Calculate OH- and pH for 1.5 x 10-3 M Sr(OH)2
Use this premium chemistry calculator to find hydroxide concentration, pOH, and pH for aqueous strontium hydroxide. The default setup solves the exact problem: 1.5 x 10-3 M Sr(OH)2, a strong base that dissociates into one Sr2+ ion and two OH- ions.
Calculated Results
Click the button to compute OH-, pOH, and pH for Sr(OH)2.
How to calculate OH- and pH for 1.5 x 10-3 M Sr(OH)2
To calculate OH- and pH for 1.5 x 10-3 M Sr(OH)2, the key idea is that strontium hydroxide is treated as a strong base in introductory and general chemistry. That means it dissociates essentially completely in dilute aqueous solution:
Sr(OH)2(aq) -> Sr2+(aq) + 2OH-(aq)
The stoichiometry matters more than anything else. Every 1 mole of Sr(OH)2 produces 2 moles of hydroxide ions. So if the strontium hydroxide concentration is 1.5 x 10-3 M, then the hydroxide concentration is:
- Start with the base concentration: [Sr(OH)2] = 1.5 x 10-3 M
- Use the 1:2 dissociation ratio: [OH-] = 2 x 1.5 x 10-3
- Calculate hydroxide concentration: [OH-] = 3.0 x 10-3 M
- Find pOH using pOH = -log[OH-]
- Find pH using pH = 14.00 – pOH at 25 degrees C
Performing the logarithm gives pOH approximately 2.52. Subtracting from 14.00 gives a pH of about 11.48. Those are the standard textbook answers for this problem at room temperature. Because the hydroxide concentration is much larger than 1.0 x 10-7 M, the contribution of water autoionization is negligible, so the strong-base approximation works extremely well.
Why Sr(OH)2 changes pH so strongly
The reason strontium hydroxide creates a basic solution is simple: it supplies hydroxide ions directly. In water chemistry, pH and pOH are logarithmic measures of hydronium and hydroxide concentration. Since Sr(OH)2 releases two hydroxide ions per dissolved formula unit, its effect on basicity is stronger than a monohydroxide base of the same molarity. For example, 1.5 x 10-3 M NaOH would yield 1.5 x 10-3 M OH-, while 1.5 x 10-3 M Sr(OH)2 yields double that amount, 3.0 x 10-3 M OH-.
This is exactly why students are often tested on compounds such as Ca(OH)2, Ba(OH)2, and Sr(OH)2. The concept checks whether you remember to multiply by the number of hydroxides released during dissociation. Missing that multiplier is one of the most common mistakes in acid-base calculations.
Core equations used in this calculator
- [OH-] = n x C, where n is the number of hydroxide ions released and C is the molarity of the base
- pOH = -log[OH-]
- pH = 14.00 – pOH at 25 degrees C
- pH + pOH = 14.00 for standard classroom conditions at 25 degrees C
Step-by-step worked example
Step 1: Identify the base and its dissociation pattern
Strontium hydroxide has the formula Sr(OH)2. The subscript 2 after OH tells you that each formula unit contains two hydroxide groups. In water, the dissociation is represented as:
Sr(OH)2 -> Sr2+ + 2OH-
Therefore, one mole of dissolved Sr(OH)2 contributes two moles of OH-.
Step 2: Convert base concentration into hydroxide concentration
Given 1.5 x 10-3 M Sr(OH)2:
[OH-] = 2 x 1.5 x 10-3 = 3.0 x 10-3 M
This is the single most important step in the problem. If you forget the factor of 2, the rest of your answer will be wrong.
Step 3: Calculate pOH
pOH is the negative base-10 logarithm of the hydroxide concentration:
pOH = -log(3.0 x 10-3)
Splitting the logarithm can help:
- log(3.0) ≈ 0.4771
- log(10-3) = -3
- So log(3.0 x 10-3) = 0.4771 – 3 = -2.5229
- Therefore pOH = 2.5229
Rounded properly, pOH = 2.52.
Step 4: Calculate pH
At 25 degrees C, use the familiar relationship:
pH = 14.00 – 2.5229 = 11.4771
Rounded to two decimal places, pH = 11.48.
Common mistakes students make
- Forgetting the coefficient 2 for OH-. Sr(OH)2 does not produce just one hydroxide ion. It produces two.
