Solubility Product: Calculate Solubility Product Equilibrium Constant
Question
From the solubility data given, calculate Ksp:
SrF2, 1.2 x 10-2 g/100 mL
Show/Hide Answer
Ksp = 3.66 × 10-9
Refer to Section 7.5: Solubility Equilibria (1).
Strategy Map
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| Strategy Map Steps |
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| 1. This species is a slightly soluble salt. The solubility is a small number, so we expect a small value for the solubility product equilibrium constant (Ksp). |
2. Write the Ksp reaction showing the solubility product equilibrium.
Show/Hide HintBreak the salt into its cation and anion, making sure the reaction is balanced. Use equilibrium arrows ⇌ to indicate that the reaction does not go fully to products. |
3. Write the Ksp expression.
Show/Hide HintRecall when writing expressions for K that solids are not included. Remember where AND how many times you represent the stoichiometry of each solute |
4. Using your Ksp expression, relate the concentrations of ions in terms of a variable (‘x’).
Show/Hide HintFor every mole of a salt that dissolves, look at how many moles of each ion are formed. |
| 5. Solubility is given in g/100 mL; convert it to M (moles per litre). |
| 6. Plug in your given solubility and solve for ‘x’ algebraically. |
Solution
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[latex]\mathrm{SrF}_2\,(\mathrm{s})\rightleftharpoons\mathrm{Sr}^{2+}(\mathrm{aq})+2\mathrm{F}^-(\mathrm{aq})[/latex]
Convert solubility:
[latex]\begin{gathered} \frac{1.22\times 10^{-2}\mathrm{~g}}{100\mathrm{~ml}}\times\frac{1\mathrm{~mol}}{125.616\mathrm{~g}}\times\frac{1000\mathrm{~ml}}{1\mathrm{~L}}=9.71\times 10^{-4} \mathrm{~M} \end{gathered}[/latex]
[latex]\begin{aligned} \mathrm{K}_\mathrm{sp}&=[\mathrm{Sr}^{2+}][\mathrm{F}^{-}]^2 \\ \mathrm{K}_\mathrm{sp}&=(\mathrm{x})(2\mathrm{x})^2 \\ \mathrm{K}_\mathrm{sp}&=(\mathrm{x})(4\mathrm{x}^2) \\ \mathrm{K}_\mathrm{sp}&=4\mathrm{x}^3 \end{aligned}[/latex]
[latex]\begin{gathered} \mathrm{x}=9.71\times10^{-4}\mathrm{~M}\\ \mathrm{K}_\mathrm{sp}=4(9.71\times10^{-4})^3=3.66\times10^{-9} \end{gathered}[/latex]
Answer: The Ksp of SrF2 is 3.66 × 10-9
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
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| Guided Solution Ideas |
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This question is a calculation problem where you must set up the Ksp expression for the given slightly soluble salt and solve for the Ksp given the solubility of the slightly soluble salt.
Show/Hide ResourceRefer to Section 7.5: Solubility Equilibria (1). |
| When a salt dissolves in water, it breaks into its ions. This is a slightly soluble salt, so the concentrations of each ion will be small but still present in the solution.
A Ksp reaction will have the solid salt as the reactant and the cation and anion as the product. Show/Hide Think About This!When the salt dissolves, the cation and anion will be in the solution. Make sure you recognize how many of each ion ends up in the solution. |
Looking at your reaction, write the Ksp expression.
Show/Hide Think About This!A Ksp expression is the equilibrium constant expression for a slightly soluble salt. Since the reactant is a solid, it does not appear in the Ksp expression. [latex]\begin{aligned} \mathrm{AB}(\mathrm{s})&\rightleftharpoons \mathrm{A}^{+}(\mathrm{aq})+\mathrm{B}^{-}(\mathrm{aq})\\ \mathrm{K}_{\mathrm{sp}}&=[\mathrm{A}^{+}][\mathrm{B}^{-}] \end{aligned}[/latex] [latex]\begin{aligned} \mathrm{SrF_2}(\mathrm{s})&\rightleftharpoons \mathrm{Sr}^{2+}(\mathrm{aq})+\mathrm{2F}^{-}(\mathrm{aq})\\ \mathrm{K}_{\mathrm{sp}}&=[\mathrm{Sr}^{2+}]([\mathrm{F}^{-}])^2 \end{aligned}[/latex] |
You were provided with a value of 1.22 × 10−2 g/100 mL. Watch your units!
