Calculate The Ph Of A 0.186 M Nabro Solution

This premium calculator solves the pH of sodium hypobromite solutions by treating NaBrO as the salt of a strong base and a weak acid. Enter the concentration, select a Ka value for hypobromous acid, and the calculator will determine Kb, hydroxide concentration, pOH, and final pH using the equilibrium relationship for BrO^– hydrolysis.

NaBrO pH Calculator

Equilibrium Setup

Sodium hypobromite dissociates completely in water to give Na⁺ and BrO^–. The sodium ion is a spectator ion, while the hypobromite ion acts as a weak base:

Core Equations

NaBrO → Na⁺ + BrO^–

BrO^– + H₂O ⇌ HOBr + OH^–

K_b = K_w / K_a

K_b = x² / (C – x)

pOH = -log[OH^–], pH = 14 – pOH

How to calculate the pH of a 0.186 M NaBrO solution

To calculate the pH of a 0.186 M NaBrO solution, you need to recognize what kind of salt sodium hypobromite is. NaBrO comes from the strong base sodium hydroxide, NaOH, and the weak acid hypobromous acid, HOBr. Because the anion BrO^– is the conjugate base of a weak acid, it reacts with water to produce hydroxide ions. That means the solution is basic, so its pH must be greater than 7 at 25 degrees C.

The chemistry is straightforward once you classify the salt correctly. Many students initially look at sodium salts and assume they are neutral because sodium is a spectator ion. That part is true. However, neutrality depends on the behavior of the accompanying anion. In this case, BrO^– hydrolyzes in water. The pH therefore depends on the base dissociation constant of BrO^–, which is obtained from the acid dissociation constant of HOBr using the relation K_b = K_w / K_a.

For the default value Ka(HOBr) = 2.3 × 10^-9 at 25 degrees C, the calculated pH of a 0.186 M NaBrO solution is approximately 10.95.

Step 1: Write the dissolution and hydrolysis reactions

First, sodium hypobromite dissociates completely:

NaBrO → Na⁺ + BrO^–

Next, the hypobromite ion behaves as a weak base in water:

BrO^– + H₂O ⇌ HOBr + OH^–

This second equation is the one that controls the pH. Every small amount of BrO^– that reacts forms hydroxide, and that hydroxide raises the pH.

Step 2: Convert Ka of HOBr into Kb of BrO-

At 25 degrees C, water has K_w = 1.0 × 10^-14. If you use Ka(HOBr) = 2.3 × 10^-9, then:

K_b = (1.0 × 10^-14) / (2.3 × 10^-9) = 4.35 × 10^-6

This tells you BrO^– is a weak base, but strong enough to make the solution clearly basic at a concentration of 0.186 M.

Step 3: Set up the ICE table

For the hydrolysis equilibrium:

BrO^– + H₂O ⇌ HOBr + OH^–

• Initial [BrO^–] = 0.186 M
• Initial [HOBr] = 0
• Initial [OH^–] ≈ 0

Let x be the amount of BrO^– that reacts:

• Equilibrium [BrO^–] = 0.186 – x
• Equilibrium [HOBr] = x
• Equilibrium [OH^–] = x

Now substitute into the base equilibrium expression:

K_b = x² / (0.186 – x)

Step 4: Solve for hydroxide concentration

Because K_b is much smaller than the initial concentration, you can often use the weak-base approximation and assume x is much smaller than 0.186. Then:

x ≈ √(K_b × C) = √[(4.35 × 10^-6)(0.186)]

x ≈ 8.99 × 10^-4 M

So [OH^–] ≈ 8.99 × 10^-4 M. The exact quadratic solution gives essentially the same result, confirming that the approximation is valid here.

Step 5: Find pOH and pH

Using the hydroxide concentration:

1. pOH = -log(8.99 × 10^-4) ≈ 3.05
2. pH = 14.00 – 3.05 = 10.95

That is the expected pH for a 0.186 M NaBrO solution under the default assumptions.

Why NaBrO gives a basic solution

Understanding the acid-base identity of ions is more important than memorizing individual examples. Na⁺ is neutral in water because it comes from the strong base NaOH. BrO^–, however, is the conjugate base of HOBr, which is a weak acid. Conjugate bases of weak acids have measurable basicity. They pull a proton from water, producing OH^–. That simple pattern lets you classify many salt solutions quickly:

• Strong acid + strong base salt: usually neutral
• Strong acid + weak base salt: acidic
• Weak acid + strong base salt: basic
• Weak acid + weak base salt: depends on relative Ka and Kb

NaBrO clearly belongs in the weak acid + strong base category, so a pH above 7 is the correct expectation even before doing any math.

