1.54 × 10-4 M Sr(OH)2: Calculate the pH
Use this premium chemistry calculator to find pOH, pH, and hydroxide concentration for strontium hydroxide solutions, including the classic 1.54 × 10-4 M example.
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Enter the concentration for Sr(OH)2 and click Calculate pH. The default example is 1.54 × 10-4 M, which is commonly used in general chemistry practice.
How to calculate the pH of 1.54 × 10-4 M Sr(OH)2
When a chemistry problem asks you to determine the pH of 1.54 × 10-4 M Sr(OH)2, the key is to recognize that strontium hydroxide is treated as a strong base in standard introductory chemistry. That means it dissociates essentially completely in dilute aqueous solution into one strontium ion and two hydroxide ions:
Sr(OH)2 → Sr2+ + 2OH–
This stoichiometric relationship tells you that every mole of Sr(OH)2 produces two moles of OH–. Since pH is not calculated directly from the base concentration in this case, you first find the hydroxide concentration, then calculate pOH, and finally convert pOH to pH.
Step 1: Write the given concentration
The problem gives:
[Sr(OH)2] = 1.54 × 10-4 M
Step 2: Convert base concentration to hydroxide concentration
Because strontium hydroxide yields two hydroxide ions per formula unit:
[OH–] = 2 × 1.54 × 10-4 = 3.08 × 10-4 M
Step 3: Calculate pOH
Use the standard definition:
pOH = -log[OH–]
Substitute the hydroxide concentration:
pOH = -log(3.08 × 10-4) ≈ 3.51
Step 4: Convert pOH to pH
At 25°C, the common classroom relationship is:
pH + pOH = 14.00
So:
pH = 14.00 – 3.51 = 10.49
Final answer: The pH of 1.54 × 10-4 M Sr(OH)2 is approximately 10.49.
Why the factor of 2 matters
A very common student error is to use the original concentration of Sr(OH)2 directly in the logarithm without accounting for the two hydroxide ions released during dissociation. If you ignore the factor of 2, your hydroxide concentration becomes too small, your pOH becomes too large, and your final pH becomes too low. The distinction matters because pH is logarithmic. Even a small stoichiometric mistake can create a noticeably different answer.
For example, if a student incorrectly used [OH–] = 1.54 × 10-4 M, then:
- Incorrect pOH = -log(1.54 × 10-4) ≈ 3.81
- Incorrect pH = 14.00 – 3.81 = 10.19
That is about 0.30 pH units lower than the correct answer, which is a substantial difference for a basic chemistry problem.
Complete method for strong base pH problems
The general method for compounds like Sr(OH)2, Ba(OH)2, or Ca(OH)2 is straightforward when you organize the work correctly:
- Identify whether the compound behaves as a strong base in the problem context.
- Write the dissociation equation.
- Determine how many OH– ions each formula unit contributes.
- Multiply the base molarity by that stoichiometric factor.
- Compute pOH using -log[OH–].
- Use pH = 14 – pOH at 25°C.
This workflow applies to a broad range of classroom examples and is especially useful when a polyhydroxide base is involved.
Comparison table: Correct vs incorrect setup for 1.54 × 10-4 M Sr(OH)2
| Method | [OH–] Used | pOH | pH | Comment |
|---|---|---|---|---|
| Correct stoichiometric method | 3.08 × 10-4 M | 3.51 | 10.49 | Accounts for 2 OH– per Sr(OH)2 |
| Common mistake | 1.54 × 10-4 M | 3.81 | 10.19 | Fails to multiply by 2 |
Interpreting the answer chemically
A pH of about 10.49 means the solution is clearly basic, but not as extreme as concentrated strong base solutions. On the pH scale, this puts the sample well above neutral water and into a moderately basic region. Since the pH scale is logarithmic, a solution with pH 10.49 has a much greater hydroxide ion concentration than a solution near pH 8 or 9.
Many students benefit from remembering that every whole pH unit corresponds to a tenfold change in hydrogen ion concentration. That means a pH 10.49 solution is not just “a little more basic” than pH 9.49; it is about 10 times more basic in terms of the inverse hydrogen ion scale under standard assumptions.
