is very small, x is much smaller than 0.20. The standard approximation is:

x ≈ √(Ka × C) = √[(5.56 × 10^-10)(0.20)] = 1.05 × 10^-5

Then pH is:

pH = -log10(1.05 × 10^-5) ≈ 4.98

So the pH of a 0.20 m solution of NH4Br is approximately 4.98 at 25°C, assuming the usual dilute solution treatment and a Kb for ammonia of 1.8 × 10^-5.

Why NH4Br Produces an Acidic Solution

Many students first try to classify salts only by whether they are ionic, but the more useful approach is to identify the parent acid and parent base. NH4Br comes from:

• NH3, a weak base
• HBr, a strong acid

When a salt is formed from a strong acid and a weak base, the cation is usually acidic. In this case, NH4+ can donate a proton to water, generating H3O+. Bromide remains effectively neutral because it is the conjugate base of a strong acid and has negligible basicity in water. This is the central reason the pH falls below 7.

Exact Method vs Approximation

For most textbook problems, the square root approximation is accepted because the percent ionization is tiny. However, a more exact solution uses the quadratic equation. Starting from:

Ka = x² / (C – x)

Rearrange:

x² + Kax – KaC = 0

The physically meaningful root is:

x = [-Ka + √(Ka² + 4KaC)] / 2

Using C = 0.20 and Ka = 5.56 × 10^-10 gives essentially the same value, x ≈ 1.05 × 10^-5, and therefore pH ≈ 4.98. The tiny difference confirms that the approximation is valid.

Step by Step Summary

1. Write the dissociation of NH4Br into NH4+ and Br-.
2. Identify Br- as a spectator ion because it is the conjugate base of strong acid HBr.
3. Focus on the hydrolysis of NH4+ as a weak acid.
4. Calculate Ka of NH4+ from Kw/Kb.
5. Set up an ICE table using the initial concentration of 0.20.
6. Solve for x using either the exact quadratic formula or the approximation x ≈ √(KaC).
7. Compute pH from pH = -log10[H3O+].

Worked Numerical Example for 0.20 m NH4Br

Let us carry out the entire process cleanly with numbers:

1. Given concentration: C = 0.20
2. Given Kb for NH3: 1.8 × 10^-5
3. Given Kw at 25°C: 1.0 × 10^-14
4. Find Ka: 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10
5. Approximate [H3O+]: √[(5.56 × 10^-10)(0.20)] = 1.05 × 10^-5
6. Find pH: -log10(1.05 × 10^-5) = 4.98

The solution is moderately acidic, but not strongly acidic. It is much less acidic than a strong acid solution at the same concentration because only a small fraction of NH4+ donates a proton.

Comparison Table: Salt Type and Expected pH Behavior

Salt Example	Parent Acid	Parent Base	Dominant Hydrolysis	Typical pH Trend
NH4Br	Strong acid HBr	Weak base NH3	NH4+ acts as weak acid	Acidic, below 7
NaCl	Strong acid HCl	Strong base NaOH	Negligible	Near neutral, about 7
CH3COONa	Weak acid CH3COOH	Strong base NaOH	CH3COO- acts as weak base	Basic, above 7
NH4CH3COO	Weak acid CH3COOH	Weak base NH3	Depends on Ka vs Kb	Variable

Comparison Table: pH of NH4Br at Different Concentrations

The following estimates use Kb for NH3 = 1.8 × 10^-5 and Kw = 1.0 × 10^-14 at 25°C. These values are calculated using the standard weak acid approximation and illustrate how concentration affects pH.

NH4Br Concentration	Ka of NH4+	Estimated [H3O+]	Estimated pH	Percent Ionization
0.010	5.56 × 10^-10	2.36 × 10^-6	5.63	0.0236%
0.050	5.56 × 10^-10	5.27 × 10^-6	5.28	0.0105%
0.20	5.56 × 10^-10	1.05 × 10^-5	4.98	0.0053%
0.50	5.56 × 10^-10	1.67 × 10^-5	4.78	0.0033%
1.00	5.56 × 10^-10	2.36 × 10^-5	4.63	0.0024%

Common Mistakes When Solving NH4Br pH Problems

• Treating NH4Br as neutral. It is not neutral because NH4+ is acidic.
• Using Kb directly instead of converting to Ka. The reacting species is NH4+, not NH3.
• Forgetting that Br- is a spectator ion. Bromide does not appreciably hydrolyze in water.
• Mixing molality and molarity without stating an assumption. In many classroom problems, a dilute approximation is used, but in rigorous physical chemistry that distinction matters.
• Ignoring temperature. Kb and Kw both vary with temperature, so pH changes if temperature changes.

How Reliable Is the Approximation?

The five percent rule is a standard check in acid base equilibrium work. If x is less than 5% of the initial concentration, replacing 0.20 – x with 0.20 is acceptable. Here, x is about 1.05 × 10^-5, which is only about 0.0053% of 0.20. That is far below 5%, so the approximation is extremely safe.

In other words, both the exact and approximate methods lead to almost the same pH. For educational purposes, this is an excellent example of when the square root shortcut is not only convenient but also highly accurate.

Interpretation of the Final pH

A pH near 4.98 means the solution is mildly acidic. It is nowhere near as acidic as a 0.20 M strong acid solution, which would have pH around 0.70. The difference arises because NH4+ is only a weak acid. The generated hydronium concentration, around 1.05 × 10^-5 M, is many orders of magnitude below the formal salt concentration. That tells you most ammonium ions remain un-ionized with respect to acid donation.

Authoritative Chemistry References

For deeper review of acid base equilibria, ionic product of water, and ammonium chemistry, consult these high quality educational sources:

• LibreTexts Chemistry educational library
• U.S. EPA overview of pH and aqueous chemistry
• University of California, Berkeley Chemistry resources

Final Answer

If you are asked to calculate the pH of a 0.20 m solution of NH4Br under standard 25°C textbook conditions, the accepted result is:

pH ≈ 4.98

This result comes from recognizing NH4+ as a weak acid, converting Kb of NH3 to Ka of NH4+, and solving the acid hydrolysis equilibrium. Use the calculator above if you want to test different concentrations, Kb values, or compare the exact and approximate methods.