Calculate Solubility from the Following Ksp Values Sr3(PO4)2
Use this premium calculator to convert the solubility product constant, Ksp, of strontium phosphate into molar solubility and equilibrium ion concentrations using the correct stoichiometric relationship for Sr3(PO4)2.
Sr3(PO4)2 Solubility Calculator
Ksp = [Sr2+]3[PO43-]2 = (3s)3(2s)2 = 108s5
Therefore, molar solubility s = (Ksp / 108)1/5
Equilibrium Concentration Chart
Expert Guide: How to Calculate Solubility from the Ksp of Sr3(PO4)2
When students and lab professionals search for how to calculate solubility from the following Ksp values Sr3(PO4)2, they are usually trying to convert a very small solubility product constant into a meaningful chemical quantity. The quantity most often requested is molar solubility, which tells you how many moles of strontium phosphate dissolve per liter of water at equilibrium. This is a classic equilibrium problem in general chemistry, analytical chemistry, and water chemistry because strontium phosphate is a sparingly soluble ionic solid. The key to solving it correctly is not simply plugging the Ksp into a calculator. You must first write the balanced dissolution equation, define the molar solubility variable, and apply the exponents that come from the stoichiometric coefficients.
Strontium phosphate has the formula Sr3(PO4)2. When it dissolves, each formula unit releases three strontium ions and two phosphate ions. That means the equilibrium expression is more involved than for a 1:1 salt such as AgCl. The correct dissolution reaction is:
Because solids do not appear in the equilibrium expression, the Ksp expression includes only the aqueous ions:
Now let the molar solubility of Sr3(PO4)2 be s mol/L. At equilibrium, the concentration of strontium ions will be 3s and the concentration of phosphate ions will be 2s. Substitute those expressions into the Ksp formula:
This simplifies to the most important working equation for this compound:
Why stoichiometry matters so much
The biggest source of errors in Sr3(PO4)2 Ksp problems is forgetting that the coefficients become exponents in the equilibrium expression. If someone uses Ksp = s2 or Ksp = s3, the result can be off by many orders of magnitude. For compounds like strontium phosphate, calcium phosphate, or aluminum phosphate, stoichiometry controls the shape of the final equation. Since Ksp values are often extremely small, even a minor algebra mistake produces a very large relative error in the solubility estimate.
For example, if Ksp = 1.0 x 10-31, then:
- Write the balanced dissolution equation.
- Set [Sr2+] = 3s and [PO43-] = 2s.
- Substitute into Ksp: Ksp = (3s)3(2s)2 = 108s5.
- Rearrange: s5 = Ksp / 108.
- Take the fifth root: s = (Ksp / 108)1/5.
This process gives the molar solubility in mol/L. Once you know s, you can immediately find each ion concentration:
- [Sr2+] = 3s
- [PO43-] = 2s
Worked example using a typical tiny Ksp value
Suppose your chemistry homework or lab sheet gives a Ksp for strontium phosphate of 1.0 x 10-31 at 25 C. Start with the formula:
First divide the Ksp by 108. The result is approximately 9.26 x 10-34. Now take the fifth root. The molar solubility is about 2.47 x 10-7 M. From that:
- [Sr2+] = 3s ≈ 7.41 x 10-7 M
- [PO43-] = 2s ≈ 4.94 x 10-7 M
This example illustrates something important about low Ksp salts: a very tiny Ksp does not mean zero dissolution. It means the equilibrium concentration of ions is small, but still measurable and chemically meaningful. In precipitation reactions, these small concentrations can strongly influence whether a solid forms, whether scaling occurs, or whether an analytical separation is successful.
Comparison table: Ksp versus molar solubility for Sr3(PO4)2
The relationship between Ksp and solubility is not linear here because the equation includes the fifth root. That means a dramatic increase in Ksp leads to a more moderate increase in solubility. The comparison below shows this behavior.
| Ksp value | Calculated molar solubility, s (mol/L) | [Sr2+] at equilibrium (mol/L) | [PO43-] at equilibrium (mol/L) |
|---|---|---|---|
| 1.0 x 10-31 | 2.47 x 10-7 | 7.41 x 10-7 | 4.94 x 10-7 |
| 5.0 x 10-33 | 1.40 x 10-7 | 4.20 x 10-7 | 2.80 x 10-7 |
| 2.5 x 10-29 | 7.11 x 10-7 | 2.13 x 10-6 | 1.42 x 10-6 |
Common mistakes in Sr3(PO4)2 solubility calculations
- Using the wrong dissolution equation. The correct reaction produces 3 Sr2+ and 2 PO43-. If you write the reaction incorrectly, every later step is wrong.
