7.54 × 10-4 M Sr(OH)2 pH Calculator
Quickly calculate pH, pOH, hydroxide concentration, and dissociation-based values for strontium hydroxide solutions using a premium chemistry calculator and expert guide.
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Enter the concentration and click Calculate pH to see the answer for 7.54 × 10-4 M Sr(OH)2.
How to calculate the pH of 7.54 × 10-4 M Sr(OH)2
If you are trying to solve the chemistry question “7.54 10 4 m sr oh 2 calculate ph,” the intended expression is almost always 7.54 × 10-4 M Sr(OH)2. This means the solution concentration of strontium hydroxide is 0.000754 moles per liter. Because strontium hydroxide is a strong base in standard general chemistry treatment, it dissociates essentially completely in water:
Sr(OH)2 → Sr2+ + 2OH–
The key point is that each mole of Sr(OH)2 produces two moles of hydroxide ions. Since pH for a basic solution is found through hydroxide concentration, you first convert the formal base concentration into [OH–], then calculate pOH, and finally convert pOH to pH.
Final answer for 7.54 × 10-4 M Sr(OH)2
- Start with the base concentration: [Sr(OH)2] = 7.54 × 10-4 M
- Because each unit gives 2 hydroxide ions: [OH–] = 2 × 7.54 × 10-4 = 1.508 × 10-3 M
- Calculate pOH: pOH = -log(1.508 × 10-3) ≈ 2.82
- Use pH + pOH = 14.00 at 25 degrees C
- Therefore: pH = 14.00 – 2.82 = 11.18
Why Sr(OH)2 changes the pH so much
Strontium hydroxide belongs to the family of metal hydroxides that produce hydroxide ions in water. In aqueous chemistry, hydroxide concentration directly controls how basic a solution is. The more OH– present, the lower the pOH and the higher the pH.
Sr(OH)2 is especially important for students because it is a dibasic strong base. “Dibasic” means each dissolved formula unit contributes two hydroxide ions. Compare that with sodium hydroxide, NaOH, which contributes only one hydroxide ion per formula unit. At the same molarity, Sr(OH)2 creates approximately twice the hydroxide concentration of NaOH, assuming complete dissociation.
Step-by-step breakdown of the chemistry
Let’s unpack the method in a way you can reuse on quizzes, homework, labs, and exam problems.
- Step 1: Identify whether the substance is an acid or a base.
- Step 2: Decide whether it dissociates completely or partially under the level of chemistry you are studying.
- Step 3: Determine how many H+ or OH– ions are released per formula unit.
- Step 4: Convert the compound molarity into ion concentration.
- Step 5: Use logarithms to compute pH or pOH.
- Step 6: Check whether the final answer makes chemical sense.
For this problem, the logic is straightforward because Sr(OH)2 is treated as a strong base in general chemistry. That means no ICE table is usually required, and you can proceed directly from concentration to hydroxide ion concentration.
Detailed formula set
Here are the exact equations used in this calculator:
- [OH–] = n × C, where n is the number of hydroxide ions per formula unit and C is the base molarity
- pOH = -log10[OH–]
- pH = 14.00 – pOH at 25 degrees C
For the query value:
- C = 7.54 × 10-4 M
- n = 2
- [OH–] = 2 × 7.54 × 10-4 = 1.508 × 10-3 M
- pOH = -log(1.508 × 10-3) = 2.8216
- pH = 14 – 2.8216 = 11.1784
Comparison table: common strong bases and OH– release
| Base | Formula | OH- Released per Formula Unit | At 1.00 × 10^-3 M Base, [OH-] | Approx. pOH | Approx. pH at 25 degrees C |
|---|---|---|---|---|---|
| Sodium hydroxide | NaOH | 1 | 1.00 × 10^-3 M | 3.00 | 11.00 |
| Potassium hydroxide | KOH | 1 | 1.00 × 10^-3 M | 3.00 | 11.00 |
| Calcium hydroxide | Ca(OH)2 | 2 | 2.00 × 10^-3 M | 2.70 | 11.30 |
| Strontium hydroxide | Sr(OH)2 | 2 | 2.00 × 10^-3 M | 2.70 | 11.30 |
| Barium hydroxide | Ba(OH)2 | 2 | 2.00 × 10^-3 M | 2.70 | 11.30 |
This comparison helps explain why Sr(OH)2 gives a higher pH than a monohydroxide base of the same molarity. The chemistry is not mysterious: it simply injects more hydroxide into the solution.
