Calculate pH of 0.25 M Sr(OH)2
Use this interactive calculator to determine hydroxide concentration, pOH, and pH for strontium hydroxide solutions. The default setup solves the exact problem: calculate the pH of 0.25 M Sr(OH)2 at 25 C, assuming complete dissociation.
How to calculate the pH of 0.25 M Sr(OH)2
To calculate the pH of 0.25 M strontium hydroxide, you start by identifying the compound type and its dissociation behavior in water. Strontium hydroxide, written as Sr(OH)2, is a strong base in introductory and most intermediate aqueous chemistry calculations. That means it is typically treated as fully dissociated in solution. One formula unit of Sr(OH)2 releases one strontium ion, Sr2+, and two hydroxide ions, OH-. Because the hydroxide concentration controls pOH and pH in a basic solution, the key step is multiplying the base concentration by 2.
The dissociation equation is:
Sr(OH)2(aq) → Sr2+(aq) + 2OH-(aq)
If the initial concentration of Sr(OH)2 is 0.25 M, then the hydroxide ion concentration becomes:
[OH-] = 2 × 0.25 = 0.50 M
Next, calculate pOH using the base ten logarithm:
pOH = -log10[OH-] = -log10(0.50) = 0.3010
At 25 C, the standard relationship between pH and pOH is:
pH + pOH = 14.00
So:
pH = 14.00 – 0.3010 = 13.699
Why Sr(OH)2 gives two hydroxide ions
Students often make the mistake of using the concentration of Sr(OH)2 directly as the hydroxide concentration. That would be correct for a monohydroxide strong base such as NaOH or KOH, because each formula unit provides only one OH- ion. Sr(OH)2 is different. The subscript 2 after the hydroxide group means each dissolved unit produces two hydroxide ions. This doubling is the reason 0.25 M Sr(OH)2 behaves like a 0.50 M hydroxide source for pOH and pH calculations.
Here is the logic in a compact sequence:
- Recognize Sr(OH)2 as a strong ionic base.
- Write the dissociation stoichiometry.
- Multiply the molarity by 2 to get [OH-].
- Compute pOH with the negative logarithm.
- Subtract pOH from 14.00 at 25 C to obtain pH.
Worked example in full detail
Let us walk through the complete calculation in a classroom style format. Suppose a problem asks: “Calculate the pH of 0.25 M Sr(OH)2.” The chemical formula reveals one strontium cation and two hydroxide anions per formula unit. If the substance dissociates completely, then every 0.25 mole of dissolved Sr(OH)2 in 1 liter contributes 0.50 mole of OH-. Since pOH is based on hydroxide concentration, you insert 0.50 into the logarithmic expression. The result is a pOH of about 0.301. Basic solutions have high pH values, so subtracting this small pOH from 14.00 yields a strongly basic pH near 13.70. That is a sensible answer because the hydroxide concentration is very large compared with neutral water.
It is helpful to check the answer qualitatively. Neutral water at 25 C has pH 7. A solution with [OH-] = 1.0 × 10^-7 M is also neutral by the pH and pOH relationship. Here, [OH-] = 0.50 M, which is millions of times more basic than neutral water. A pH near 13.70 is therefore reasonable.
Comparison table: strong bases and hydroxide production
| Base | Dissociation pattern | OH- ions released per formula unit | [OH-] from a 0.25 M solution | pOH at 25 C | pH at 25 C |
|---|---|---|---|---|---|
| NaOH | NaOH → Na+ + OH- | 1 | 0.25 M | 0.6021 | 13.3979 |
| KOH | KOH → K+ + OH- | 1 | 0.25 M | 0.6021 | 13.3979 |
| Ca(OH)2 | Ca(OH)2 → Ca2+ + 2OH- | 2 | 0.50 M | 0.3010 | 13.6990 |
| Sr(OH)2 | Sr(OH)2 → Sr2+ + 2OH- | 2 | 0.50 M | 0.3010 | 13.6990 |
| Ba(OH)2 | Ba(OH)2 → Ba2+ + 2OH- | 2 | 0.50 M | 0.3010 | 13.6990 |
This table highlights a major stoichiometric principle. A 0.25 M solution of a dibasic strong hydroxide such as Sr(OH)2 produces the same hydroxide concentration as a 0.50 M solution of a monohydroxide strong base. That is why the pH is higher than the pH of 0.25 M NaOH or KOH.
Important assumptions behind the calculation
- Complete dissociation: Sr(OH)2 is treated as a strong base, so all dissolved formula units separate into ions.
- 25 C water constant: Most textbook problems use pKw = 14.00 at 25 C.
- Ideal solution behavior: Introductory calculations usually use molarity directly rather than activity corrections.
