Calculate the pH of a 1.6 m KBrO Solution
Use this premium chemistry calculator to estimate the pH, pOH, hydroxide concentration, and base hydrolysis behavior of potassium hypobromite, KBrO. The default setup is preloaded for a 1.6 m solution at 25 degrees Celsius, with options to use a dilute approximation or convert molality to molarity using density.
KBrO pH Calculator
Calculated Results
Select your assumptions and click Calculate pH to update the equilibrium results.
How to calculate the pH of a 1.6 m KBrO solution
To calculate the pH of a 1.6 m KBrO solution, you first identify what kind of salt KBrO is and how it behaves in water. Potassium hypobromite, KBrO, dissociates essentially completely into K+ and BrO–. The potassium ion is a spectator ion because it comes from the strong base KOH and does not significantly affect pH. The important species is BrO–, which is the conjugate base of hypobromous acid, HOBr. Since HOBr is a weak acid, BrO– acts as a weak base in water and generates OH– through hydrolysis. That is why a KBrO solution is basic.
The central equilibrium is:
BrO– + H2O ⇌ HOBr + OH–
Once that reaction is recognized, the rest of the calculation becomes a standard weak-base equilibrium problem. If the problem gives only the molality, 1.6 m, and no density, many general chemistry problems use the simplifying assumption that 1.6 m is close enough to 1.6 M for pH estimation. Strictly speaking, molality and molarity are not identical, but for educational work this approximation is often accepted unless you are asked to account for density or nonideal activity effects.
Step 1: Determine the relevant equilibrium constant
For the base BrO–, you usually are not given Kb directly. Instead, you may be given the acid dissociation constant for HOBr, Ka. At 25 degrees Celsius, a commonly used textbook value is:
- Ka(HOBr) ≈ 2.0 × 10-9
- Kw = 1.0 × 10-14
Then:
Kb = Kw / Ka = (1.0 × 10-14) / (2.0 × 10-9) = 5.0 × 10-6
Step 2: Write the ICE setup
Assuming the 1.6 m solution is approximated as 1.6 M in BrO–, the initial concentration of base is 1.6 M. Let x be the amount of OH– produced:
- Initial: [BrO–] = 1.6, [HOBr] = 0, [OH–] = 0
- Change: [BrO–] = -x, [HOBr] = +x, [OH–] = +x
- Equilibrium: [BrO–] = 1.6 – x, [HOBr] = x, [OH–] = x
Insert into the base equilibrium expression:
Kb = x2 / (1.6 – x)
With Kb = 5.0 × 10-6, solving gives x ≈ 2.82 × 10-3 M. Since x is much smaller than 1.6, the weak-base approximation is valid and the square-root shortcut works very well here.
Step 3: Calculate pOH and pH
Because x represents the hydroxide concentration:
- [OH–] ≈ 2.82 × 10-3 M
- pOH = -log(2.82 × 10-3) ≈ 2.55
- pH = 14.00 – 2.55 = 11.45
So the expected answer is:
Why KBrO makes the solution basic
Students sometimes get confused because salts can be neutral, acidic, or basic depending on the strengths of the parent acid and base. KBrO comes from KOH, a strong base, and HOBr, a weak acid. The cation from the strong base, K+, is neutral in water. The anion from the weak acid, BrO–, is basic. That is the pattern to remember: a salt containing the conjugate base of a weak acid usually raises pH.
This principle is broadly useful in acid-base chemistry. If you can classify the ions, you can often predict whether a salt solution will be acidic, neutral, or basic before doing any calculation. For KBrO, the prediction and the math agree: the solution is clearly basic, with a pH well above 7.
Molality vs molarity in a 1.6 m KBrO problem
The notation 1.6 m means molality, not molarity. Molality is moles of solute per kilogram of solvent. Molarity is moles of solute per liter of solution. To convert molality to molarity, you need the solution density and molar mass. This calculator includes that feature because more advanced chemistry and engineering work may require it. The exact conversion is:
M = (1000 × m × density) / (1000 + m × molar mass)
where density is in g/mL and molar mass is in g/mol. If the solution density is near 1.00 g/mL, the converted molarity for 1.6 m KBrO is lower than 1.6 M because the dissolved salt adds mass and volume. If the density is higher, the molarity can increase. That means your final pH depends slightly on whether you treat the concentration as approximate or do the conversion more carefully.