Show/Hide Don’t Forget!Convert solubility to molarity (M). You will plug this value in for ‘x’ to solve for your Ksp. Show/Hide Don’t Forget!Remember that the stoichiometric coefficient of each species will become its exponent. |
| Remember that your stoichiometric coefficient will show up in 2 places. Each SrF2 that dissolves produces 2 fluoride ions in the solution. It is also the exponent in the Ksp expression for fluoride. |
| Complete Solution |
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| The solubility product reaction:
[latex]\mathrm{SrF}_2\,(\mathrm{s})\rightleftharpoons\mathrm{Sr}^{2+}(\mathrm{aq})+2\mathrm{F}^-(\mathrm{aq})[/latex] |
| The Ksp expression:
[latex]\begin{gathered} \mathrm{K}_{\mathrm{sp}}=[\mathrm{Sr}^{2+}][\mathrm{2F}^{-}] \end{gathered}[/latex] |
| Convert solubility from g/100 mL to moles per L (M).
The molar mass of SrF2 is: [latex]87.62\mathrm{~g/mol}+2(19.00\mathrm{~g/mol})=125.62\mathrm{~g/mol}[/latex] [latex]\begin{gathered} \frac{1.22\times 10^{-2}\mathrm{~g}}{100\mathrm{~ml}}\times\frac{1\mathrm{~mol}}{125.616\mathrm{~g}}\times\frac{1000\mathrm{~ml}}{1\mathrm{~L}}=9.71\times 10^{-4} \mathrm{~M} \end{gathered}[/latex] |
| Write the Ksp expression and replace your solute species with ‘x,’ representing solubility:
[latex]\begin{aligned} \mathrm{K}_\mathrm{sp}&=[\mathrm{Sr}^{2+}][\mathrm{F}^{-}]^2 \\ \mathrm{K}_\mathrm{sp}&=(\mathrm{x})(2\mathrm{x})^2 \\ \mathrm{K}_\mathrm{sp}&=(\mathrm{x})(4\mathrm{x}^2) \\ \mathrm{K}_\mathrm{sp}&=4\mathrm{x}^3 \end{aligned}[/latex] Show/Hide Watch Out!Remember that the stoichiometric coefficient of F shows up twice! For each SrF2 that dissolves, we get 2 F– in the solution; so, the solubility of F– is 2x. It also shows up as the exponent for F– in the Ksp expression. [latex]\mathrm{x}=\text{solubility}=9.71\times10^{-4}\mathrm{~M}[/latex] Plug in value of x and solve for Ksp: [latex]\mathrm{K}_\mathrm{sp}=4(9.71\times10^{-4})^3=3.66\times10^{-9}[/latex] Answer: The Ksp of SrF2 is 3.66 × 10-9 |
Check Your Work
Since this is a slightly soluble salt, we expect the Ksp to be a small numerical value.
The equilibrium constant for a dissolution reaction, known as the solubility product (Ksp), is defined in terms of the molar concentrations of the component ions raised to the appropriate stoichiometry. The Ksp expression also takes into account the mole ratio of the ions. When given the slightly soluble salt solubility, the Ksp can be calculated.
Does your answer make chemical sense?
Show/Hide Answer
The value provided in the question was the concentration of solute or solubility. When the resulting concentration of cation and anion is plugged into the Ksp expression, you can calculate the Ksp.
In chemistry, we use this to identify how soluble a salt solution will be. Slightly soluble salts favour the solid reactant when at equilibrium in water.
We expected a small value (much less than 1) for the Ksp and our number is small.
PASS Attribution
- LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (2)
- Question 7.E.25 (b) from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (3) is used under a CC BY-NC-SA 3.0 license.
- Question 7.E.25(b) is question 12b (17.4 Solubility Equilibria, Numerical Problems) from LibreTexts Exercises: Brown et al. (4), which is under a CC BY-NC-SA 4.0 license.
References
1. Thompson Rivers University. 7.5: Solubility Equilibria. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/07%3A_Buffers_Titrations_and_Solubility_Equilibria/7.05%3A_Solubility_Equilibria.
2. Thompson Rivers University. TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520).
3. Thompson Rivers University. 7.E: Buffers, Titrations and Solubility Equilibria (Exercises). In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/07%3A_Buffers_Titrations_and_Solubility_Equilibria/7.E%3A_Buffers_Titrations_and_Solubility_Equilibria_(Exercises).
4. LibreTexts. 17.E: Additional Aspects of Aqueous Equilibria (Exercises). In Exercises: Brown et al; LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./17.E%3A_Additional_Aspects_of_Aqueous_Equilibria_(Exercises).