- No competing equilibria: Carbon dioxide absorption, precipitation, or common ion effects are ignored.
These assumptions are standard for general chemistry. In advanced analytical chemistry, highly concentrated ionic solutions can deviate from ideality, and very precise pH values may involve activity coefficients rather than simple concentration. However, for a problem framed as “calculate pH of 0.25 M Sr(OH)2,” the accepted answer is almost always 13.70.
Where students lose points on this problem
There are several common errors. The first is forgetting to multiply by 2 for hydroxide production. The second is taking pH directly from the Sr(OH)2 concentration rather than from [OH-]. The third is confusing pH and pOH. The fourth is using natural logarithms rather than base ten logarithms in the pH formulas. Finally, some students report too few significant figures or round too early. Since 0.25 has two significant figures, a final pH of 13.70 is usually acceptable, though many instructors also show the intermediate unrounded value 13.699.
Step-by-step formula summary
- Write the base dissociation: Sr(OH)2 → Sr2+ + 2OH-
- Calculate hydroxide concentration: [OH-] = 2 × 0.25 = 0.50 M
- Find pOH: pOH = -log10(0.50) = 0.3010
- Find pH: pH = 14.00 – 0.3010 = 13.699
- Round appropriately: pH ≈ 13.70
Reference data table: pH as Sr(OH)2 concentration changes
| Sr(OH)2 concentration | Calculated [OH-] | pOH | pH at 25 C | Interpretation |
|---|---|---|---|---|
| 0.001 M | 0.002 M | 2.6990 | 11.3010 | Clearly basic |
| 0.010 M | 0.020 M | 1.6990 | 12.3010 | Strongly basic |
| 0.050 M | 0.100 M | 1.0000 | 13.0000 | Very strongly basic |
| 0.250 M | 0.500 M | 0.3010 | 13.6990 | Default problem value |
| 0.500 M | 1.000 M | 0.0000 | 14.0000 | Idealized upper textbook case at 25 C |
The numbers in this table are computed using the standard strong base model and pKw = 14.00. They demonstrate how quickly pH rises when hydroxide concentration increases. Because Sr(OH)2 yields two hydroxide ions, its pH climbs faster than that of a one to one strong base at the same formal molarity.
Real chemistry context: concentration, solubility, and measurement
Although classroom pH problems are designed to emphasize stoichiometry and logarithms, actual pH measurement in the laboratory can involve additional factors. Activity effects become more relevant at higher ionic strength. Temperature alters pKw, so the relationship pH + pOH = 14.00 strictly applies only near 25 C. Carbon dioxide from air can dissolve into basic solutions and reduce pH over time by forming carbonate and bicarbonate species. In practice, freshly prepared solutions and calibrated pH electrodes are important for quality measurements.
For educational purposes, however, the 0.25 M Sr(OH)2 problem is a classic strong base calculation. It reinforces the idea that chemical formulas matter. The subscript on hydroxide determines stoichiometric yield, and stoichiometric yield determines hydroxide concentration. Once [OH-] is known, pOH and pH follow directly.
Authoritative sources for deeper study
For supporting chemistry fundamentals and water chemistry references, review: U.S. Environmental Protection Agency on pH, LibreTexts Chemistry educational content, National Institute of Standards and Technology, and U.S. Geological Survey on pH and water.
Frequently asked questions
Is Sr(OH)2 always treated as a strong base? In standard aqueous chemistry problems, yes. The dissolved portion is treated as completely dissociated. For very advanced work, solubility, ionic strength, and activity corrections may be considered, but they are not usually part of introductory pH questions.
Why is the pH not exactly 14? Because the hydroxide concentration is 0.50 M, not 1.00 M. Since pOH = -log10(0.50) = 0.3010, the pOH is positive, so pH must be slightly less than 14 under the 25 C convention.
Can pH be above 14? In concentrated real solutions, measured pH values can extend beyond 14 depending on the scale and activity considerations. However, in elementary chemistry with the simplified 25 C concentration based model, the familiar range is usually presented as 0 to 14.
What is the quickest way to solve this problem on an exam? Memorize the stoichiometric shortcut: for Sr(OH)2, double the molarity to get [OH-]. Then do pOH = -log[OH-], and pH = 14 – pOH.
Bottom line
If you need to calculate the pH of 0.25 M Sr(OH)2, the process is straightforward once you account for the two hydroxide ions released by each formula unit. Multiply 0.25 M by 2 to get 0.50 M OH-. Calculate pOH as 0.3010. Subtract from 14.00 at 25 C. The final pH is 13.70. This result is the standard textbook answer and the correct one for most academic, homework, quiz, and exam contexts.