6 calculate the ph of 0.100 m kbro solution k
Use this interactive calculator to find the pH of a potassium hypobromite solution, show the full weak-base equilibrium setup, and visualize how concentration affects final pH.
KBrO Solution pH Calculator
How to calculate the pH of a 0.100 M KBrO solution
If your assignment asks you to solve “6 calculate the ph of 0.100 m kbro solution k,” the chemistry idea behind it is straightforward once you identify what kind of salt KBrO is. KBrO is potassium hypobromite. It dissociates essentially completely in water into K+ and BrO–. The potassium ion comes from the strong base KOH and does not hydrolyze appreciably, so it is a spectator ion in the pH calculation. The hypobromite ion, however, is the conjugate base of hypobromous acid, HBrO, which is a weak acid. Because BrO– is a weak base, the solution becomes basic.
The key hydrolysis reaction is:
BrO– + H2O ⇌ HBrO + OH–
Since hydroxide is produced, the pH rises above 7. The whole calculation depends on converting the acid dissociation constant of HBrO to the base dissociation constant of BrO–. At 25°C, many general chemistry texts use a Ka for HBrO of about 2.3 × 10-9. From this:
Kb = Kw / Ka = (1.0 × 10-14) / (2.3 × 10-9) = 4.35 × 10-6
Then use the weak-base equilibrium expression:
Kb = [HBrO][OH–] / [BrO–]
Starting with 0.100 M BrO– and letting x represent the amount that reacts:
- [BrO–] starts at 0.100 and changes to 0.100 – x
- [HBrO] starts at 0 and changes to x
- [OH–] starts at 0 and changes to x
So the equilibrium expression becomes:
Kb = x2 / (0.100 – x)
Because Kb is small compared with 0.100, the common approximation is 0.100 – x ≈ 0.100. Then:
x ≈ √(Kb × C) = √(4.35 × 10-6 × 0.100) = 6.60 × 10-4 M
This x value is the hydroxide concentration. Then:
- pOH = -log(6.60 × 10-4) ≈ 3.180
- pH = 14.000 – 3.180 = 10.820
Therefore, the pH of a 0.100 M KBrO solution is approximately 10.82 at 25°C when Ka(HBrO) = 2.3 × 10-9.
Why KBrO gives a basic solution
Many pH problems become easier once you classify the species in solution. KBrO is a salt made from:
- KOH, a strong base
- HBrO, a weak acid
A salt produced from a strong base and a weak acid always tends to produce a basic solution because the anion hydrolyzes water. This is the same general pattern seen with sodium acetate, sodium fluoride, and sodium hypochlorite. In each case, the anion is the conjugate base of a weak acid and accepts a proton from water, forming OH–.
By contrast, if a salt comes from a strong acid and a strong base, such as NaCl or KNO3, the solution is generally neutral. If the salt comes from a strong acid and a weak base, such as NH4Cl, the solution is acidic. This classification step is one of the fastest ways to avoid mistakes on exams.
Exact method versus approximation
In weak acid and weak base problems, instructors often accept the square root approximation when the degree of ionization is small. Still, the exact quadratic result is more rigorous. For KBrO:
x = (-Kb + √(Kb2 + 4KbC)) / 2
Plugging in Kb = 4.35 × 10-6 and C = 0.100 gives x ≈ 6.57 × 10-4 M. This produces a pH very close to the approximation result. The percent ionization is only about 0.66%, which is well under the common 5% threshold used to justify the approximation.
| Parameter | Value | Meaning in the calculation |
|---|---|---|
| Initial KBrO concentration | 0.100 M | Sets the starting concentration of BrO– |
| Ka of HBrO | 2.3 × 10-9 | Weak acid strength used to derive Kb |
| Kw at 25°C | 1.0 × 10-14 | Relates conjugate acid and base constants |
| Kb of BrO– | 4.35 × 10-6 | Controls OH– formation |
| [OH–] | 6.6 × 10-4 M | Found from equilibrium calculation |
| Final pH | 10.82 | Basic solution |
Step by step exam solution format
If you need to show full work in a homework or quiz response, this structure is both complete and efficient:
- Write the dissociation: KBrO → K+ + BrO–
- State that K+ is a spectator ion.