Exact answer for 0.186 M NaBrO

If you solve the equilibrium expression exactly using the quadratic formula, you start from:

K_b = x² / (0.186 – x)

which rearranges to:

x² + K_bx – K_b(0.186) = 0

With K_b = 4.35 × 10^-6, the positive root gives x ≈ 8.97 × 10^-4 M. Then:

• [OH^–] = 8.97 × 10^-4 M
• pOH ≈ 3.05
• pH ≈ 10.95

The difference between the exact and approximate methods is tiny because x is far below 5 percent of the starting concentration. In fact, the percent ionization is only about 0.48 percent, which makes the square-root approximation reliable.

Quantity	Value for 0.186 M NaBrO	Meaning
Ka of HOBr	2.3 × 10^-9	Acid strength of hypobromous acid
Kb of BrO^–	4.35 × 10^-6	Base strength of hypobromite ion
[OH^–]	8.97 × 10^-4 M	Hydroxide formed by hydrolysis
pOH	3.05	Measure of hydroxide level
pH	10.95	Final basicity of solution
Percent hydrolysis	0.48%	Fraction of BrO^– reacting with water

Approximation vs exact method

For weak acid and weak base equilibria, students often wonder whether the quick square-root formula is acceptable. In this specific NaBrO example, it is. Since K_b is small and the starting concentration is moderate, the amount reacting is tiny relative to 0.186 M. Here is how the two methods compare.

Method	[OH^–] (M)	pOH	pH	Difference from exact pH
Weak-base approximation	8.99 × 10^-4	3.046	10.954	About 0.001 pH units
Exact quadratic solution	8.97 × 10^-4	3.047	10.953	Reference value

In practical classroom work, either method would typically be accepted if the setup is correct and significant figures are reasonable. However, the exact quadratic method is mathematically safer because it always works, even when the approximation starts to break down.

Common mistakes when calculating the pH of sodium hypobromite

• Treating NaBrO as a neutral salt. Only salts from strong acid and strong base are reliably neutral. NaBrO comes from a weak acid and is basic.
• Using Ka directly instead of converting to Kb. The species in solution is BrO^–, so you need Kb for the base hydrolysis reaction.
• Forgetting that sodium is a spectator ion. Na⁺ does not control the pH.
• Using pH = -log concentration directly. You do not take the negative log of 0.186 to get the pH because 0.186 is not [H⁺] or [OH^–].
• Ignoring significant figures. Input values such as 0.186 M imply about three significant figures.

How concentration affects the pH of NaBrO

As the concentration of NaBrO increases, the hydroxide concentration generated by hydrolysis also increases, so the pH rises. However, because the equilibrium is governed by a square-root relationship in the weak-base approximation, the pH does not increase linearly with concentration. Doubling concentration does not double pH. Instead, it changes [OH^–] by a factor related to the square root of the concentration change.

This is an important conceptual point. pH is a logarithmic scale, and weak electrolyte equilibria often involve square-root dependence. That combination makes pH behavior more subtle than it appears at first glance. A graph of concentration versus pH is therefore a useful learning tool, which is why the calculator includes a chart.

Does molality vs molarity matter here?

The problem statement sometimes appears as “0.186 m NaBrO,” where lowercase m indicates molality. Strictly speaking, molality is moles of solute per kilogram of solvent, while molarity is moles per liter of solution. In rigorous physical chemistry, these are not identical. However, for many aqueous homework problems at moderate concentration, they are often treated as numerically close enough for a standard equilibrium calculation unless density information is provided.

This calculator makes that assumption when molality is selected. If you needed a more precise answer from an experimental context, you would convert molality to molarity using solution density and molar mass data before solving the equilibrium expression.

Reference chemistry and authoritative sources

For deeper review of aqueous acid-base chemistry, equilibrium constants, and pH scales, the following authoritative resources are useful:

• LibreTexts Chemistry for worked acid-base equilibrium explanations
• U.S. Environmental Protection Agency (.gov): pH overview and significance
• Massachusetts Institute of Technology Chemistry (.edu) for university-level chemistry references
• U.S. Geological Survey (.gov): pH and water science

Final answer

Using Ka(HOBr) = 2.3 × 10^-9 and K_w = 1.0 × 10^-14 at 25 degrees C:

1. K_b(BrO^–) = 4.35 × 10^-6
2. [OH^–] ≈ 8.97 × 10^-4 M
3. pOH ≈ 3.05
4. pH ≈ 10.95

So, the pH of a 0.186 M NaBrO solution is approximately 10.95 under standard 25 degrees C conditions and the commonly used Ka value for hypobromous acid.

Educational note: reported literature values for Ka of HOBr can vary slightly by source and temperature, so small differences in the final pH are normal.