Calculate pH for 1.3 × 10-3 M Sr(OH)2
Use this premium calculator to find hydroxide concentration, pOH, and pH for strontium hydroxide solutions. The default setup matches 1.3 × 10-3 M Sr(OH)2, a strong base that dissociates into one Sr2+ ion and two OH– ions.
Results
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 water under standard textbook conditions. That means every mole of Sr(OH)2 produces two moles of hydroxide ions, OH–. Once you know the hydroxide concentration, the rest is straightforward: calculate pOH using the negative base-10 logarithm, then convert pOH to pH with the standard relationship pH + pOH = 14 at 25°C.
The exact sequence is:
- Write the dissociation equation for strontium hydroxide.
- Convert the formula concentration into hydroxide concentration.
- Calculate pOH.
- Convert pOH into pH.
Step 1: Write the dissociation equation
Strontium hydroxide is represented as Sr(OH)2. In water, the idealized strong-base dissociation is:
Sr(OH)2 → Sr2+ + 2OH–
This is the most important chemistry fact in the problem. The coefficient of 2 in front of OH– tells you that one dissolved formula unit gives two hydroxide ions. Students often lose points here by forgetting to double the base concentration. If the molarity of Sr(OH)2 is 1.3 × 10-3 M, then the hydroxide concentration is not 1.3 × 10-3 M. It is twice that amount.
Step 2: Find hydroxide concentration
Given:
- Sr(OH)2 concentration = 1.3 × 10-3 M
- Hydroxide ions per formula unit = 2
Therefore:
[OH–] = 2 × 1.3 × 10-3 = 2.6 × 10-3 M
That value, 2.6 × 10-3 M, is the quantity that must be inserted into the pOH formula.
Step 3: Calculate pOH
Use the definition:
pOH = -log[OH–]
Substitute 2.6 × 10-3:
pOH = -log(2.6 × 10-3)
Evaluating this gives:
pOH ≈ 2.585
Step 4: Convert pOH to pH
At 25°C, use:
pH = 14.00 – pOH
So:
pH = 14.00 – 2.585 = 11.415
Rounded appropriately, the pH is 11.42.
Why Sr(OH)2 changes pH so effectively
Strontium hydroxide belongs to the family of metal hydroxides that strongly increase pH by releasing hydroxide ions into water. Because it contributes two OH– ions per formula unit, its effect on pH is stronger than a monohydroxide solution of the same formal molarity. For example, a 1.3 × 10-3 M NaOH solution would yield 1.3 × 10-3 M hydroxide, while the same concentration of Sr(OH)2 yields 2.6 × 10-3 M hydroxide.
This distinction matters in stoichiometry and in acid-base calculations. Chemistry courses often test whether you can distinguish between:
- formula concentration, which is the molarity of the dissolved compound
- ion concentration, which is the actual concentration of species produced after dissociation
- pOH and pH, which are logarithmic measures derived from ion concentration
Comparison table: equal molarity, different hydroxide output
The table below shows how several common strong bases compare when each has the same formal concentration of 1.3 × 10-3 M. The main difference is the number of hydroxide ions released per formula unit.
| Base | Formal concentration (M) | OH- ions released | [OH-] produced (M) | Calculated pOH | Calculated pH at 25°C |
|---|---|---|---|---|---|
| NaOH | 1.3 × 10-3 | 1 | 1.3 × 10-3 | 2.886 | 11.114 |
| KOH | 1.3 × 10-3 | 1 | 1.3 × 10-3 | 2.886 | 11.114 |
| Sr(OH)2 | 1.3 × 10-3 | 2 | 2.6 × 10-3 | 2.585 | 11.415 |
| Ba(OH)2 | 1.3 × 10-3 | 2 | 2.6 × 10-3 | 2.585 | 11.415 |
| Al(OH)3 textbook stoichiometric model | 1.3 × 10-3 | 3 | 3.9 × 10-3 | 2.409 | 11.591 |
Important concept: pH is logarithmic, not linear
One of the most useful ideas in acid-base chemistry is that pH and pOH use logarithmic scales. That means a small numerical shift can correspond to a major concentration change. If hydroxide concentration doubles, pOH does not simply halve, and pH does not rise by the same arithmetic amount every time. Instead, the shift is tied to logarithms.
