Calculate pH for 1.3 × 10-3 M Sr(OH)2
This interactive calculator computes hydroxide concentration, pOH, and pH for strontium hydroxide solutions. It is prefilled for the classic chemistry problem of finding the pH of a 1.3 × 10-3 M Sr(OH)2 solution, while also letting you test other concentrations and ionization assumptions.
- Strong base model
- 2 OH– per formula unit
- Instant pH and pOH output
- Visual chart included
Sr(OH)2 pH Calculator
Default example: 1.3 × 10-3 M Sr(OH)2. For this strong base, each mole of Sr(OH)2 contributes 2 moles of OH–.
Results
Enter your values and click Calculate pH to see the answer, steps, and a visual breakdown.
How to calculate pH for 1.3 × 10-3 M Sr(OH)2
To calculate pH for 1.3 × 10-3 M Sr(OH)2, you treat strontium hydroxide as a strong base that dissociates essentially completely in introductory chemistry problems. That means each formula unit of Sr(OH)2 releases one Sr2+ ion and two OH– ions into solution. Because pH is tied to hydrogen ion concentration and pOH is tied to hydroxide ion concentration, the quickest path is to first determine the hydroxide concentration, then compute pOH, and finally convert that value to pH using the standard room-temperature relationship pH + pOH = 14.
Short answer: For a 1.3 × 10-3 M Sr(OH)2 solution, [OH–] = 2.6 × 10-3 M, pOH ≈ 2.59, and pH ≈ 11.41 at 25°C.
Step-by-step solution
- Write the dissociation equation:
Sr(OH)2 → Sr2+ + 2OH– - Use the stoichiometric ratio:
Every 1 mole of Sr(OH)2 produces 2 moles of OH–. - Calculate hydroxide concentration:
[OH–] = 2 × 1.3 × 10-3 = 2.6 × 10-3 M - Calculate pOH:
pOH = -log(2.6 × 10-3) ≈ 2.585 - Convert to pH:
pH = 14.00 – 2.585 = 11.415 - Round properly:
pH ≈ 11.41
This is the standard answer expected in general chemistry unless the problem specifically asks you to account for non-ideal behavior, activity coefficients, or a temperature other than 25°C. In most classroom settings, Sr(OH)2 is treated as a fully dissociated strong base and the calculation is entirely stoichiometric.
Why Sr(OH)2 changes the pH so strongly
Strontium hydroxide belongs to the class of metal hydroxides that act as strong bases in water. The most important detail is not only that it dissociates extensively, but also that it contributes two hydroxide ions per formula unit. That doubles the OH– concentration relative to a monohydroxide base such as NaOH at the same molarity. For example, a 1.3 × 10-3 M NaOH solution produces 1.3 × 10-3 M OH–, but a 1.3 × 10-3 M Sr(OH)2 solution produces 2.6 × 10-3 M OH–.
That distinction matters because the pH scale is logarithmic. A doubling of hydroxide concentration does not simply add a fixed arithmetic amount to pH under every circumstance; instead, it shifts pOH by the logarithm of the concentration ratio. In practical terms, a divalent hydroxide can create a noticeably more basic solution than a monovalent hydroxide at equal molarity.
Core formulas you need
- Dissociation: Sr(OH)2 → Sr2+ + 2OH–
- Hydroxide concentration: [OH–] = 2 × CSr(OH)2
- pOH: pOH = -log[OH–]
- pH at 25°C: pH = 14 – pOH
- Water ion product reference: Kw = 1.0 × 10-14 at 25°C
Worked example with full interpretation
Suppose your instructor asks: “Calculate the pH of a 1.3 × 10-3 M Sr(OH)2 solution.” The first thing to recognize is the notation. The concentration is not 1.3, and it is not 10-3 M separately. The full concentration is 1.3 × 10-3 M, which equals 0.0013 M.
Since Sr(OH)2 delivers two hydroxide ions per mole, multiply the base concentration by 2:
[OH–] = 2(0.0013) = 0.0026 M
Now calculate pOH:
pOH = -log(0.0026) ≈ 2.59
Finally, use the room-temperature relationship:
pH = 14.00 – 2.59 = 11.41
That pH tells you the solution is clearly basic. It is nowhere near neutral, which is pH 7 at 25°C, and it is also not near the extreme upper end of the pH scale, because the hydroxide concentration is still in the millimolar range rather than the tenths-of-a-molar or molar range.