- Forgetting exponents. Ksp is [Sr2+]3[PO43-]2, not [Sr2+][PO43-].
- Using s instead of 3s and 2s. For this compound, the ion concentrations are not equal to the molar solubility.
- Ignoring temperature. Ksp values are temperature dependent. Always match your calculation to the stated temperature when possible.
- Mixing molar solubility with mass solubility. Molar solubility is in mol/L. If you need g/L, multiply by the molar mass of Sr3(PO4)2.
Mass solubility and practical interpretation
Sometimes a teacher or technician wants the answer in grams per liter instead of moles per liter. In that case, after calculating s, multiply by the molar mass of strontium phosphate. Using approximate atomic masses, the molar mass of Sr3(PO4)2 is about 452.8 g/mol. If the molar solubility were 2.47 x 10-7 mol/L, the mass solubility would be roughly:
That is a very small mass concentration, which is consistent with the notion that strontium phosphate is sparingly soluble in pure water. In applied settings such as water treatment or geochemistry, low-solubility phosphates can contribute to mineral precipitation, scale formation, and nutrient immobilization.
How common ions affect the apparent solubility
The formula s = (Ksp / 108)1/5 applies most directly in pure water, where the only significant ion source is the dissolving solid itself. If a solution already contains Sr2+ or PO43- from another source, the solubility decreases because of the common ion effect. In those cases, the simple substitution [Sr2+] = 3s and [PO43-] = 2s may no longer hold. Instead, you include the initial ion concentration in an ICE table and solve the modified equilibrium expression. This is especially relevant in buffer systems, biological media, hard water, and phosphate-rich solutions.
Comparison table: Sr3(PO4)2 versus simpler salt behavior
One reason this problem feels difficult is that many introductory examples begin with 1:1 salts. The table below compares how the stoichiometric form changes the Ksp math.
| Salt | Dissolution reaction | Ksp expression | Solubility form in pure water |
|---|---|---|---|
| AgCl | AgCl(s) ⇌ Ag+ + Cl- | Ksp = [Ag+][Cl-] | s = (Ksp)1/2 |
| CaF2 | CaF2(s) ⇌ Ca2+ + 2F- | Ksp = [Ca2+][F-]2 | s = (Ksp / 4)1/3 |
| Sr3(PO4)2 | Sr3(PO4)2(s) ⇌ 3Sr2+ + 2PO43- | Ksp = [Sr2+]3[PO43-]2 | s = (Ksp / 108)1/5 |
What authoritative sources say about Ksp and solubility
For reliable chemical data and instructional support, it is best to cross-check equilibrium concepts with authoritative educational and government resources. The following sources are particularly useful for understanding solubility, ionic equilibria, and the role of Ksp in chemical systems:
- Chemistry LibreTexts educational resource for step by step treatment of solubility product calculations.
- U.S. Environmental Protection Agency for water chemistry context, ionic contaminants, and precipitation relevance in environmental systems.
- NIST Chemistry WebBook for broader chemical property reference and standards-oriented data support.
How to check whether your final answer makes sense
A good chemistry calculation does not stop at arithmetic. You should also sanity check the result. If Ksp is extremely small, the molar solubility should also be small. If your answer comes out larger than 0.01 M for a Ksp around 10-31, something is definitely wrong. Also verify that your ion concentrations satisfy the stoichiometric ratio of 3:2. If your calculated [Sr2+] is not exactly 1.5 times [PO43-], then the setup likely contains an error.
Another useful check is to plug your calculated values back into the Ksp expression:
If the product matches the original Ksp to reasonable rounding precision, your answer is internally consistent. This reverse check is simple, fast, and highly effective during exams and lab work.
Final takeaway
To calculate solubility from the following Ksp values Sr3(PO4)2, always begin with the balanced dissolution equation. For strontium phosphate, the stoichiometry gives [Sr2+] = 3s and [PO43-] = 2s, which leads to Ksp = 108s5. Solving for s gives the compact formula s = (Ksp / 108)1/5. Once you know s, the equilibrium ion concentrations follow immediately. This approach is the correct, professional way to solve pure-water Ksp problems for Sr3(PO4)2, and it scales well to homework, laboratory analysis, and environmental chemistry interpretation.