Calculate The Ph Of A 1.60 M Kbro Solution

This premium calculator estimates the pH of a potassium hypobromite solution by treating BrO^– as the conjugate base of hypobromous acid, HOBr. Enter or confirm the default values below to compute pH, pOH, K_b, hydroxide concentration, and percent hydrolysis using the exact equilibrium expression.

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How to Calculate the pH of a 1.60 m KBrO Solution

To calculate the pH of a 1.60 m KBrO solution, you need to recognize what potassium hypobromite does in water. KBrO is a soluble ionic compound made from K⁺ and BrO^–. The potassium ion is essentially pH-neutral because it comes from the strong base KOH. The hypobromite ion, however, is the conjugate base of hypobromous acid, HOBr, which is a weak acid. That means BrO^– reacts with water to generate OH^–, making the solution basic.

The central chemistry is a weak-base hydrolysis equilibrium:

BrO- + H2O ⇌ HOBr + OH-

Since this reaction produces hydroxide ions, the pOH goes down and the pH rises above 7. The exact value depends mainly on two things: the concentration of KBrO and the acid dissociation constant of HOBr, K_a. Once K_a is known, you can determine the base dissociation constant of BrO^– using:

Kb = Kw / Ka

At 25 degrees Celsius, K_w is typically taken as 1.0 × 10^-14. A common textbook value for hypobromous acid is K_a = 2.0 × 10^-9. Using that value:

Kb = (1.0 × 10^-14) / (2.0 × 10^-9) = 5.0 × 10^-6

Step-by-Step Setup

Start with a formal concentration of 1.60 for BrO^–. Because KBrO dissociates essentially completely, the initial concentration of BrO^– equals the concentration of the salt. The equilibrium table is:

• Initial: [BrO^–] = 1.60, [HOBr] = 0, [OH^–] = 0
• Change: [BrO^–] = -x, [HOBr] = +x, [OH^–] = +x
• Equilibrium: [BrO^–] = 1.60 – x, [HOBr] = x, [OH^–] = x

Substitute those into the equilibrium expression for a weak base:

Kb = [HOBr][OH-] / [BrO-] = x^2 / (1.60 – x)

With K_b = 5.0 × 10^-6, the equation becomes:

5.0 × 10^-6 = x^2 / (1.60 – x)

Because K_b is small relative to the concentration, many students use the common approximation 1.60 – x ≈ 1.60:

x ≈ √(KbC) = √[(5.0 × 10^-6)(1.60)] ≈ 2.83 × 10^-3

That x value is the hydroxide concentration. Next:

1. pOH = -log(2.83 × 10^-3) ≈ 2.55
2. pH = 14.00 – 2.55 ≈ 11.45

So the pH of a 1.60 m KBrO solution is approximately 11.45 when K_a(HOBr) = 2.0 × 10^-9 at 25 degrees Celsius. If you solve the quadratic exactly, the answer is essentially the same to normal reporting precision, because x is tiny compared with 1.60.

Why the Solution Is Basic

This is one of the most important acid-base patterns in general chemistry: salts from a strong base and a weak acid form basic solutions. KBrO fits that pattern exactly. Potassium comes from KOH, a strong base, so K⁺ does not hydrolyze in any meaningful way. The hypobromite ion, though, comes from HOBr, a weak acid, so it acts as a weak base in water. That is why the solution is not neutral even though it is a salt.

If you are trying to solve similar problems quickly, classify the salt first:

• 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 K_a and K_b

KBrO belongs in the third category. Once that is clear, the rest is a standard weak-base equilibrium problem.

Exact vs Approximate Calculation

For high-quality chemistry work, it is helpful to compare the exact and approximate approaches. The exact quadratic comes from rearranging:

Kb = x^2 / (C – x) → x^2 + Kb x – Kb C = 0

Solving gives:

x = [-Kb + √(Kb^2 + 4KbC)] / 2

For a 1.60 concentration and K_b = 5.0 × 10^-6, the exact hydroxide concentration is very close to the approximation. This is because the percent hydrolysis is small, roughly:

% hydrolysis = (x / 1.60) × 100 ≈ 0.18%

Since the change is well below 5%, the approximation is justified. However, in a calculator and in more advanced chemistry, solving exactly is usually better because it avoids approximation error and builds good habits.