Is autoionization of water important here?
Water naturally self-ionizes to a very small extent, producing about 1.0 × 10-7 M each of H+ and OH– at 25 degrees C. In this problem, the hydroxide concentration coming from Sr(OH)2 is 1.508 × 10-3 M, which is much larger than 1.0 × 10-7 M. Because the base contribution is over ten thousand times larger than water’s contribution, autoionization is negligible for this calculation.
That is why the simple strong-base method works so well here.
Table of pH reference points from real standard chemistry values
| Quantity | Standard Value at 25 degrees C | Why It Matters |
|---|---|---|
| pH of pure water | 7.00 | Neutral reference point under standard conditions |
| pOH of pure water | 7.00 | Shows pH + pOH = 14.00 |
| Kw | 1.0 × 10^-14 | Defines the relationship between H+ and OH– |
| [H+] in pure water | 1.0 × 10^-7 M | Used to derive neutral pH |
| [OH–] in pure water | 1.0 × 10^-7 M | Used to derive neutral pOH |
| [OH–] for this Sr(OH)2 problem | 1.508 × 10^-3 M | Confirms the solution is strongly basic relative to pure water |
Common mistakes students make
- Forgetting the coefficient of 2. Sr(OH)2 produces two hydroxide ions, not one.
- Using pH = -log[OH-]. That gives pOH, not pH.
- Dropping the negative exponent incorrectly. 10-4 means move the decimal four places left.
- Rounding too early. Keep several digits until the final step.
- Ignoring temperature assumptions. The relation pH + pOH = 14.00 is for 25 degrees C in standard introductory problems.
How to check your answer quickly
A good chemistry student always performs a reasonableness check. Here is a fast way to see if your answer is plausible:
- Sr(OH)2 is a base, so the pH must be greater than 7.
- The concentration is on the order of 10-4 M, but because there are 2 OH– ions, the hydroxide concentration becomes about 10-3 M.
- If [OH–] is about 10-3, then pOH should be near 3.
- If pOH is near 3, then pH should be near 11.
The exact answer, 11.18, fits this estimate perfectly.
When this simple method might need refinement
In advanced chemistry, some pH calculations become more complicated because of activity effects, incomplete dissociation, temperature dependence, or solubility limits. However, for the problem “7.54 × 10-4 M Sr(OH)2 calculate pH,” the standard classroom expectation is the strong-base dissociation model used above.
If you are working in an analytical chemistry or physical chemistry context, you may also consider:
- Activity coefficients in ionic solutions
- Non-ideal behavior at higher ionic strength
- Temperature-dependent values of Kw
- Whether the dissolved concentration is constrained by solubility in a specific experimental setup
For normal homework and textbook questions, those refinements are usually unnecessary unless your instructor explicitly requests them.
Authoritative chemistry references
If you want trusted supporting information on pH, pOH, water chemistry, and laboratory safety, review these authoritative resources:
- U.S. Environmental Protection Agency: pH Overview
- Chemistry LibreTexts Educational Resource
- NIST Chemistry WebBook
Practical interpretation of the result
A pH of about 11.18 means the solution is distinctly basic. In practical laboratory terms, that means it can be corrosive to skin and eyes and should be handled using standard base safety precautions, including gloves, goggles, and appropriate spill protocols. Strong hydroxide solutions are chemically active because hydroxide ions participate readily in neutralization reactions, precipitation chemistry, and numerous inorganic equilibria.
This value also shows why polyhydroxide bases are important. Even when the molarity of the compound appears modest, the actual hydroxide concentration can be substantially higher than expected if more than one OH– is released per dissolved formula unit.
Quick recap
- Interpret the problem as 7.54 × 10-4 M Sr(OH)2
- Sr(OH)2 is treated as a strong base
- Each mole gives 2 moles of OH–
- [OH–] = 1.508 × 10-3 M
- pOH = 2.82
- pH = 11.18
If you want a fast answer, the final result is simple: the pH is 11.18. If you want to understand the chemistry behind it, remember that the decisive step is doubling the hydroxide concentration because Sr(OH)2 contains two hydroxide groups.