| Quantity | Typical value used | Role in pH calculation |
|---|---|---|
| Ka of HOBr | 2.0 × 10-9 | Used to obtain Kb for BrO– via Kb = Kw/Ka |
| Kw at 25 degrees C | 1.0 × 10-14 | Relates acid and base constants and converts pOH to pH |
| Molar mass of KBrO | 135.001 g/mol | Needed only when converting 1.6 m to molarity |
| Approximate pH of 1.6 m KBrO | 11.45 | Expected answer using the common textbook approximation |
Worked comparison: approximate vs density-corrected concentration
Many students wonder whether using 1.6 M instead of 1.6 m changes the answer much. The answer is usually: not dramatically for a weak-base estimate, but enough to matter if the assignment expects rigorous unit handling. The comparison below illustrates the effect. In each case, Ka(HOBr) is taken as 2.0 × 10-9 and the quadratic result is used.
| Assumption set | Density (g/mL) | Converted molarity of KBrO | Estimated [OH–] | Estimated pH |
|---|---|---|---|---|
| Dilute homework approximation | Not used | 1.600 M | 2.82 × 10-3 M | 11.45 |
| Density-corrected example | 1.00 | 1.302 M | 2.55 × 10-3 M | 11.41 |
| Density-corrected example | 1.10 | 1.432 M | 2.68 × 10-3 M | 11.43 |
These figures show that the pH remains strongly basic regardless of the small concentration correction. That is why many classroom solutions report a pH around 11.4 to 11.5 without worrying about density unless it is explicitly provided.
Common mistakes when solving KBrO pH problems
- Treating KBrO as a strong base. KBrO is not like KOH. It contains the weak base BrO–, so you must use an equilibrium calculation.
- Using Ka directly instead of converting to Kb. Because BrO– is a base, use Kb = Kw/Ka.
- Mixing up molality and molarity. A value given in m is not automatically M. Check whether the problem expects approximation or conversion.
- Forgetting to calculate pOH first. The hydrolysis gives [OH–], so pOH is often calculated before pH.
- Ignoring the validity of the weak-base approximation. Always compare x with the initial concentration. Here x is much less than 5 percent of 1.6, so the approximation is safe.
Practical chemistry context for hypobromite systems
Hypobromite chemistry appears in oxidation chemistry, disinfection discussions, and equilibrium problems involving bromine oxyanions. Although KBrO itself is often presented in textbook equilibrium exercises, the underlying acid-base logic extends to many real systems. Weak oxyacids and their conjugate bases commonly produce measurable pH shifts in aqueous solution, and the same Ka, Kb, pH, and pOH relationships are central to analytical chemistry, environmental chemistry, and water treatment calculations.
In more advanced work, you may also account for ionic strength and activity coefficients. At a concentration this high, nonideal behavior can become important. However, general chemistry pH calculations nearly always use concentrations rather than activities unless the problem explicitly requests thermodynamic treatment. That is why the standard answer still focuses on the equilibrium setup shown above.
Fast method for exam settings
If you are taking a timed exam, here is the quickest path:
- Identify BrO– as a weak base.
- Compute Kb = Kw/Ka = 1.0 × 10-14 / 2.0 × 10-9 = 5.0 × 10-6.
- Use x ≈ √(KbC) = √[(5.0 × 10-6)(1.6)] ≈ 2.83 × 10-3.
- Find pOH = 2.55.
- Find pH = 14.00 – 2.55 = 11.45.
This route is fast, accurate, and usually accepted for standard chemistry coursework.
Authoritative chemistry references
For further reading on acid-base equilibria, water ionization, and concentration units, consult these authoritative sources:
- University-level acid-base equilibrium calculations from LibreTexts
- NIST reference publications on chemical measurement and standard data
- U.S. EPA water quality resources relevant to pH and aqueous chemistry
Final answer summary
When asked to calculate the pH of a 1.6 m KBrO solution, the standard chemistry approach is to treat KBrO as a salt containing the weak base BrO–. Using a typical value of Ka(HOBr) = 2.0 × 10-9, you obtain Kb = 5.0 × 10-6. If the problem allows the common approximation 1.6 m ≈ 1.6 M, then the hydroxide concentration is about 2.82 × 10-3 M, the pOH is about 2.55, and the pH is about 11.45. If density is supplied and you convert molality to molarity more rigorously, the pH changes slightly but remains strongly basic.