Assumed Ka for HOBr	Derived Kb for BrO^–	Calculated [OH^–]	pOH	pH at 1.60 concentration
2.0 × 10^-9	5.0 × 10^-6	2.826 × 10^-3	2.549	11.451
2.3 × 10^-9	4.348 × 10^-6	2.637 × 10^-3	2.579	11.421
2.5 × 10^-9	4.0 × 10^-6	2.528 × 10^-3	2.597	11.403

This table shows why published K_a values matter. Different textbooks and reference compilations sometimes round or report slightly different numbers for hypobromous acid. Even so, the final pH still falls near 11.4 for a 1.60 concentration solution, which is the chemically important takeaway.

Molality vs Molarity in This Problem

The problem statement uses 1.60 m, which formally means molality, or moles of solute per kilogram of solvent. Strictly speaking, pH calculations are most naturally written in terms of activities, and introductory problems often approximate those activities using molarity. In many classroom exercises, instructors use the given concentration directly in the equilibrium expression unless density or activity data are provided. That is exactly what this calculator does.

For a highly concentrated ionic solution, the true activity-based pH can differ somewhat from the idealized textbook result. But in standard general chemistry problem solving, the accepted path is:

1. Treat the KBrO concentration as the formal concentration of BrO^–.
2. Convert K_a to K_b using K_w.
3. Solve the weak-base equilibrium.
4. Find pOH, then pH.

That method gives the familiar answer of roughly 11.45 under common assumptions.

Common Mistakes to Avoid

• Assuming the solution is neutral because KBrO is a salt. Not all salts are neutral. The anion matters.
• Using K_a directly in the hydrolysis expression. You need K_b for BrO^–, so convert with K_b = K_w/K_a.
• Forgetting that x equals [OH^–]. In this setup, the hydrolysis creates equal amounts of HOBr and OH^–.
• Stopping at pOH. Many students calculate pOH correctly but forget to convert to pH.
• Confusing BrO^– with BrO₃^–. Hypobromite and bromate are different species with different chemistry.

Comparison with Other Weak-Base Salt Concentrations

It is helpful to see how concentration changes the pH while keeping the same K_a assumption for HOBr. The numbers below use K_a = 2.0 × 10^-9 and exact equilibrium calculations at 25 degrees Celsius.

KBrO Concentration	Kb of BrO^–	Exact [OH^–]	pOH	pH
0.010	5.0 × 10^-6	2.211 × 10^-4	3.655	10.345
0.100	5.0 × 10^-6	7.046 × 10^-4	3.152	10.848
1.00	5.0 × 10^-6	2.234 × 10^-3	2.651	11.349
1.60	5.0 × 10^-6	2.826 × 10^-3	2.549	11.451

These values illustrate a classic weak-base trend: increasing concentration increases hydroxide production and raises pH, but not in a simple linear way. Because the square-root dependence dominates the approximation, the pH change becomes more gradual at higher concentration.

Interpreting the Answer Chemically

A calculated pH near 11.45 means the solution is distinctly basic. However, it is not as basic as an equally concentrated strong base such as KOH. If 1.60 concentration KOH were present, the hydroxide concentration would be close to 1.60, giving a much higher pH. KBrO only produces hydroxide through partial hydrolysis, not complete dissociation into OH^–. That is the defining difference between a weak base and a strong base.

In other words, KBrO makes solution basic because BrO^– steals a proton from water, but only a small fraction of the ions do so at equilibrium.

Authoritative Reference Reading

For deeper background on pH, aqueous equilibria, and acid-base chemistry, consult these authoritative resources:

• USGS: pH and Water
• U.S. EPA: pH Overview
• University of Wisconsin: Weak Bases and Hydrolysis

Final Answer

Under standard textbook assumptions at 25 degrees Celsius, using K_a(HOBr) = 2.0 × 10^-9 and K_w = 1.0 × 10^-14, the pH of a 1.60 m KBrO solution is:

pH ≈ 11.45

If your textbook uses a slightly different value for the acid dissociation constant of HOBr, you may see a result very close to 11.40 to 11.45. The calculator above lets you test those assumptions instantly and see how the equilibrium species change.