- Write the hydrolysis reaction: BrO– + H2O ⇌ HBrO + OH–
- Calculate Kb = Kw / Ka.
- Set up an ICE table.
- Use Kb = x2 / (0.100 – x).
- Apply the approximation or solve the quadratic.
- Find pOH from [OH–].
- Use pH = 14 – pOH.
That method demonstrates both the chemistry concept and the numerical process. Instructors usually care just as much about the setup as the final number.
How concentration changes the pH of KBrO
A very useful insight is that pH depends on concentration. If you dilute the KBrO solution, the hydroxide concentration decreases and the pH drops closer to 7, though it remains basic. For weak bases, [OH–] often scales approximately with the square root of concentration, not linearly. This means a tenfold decrease in concentration does not cause a tenfold decrease in pH. Instead, the pH changes more gradually.
For the same Ka and Kw values, the table below shows approximate pH values for several KBrO concentrations:
| KBrO concentration (M) | Approximate [OH–] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 1.000 | 2.09 × 10-3 | 2.68 | 11.32 |
| 0.100 | 6.60 × 10-4 | 3.18 | 10.82 |
| 0.010 | 2.09 × 10-4 | 3.68 | 10.32 |
| 0.001 | 6.60 × 10-5 | 4.18 | 9.82 |
This trend is especially useful if your instructor asks conceptual questions such as “What happens to the pH after dilution?” The answer is that the pH decreases, but the solution remains basic because BrO– is still acting as a weak base.
Common mistakes students make
- Using Ka directly instead of converting to Kb. Since BrO– is the base in solution, Kb is the relevant equilibrium constant.
- Treating KBrO as a strong base. KBrO is not like KOH. Only a fraction of BrO– reacts with water.
- Forgetting to calculate pOH first. Because the equilibrium gives OH–, you must get pOH before pH.
- Ignoring temperature assumptions. The common relation pH + pOH = 14.00 assumes Kw = 1.0 × 10-14, typically at 25°C.
- Confusing KBrO with KBrO3. Potassium bromate is a different compound and should not be analyzed the same way.
Chemical interpretation of the result
A pH of about 10.82 means the solution is moderately basic. It is nowhere near the alkalinity of a strong 0.100 M hydroxide solution, which would have a pH around 13.00. That difference is important because KBrO only partially hydrolyzes. In real chemical systems, weakly basic oxidizing anions such as hypobromite can be relevant in disinfection chemistry, bromine-containing equilibria, and aqueous oxidation systems.
The result also tells you that the conjugate acid HBrO is fairly weak. The weaker the acid HBrO, the stronger its conjugate base BrO–. If you changed the Ka value in the calculator above, you would immediately see the impact on Kb and therefore on final pH.
Quick comparison with other common weak-base salt systems
KBrO belongs to a broad class of salts whose pH is controlled by hydrolysis of an anion. A helpful comparison is shown below.
| Salt | Conjugate acid | Solution character | Reason |
|---|---|---|---|
| NaCl | HCl | Neutral | Cl– is the conjugate base of a strong acid and does not significantly hydrolyze |
| CH3COONa | CH3COOH | Basic | Acetate is the conjugate base of a weak acid |
| KBrO | HBrO | Basic | BrO– accepts protons from water and forms OH– |
| NH4Cl | NH4+ as weak acid | Acidic | NH4+ donates protons to water |
Authoritative chemistry and pH references
For deeper study of pH, aqueous equilibria, and water chemistry fundamentals, these authoritative resources are useful:
Final answer
Using Ka(HBrO) = 2.3 × 10-9 and Kw = 1.0 × 10-14 at 25°C, the pH of a 0.100 M KBrO solution is:
pH ≈ 10.82
If your instructor uses a slightly different Ka value for HBrO, your answer may vary slightly, usually by a few hundredths of a pH unit. The calculator above lets you test those alternative constants instantly.
Note: In chemistry notation, concentration should normally be written as 0.100 M for molarity. The phrase “0.100 m” can sometimes refer to molality, but most textbook pH problems of this type intend molarity unless stated otherwise.