Comparison table: equal molarity, different hydroxide output
| Base | Given concentration (M) | OH- ions released per formula unit | Resulting [OH-] (M) | pOH | pH at 25°C |
|---|---|---|---|---|---|
| NaOH | 1.3 × 10-3 | 1 | 1.3 × 10-3 | 2.89 | 11.11 |
| KOH | 1.3 × 10-3 | 1 | 1.3 × 10-3 | 2.89 | 11.11 |
| Sr(OH)2 | 1.3 × 10-3 | 2 | 2.6 × 10-3 | 2.59 | 11.41 |
| Ba(OH)2 | 1.3 × 10-3 | 2 | 2.6 × 10-3 | 2.59 | 11.41 |
This table shows why recognizing the formula matters. Two solutions can have the same formal molarity but different pH values if one produces more hydroxide ions per dissolved unit than the other.
Common mistakes students make
- Forgetting the coefficient 2 for OH–: This is the most common error. If you skip it, you get [OH–] too low and a pH that is too low.
- Using pH = -log[OH–]: That formula gives pOH, not pH.
- Dropping the scientific notation: 1.3 × 10-3 is 0.0013, not 1.3.
- Ignoring temperature assumptions: The equation pH + pOH = 14 is strictly tied to 25°C unless a problem states otherwise.
- Over-rounding too early: Keep extra digits until the final step to avoid a noticeable rounding drift.
What real chemistry says about ideal assumptions
In real laboratory chemistry, pH calculations can become more complex than textbook problems suggest. At very low or very high ionic strengths, chemists often use activities instead of concentrations. Likewise, the value of Kw changes with temperature, meaning pH + pOH is not always exactly 14. However, for a standard general chemistry exercise involving a dilute strong base at 25°C, the idealized method is not only acceptable but expected.
If you are studying for an exam, the safe academic approach is usually:
- Assume full dissociation for strong bases like Sr(OH)2.
- Use stoichiometry to find [OH–].
- Calculate pOH with a base-10 logarithm.
- Subtract from 14.00 to get pH.
Reference data table for nearby concentrations
| Sr(OH)2 concentration (M) | [OH-] generated (M) | pOH | pH at 25°C | Basicity interpretation |
|---|---|---|---|---|
| 1.0 × 10-4 | 2.0 × 10-4 | 3.70 | 10.30 | Mildly basic |
| 5.0 × 10-4 | 1.0 × 10-3 | 3.00 | 11.00 | Clearly basic |
| 1.3 × 10-3 | 2.6 × 10-3 | 2.59 | 11.41 | Moderately strong basic solution |
| 1.0 × 10-2 | 2.0 × 10-2 | 1.70 | 12.30 | Strongly basic |
Why authoritative references matter
When learning pH calculations, it helps to verify textbook-style assumptions with reputable scientific sources. Government and university resources explain dissociation, pH scales, water chemistry, and equilibrium constants with high reliability. For further reading, consult the following authoritative references:
- U.S. Environmental Protection Agency: pH and acid-neutralizing capacity
- Chemistry LibreTexts, university-supported chemistry resource
- U.S. Geological Survey: pH and water
Exam strategy for solving this kind of question fast
In a timed setting, pattern recognition is everything. When you see a species like Sr(OH)2, mentally label it as a strong base with two hydroxides. Immediately convert the molarity to hydroxide concentration by doubling it. Then compute pOH and subtract from 14. This sequence usually takes less than a minute once practiced.
A reliable mental checklist looks like this:
- Is the substance a strong acid or strong base?
- How many H+ or OH– ions are released per formula unit?
- Do I need pH or pOH first?
- Am I working at 25°C under standard assumptions?
- Have I kept enough significant digits until the end?
Final answer
For the question “calculate pH for 1.3 × 10-3 M Sr(OH)2”, the accepted general chemistry result is:
[OH–] = 2.6 × 10-3 M
pOH = 2.59
pH = 11.41
Use the calculator above to confirm the answer, visualize how the hydroxide concentration compares with pOH and pH, and explore how changing concentration or hydroxide stoichiometry affects the final result.