- Using pH = -log[OH-]. That equation gives pOH, not pH.
- Misreading scientific notation. 1.5 x 10-3 is 0.0015, not 0.015.
- Dropping the negative sign in the logarithm rule. pOH must be positive for this concentration range.
- Using weak-base logic. Sr(OH)2 is generally treated as a strong base in standard chemistry problems, so no ICE table is needed here.
Comparison table: Strong bases and hydroxide yield
| Base | Dissociation pattern | OH- produced per mole of base | [OH-] if base concentration is 1.5 x 10-3 M |
|---|---|---|---|
| NaOH | NaOH -> Na+ + OH- | 1 | 1.5 x 10-3 M |
| KOH | KOH -> K+ + OH- | 1 | 1.5 x 10-3 M |
| Ca(OH)2 | Ca(OH)2 -> Ca2+ + 2OH- | 2 | 3.0 x 10-3 M |
| Sr(OH)2 | Sr(OH)2 -> Sr2+ + 2OH- | 2 | 3.0 x 10-3 M |
| Ba(OH)2 | Ba(OH)2 -> Ba2+ + 2OH- | 2 | 3.0 x 10-3 M |
The table shows why recognizing the formula structure is critical. Monohydroxide bases and dihydroxide bases can have the same formal molarity but very different hydroxide ion concentrations. Because pH is logarithmic, even a factor-of-two change in hydroxide concentration causes a noticeable change in pOH and pH.
Reference values and real chemistry context
In pure water at 25 degrees C, both [H3O+] and [OH-] are 1.0 x 10-7 M, giving pH 7.00 and pOH 7.00. That benchmark helps you quickly recognize whether a solution is acidic, neutral, or basic. In the present problem, [OH-] is 3.0 x 10-3 M, which is 30,000 times larger than the hydroxide concentration in pure water. That huge increase is why the resulting pH is strongly basic at about 11.48.
| Solution | Typical [OH-] at 25 degrees C | pOH | Interpretation |
|---|---|---|---|
| Pure water | 1.0 x 10-7 M | 7.00 | Neutral benchmark |
| 1.5 x 10-3 M NaOH | 1.5 x 10-3 M | 2.82 | Basic |
| 1.5 x 10-3 M Sr(OH)2 | 3.0 x 10-3 M | 2.52 | More basic because two OH- are released |
| 0.010 M strong base with one OH- | 1.0 x 10-2 M | 2.00 | Strongly basic |
When this shortcut is valid
For most general chemistry exercises, you can directly use stoichiometric dissociation for Sr(OH)2 because it is treated as a strong base. This is especially appropriate at moderate dilute concentrations like 1.5 x 10-3 M. In more advanced physical chemistry or analytical chemistry contexts, activity corrections, temperature dependence of Kw, and non-ideal behavior can become important, but those are not usually expected for a problem phrased in this way.
If the problem specifically asks for room-temperature textbook pH, then the straightforward method used in this calculator is the correct one. The relationship pH + pOH = 14.00 assumes 25 degrees C and the standard value Kw = 1.0 x 10-14.
How to think about significant figures
The concentration 1.5 x 10-3 M has two significant figures. Multiplying by 2 gives 3.0 x 10-3 M for hydroxide, which also has two significant figures. For pH and pOH, the number of digits after the decimal point should correspond to the number of significant figures in the concentration. Since 3.0 x 10-3 has two significant figures, reporting pOH as 2.52 and pH as 11.48 is appropriate in most classroom settings.
Authority sources for pH, pOH, and water chemistry
- U.S. Environmental Protection Agency: pH overview and water chemistry context
- University-supported chemistry resources and general acid-base tutorials
- U.S. Geological Survey: pH and water science fundamentals
Quick recap
If you need to calculate OH- and pH for 1.5 x 10-3 M Sr(OH)2, remember this sequence: identify dissociation, multiply molarity by 2 to get hydroxide concentration, calculate pOH from the negative logarithm, then convert to pH using 14.00 minus pOH. That gives:
- [OH-] = 3.0 x 10-3 M
- pOH = 2.52
- pH = 11.48
Once you understand this pattern, you can solve nearly any strong-base pH problem involving metal hydroxides with confidence. The calculator above makes the process immediate, but the chemistry logic remains the same: stoichiometry first, logarithms second, and pH interpretation last.