Where pH values fit on the common aqueous scale
At 25°C, pure water has a neutral pH of about 7.00. Acidic solutions fall below 7, and basic solutions rise above 7. The pH from 1.54 × 10-4 M Sr(OH)2 lands in a region often associated with laboratory bases that are dilute enough for calculation practice but still clearly alkaline.
| Solution Type | Typical pH Range | Interpretation | Approximate [H+] in mol/L |
|---|---|---|---|
| Strongly acidic solution | 0 to 3 | High hydrogen ion concentration | 1 to 1 × 10-3 |
| Weakly acidic to near neutral | 4 to 6 | Mild acidity | 1 × 10-4 to 1 × 10-6 |
| Neutral water at 25°C | 7 | Balanced [H+] and [OH–] | 1 × 10-7 |
| Moderately basic solution | 8 to 11 | Elevated hydroxide ion concentration | 1 × 10-8 to 1 × 10-11 |
| Strongly basic solution | 12 to 14 | Very high hydroxide ion concentration | 1 × 10-12 to 1 × 10-14 |
Important assumptions used in this calculation
- Complete dissociation: Introductory chemistry usually treats Sr(OH)2 as fully dissociated in water for this type of pH problem.
- Temperature at 25°C: The equation pH + pOH = 14.00 is tied to standard classroom conditions.
- Ideal behavior: Activity effects are ignored, which is normal in general chemistry unless the problem says otherwise.
- No competing equilibria: Carbon dioxide absorption, precipitation, and ionic strength effects are not included in the basic textbook approach.
Common mistakes students make
If you are reviewing for an exam or homework set, watch for these errors:
- Forgetting the dissociation stoichiometry. Sr(OH)2 produces two hydroxides, not one.
- Taking pH directly from base molarity. You need pOH first when OH– is given or derived.
- Dropping the negative sign in the logarithm. Since the concentration is less than 1, the logarithm is negative, and pOH must be positive.
- Mixing up pH and pOH. Strong base problems usually move from [OH–] to pOH, then to pH.
- Using too few significant figures. Because the concentration has three significant figures, reporting pH as 10.49 is appropriate in most classroom settings.
How this problem relates to water autoionization
One reason the pH + pOH relationship works is the autoionization of water. At 25°C, water establishes an equilibrium where the ion product Kw = 1.0 × 10-14. That allows the widely used relationship:
[H+][OH–] = 1.0 × 10-14
Taking negative logarithms of both sides gives:
pH + pOH = 14.00
If you want an authoritative reference for water quality and pH fundamentals, the U.S. Geological Survey provides an excellent overview at usgs.gov. Additional chemistry background can be found from chem.libretexts.org and academic resources such as Michigan State University.
How strong bases with multiple hydroxides compare
Polyhydroxide bases are a favorite topic because they test whether students understand mole relationships in aqueous dissociation. Here is the pattern:
- NaOH contributes 1 OH– per mole of base.
- Ca(OH)2 contributes 2 OH– per mole of base.
- Al(OH)3 appears to contribute 3 OH– in formula terms, although its chemistry is more complex and it is not treated the same way as a simple strong base in basic practice problems.
For straightforward strong base exercises, always count the number of hydroxide groups in the formula and verify whether the compound is assumed to dissociate completely.
Worked summary for the exact problem
Let us summarize the full solution in one compact chain:
- [Sr(OH)2] = 1.54 × 10-4 M
- [OH–] = 2(1.54 × 10-4) = 3.08 × 10-4 M
- pOH = -log(3.08 × 10-4) = 3.51
- pH = 14.00 – 3.51 = 10.49
That is the complete and correct answer under standard general chemistry assumptions.
Why calculators like this help
A dedicated pH calculator is valuable because it reduces arithmetic mistakes while still showing the chemistry logic. For example, students often know the formulas but make input errors when converting scientific notation or when multiplying by the hydroxide stoichiometric factor. An interactive calculator solves that problem by making each step visible:
- It converts scientific notation to decimal molarity.
- It multiplies by the number of hydroxide ions released.
- It computes pOH automatically.
- It converts to pH with standard formatting.
- It visualizes the result on a chart for easier interpretation.
Final takeaway
If you are asked to solve “1.54 10 4 M Sr(OH)2 calculate the pH”, the intended expression is almost always 1.54 × 10-4 M Sr(OH)2. Because strontium hydroxide contributes two hydroxide ions per formula unit, the hydroxide concentration is 3.08 × 10-4 M, the pOH is 3.51, and the final pH is 10.49. Once you understand that stoichiometric doubling step, these problems become much faster and far more reliable.