For the present problem, doubling the formula concentration of Sr(OH)2 from 1.3 × 10-3 M to 2.6 × 10-3 M would double the hydroxide concentration from 2.6 × 10-3 M to 5.2 × 10-3 M, but the pH would rise by only about 0.301 units. This is a direct consequence of the log scale, because log(2) ≈ 0.301.
Quick reference table for nearby Sr(OH)2 concentrations
Students often want to know how sensitive pH is to concentration changes near the target value. The next table gives realistic textbook calculations for a few nearby concentrations of Sr(OH)2, assuming complete dissociation and 25°C.
| Sr(OH)2 concentration (M) | [OH-] = 2C (M) | pOH | pH | Interpretation |
|---|---|---|---|---|
| 1.0 × 10-4 | 2.0 × 10-4 | 3.699 | 10.301 | Mildly basic compared with stronger lab solutions |
| 5.0 × 10-4 | 1.0 × 10-3 | 3.000 | 11.000 | Convenient benchmark value for practice |
| 1.3 × 10-3 | 2.6 × 10-3 | 2.585 | 11.415 | Your target example |
| 2.0 × 10-3 | 4.0 × 10-3 | 2.398 | 11.602 | Moderately stronger basic solution |
| 1.0 × 10-2 | 2.0 × 10-2 | 1.699 | 12.301 | Strongly basic classroom example |
Common mistakes when solving this problem
1. Forgetting the stoichiometric coefficient of hydroxide
The single biggest mistake is using 1.3 × 10-3 M directly as [OH–]. That would be correct for NaOH or KOH, but not for Sr(OH)2. Since Sr(OH)2 releases two hydroxides, you must multiply by 2.
2. Mixing up pH and pOH
Another frequent error is calculating -log(2.6 × 10-3) and calling that the pH. It is actually the pOH. Once you have pOH, subtract from 14 to get pH under standard conditions.
3. Losing the negative exponent in scientific notation
1.3 × 10-3 equals 0.0013, not 1300 and not 0.13. Errors in scientific notation produce huge pH mistakes because of the logarithm.
4. Rounding too early
If you round [OH–] or pOH too aggressively in an early step, the final pH can shift slightly. Keep a few extra digits during intermediate calculations, then round your final answer based on the precision expected by your course or lab.
Exam strategy for strong base pH calculations
If you see a strong base formula and need pH quickly, use this checklist:
- Identify how many OH– ions each formula unit releases.
- Multiply the base molarity by that stoichiometric number.
- Take the negative log to find pOH.
- Subtract from 14 to get pH at 25°C.
This pattern is fast, reliable, and works for the majority of introductory chemistry questions involving soluble strong bases.
How this relates to real water quality and laboratory work
In pure academic exercises, pH calculations are often idealized and assume complete dissociation, no activity corrections, and standard temperature. In real environmental and industrial systems, pH measurement can be affected by ionic strength, temperature, dissolved gases, instrumentation, and calibration quality. Even so, the classroom framework remains essential because it establishes the stoichiometric and logarithmic principles behind acid-base behavior.
Government and university resources consistently emphasize that pH is a key indicator of chemical conditions in water and that significant pH shifts influence biological and chemical processes. If you want broader context beyond the math, these authoritative resources are useful:
Final takeaway
To calculate pH for 1.3 × 10-3 M Sr(OH)2, remember that the compound is a strong base and contributes two hydroxide ions per formula unit. Multiply the concentration by 2 to get [OH–] = 2.6 × 10-3 M, calculate pOH ≈ 2.585, and then convert to pH ≈ 11.415. If you can consistently identify the dissociation stoichiometry first, you will solve these problems accurately and much faster.