Solubility Product — Predict Solubility in Presence of a Common Ion
Question
Predict the solubility of the species below in water:
Lead(I) oxalate, PbC2O4, Ksp = 8.5 x 10-9
Show/Hide Answer
The solubility of PbC2O4 will be 9.2×10-5 M.
Refer to Section 7.5: Solubility Equilibria (1).
Strategy Map
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Check out the strategy map.
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| Strategy Map Steps |
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| 1. This species is a slightly soluble salt. We recognize that by the small value for the Ksp, the solubility product equilibrium constant. |
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. |
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. Plug in your given Ksp value and solve for ‘x’ algebraically. |
| 6. Then, relate ion concentration back to salt solubility. |
Solution
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[latex]\begin{gathered} \mathrm{PbC}_2 \mathrm{O}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+\mathrm{C}_2 \mathrm{O}_4^{2-}(\mathrm{aq}) \\ \mathrm{K}_{\mathrm{sp}}=[\mathrm{Pb}^{2+}][\mathrm{C}_2 \mathrm{O}_4^{2-}] \\\\ \mathrm{K}_{\mathrm{sp}}=8.5 \times 10^{-9}=[\mathrm{Pb}^{2+}][\mathrm{C}_2 \mathrm{O}_4^{2-}] \\ 8.5 \times 10^{-9}=(\mathrm{x})(\mathrm{x}) \\ 8.5 \times 10^{-9}=\mathrm{x}^2 \\ \sqrt{8.5 \times 10^{-9}}=\sqrt{\mathrm{x}^2} \\\\ 9.2 \times 10^{-5}=\mathrm{x} \\ \mathrm{x}=9.2 \times 10^{-5} \mathrm{M}=[\mathrm{Pb}^{2+}]=[\mathrm{C}_2 \mathrm{O}_4^{2-}] \end{gathered}[/latex]
The solubility of PbC2O4 is 9.2 x 10-5M.
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 is a calculation problem where you must set up the Ksp expression for the given slightly soluble salt and solve for the molar concentration of each species.
Show/Hide ResourceRefer to Section 7.5: Solubility Equilibria (1). |
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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 they will still be present in the solution. Break the salt into its cation and anion. Show/Hide Think About This!Look at the name of the compound. It will help you know the two ‘parts’ of the compound to break apart. |
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] |
To use a Ksp expression to solve for the concentrations of its species, relate the concentrations of the ions to each other and replace each species in the expression with a variable ‘x.’
Show/Hide Don’t Forget!Remember that the stoichiometry of each species will become the exponent for each x. (This is simple when the stoichiometric coefficient is one). |
| Then, look at the mole ratio and state the solubility of the salt to answer the question. |
| Complete Solution |
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| The solubility product reaction:
[latex]\begin{equation} \mathrm{PbC}_2 \mathrm{O}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+\mathrm{C}_2 \mathrm{O}_4^{2-}(\mathrm{aq}) \end{equation}[/latex] |
| The Ksp expression.
[latex]\begin{equation} \mathrm{K}_{\mathrm{sp}}=[\mathrm{Pb}^{2+}][\mathrm{C}_2 \mathrm{O}_4^{2-}] \end{equation}[/latex] |
| Plug in your value for the solubility product constant Ksp: [latex]\begin{equation} 8.5 \times 10^{-9}=[\mathrm{Pb}^{2+}][\mathrm{C}_2 \mathrm{O}_4^{2-}] \end{equation}[/latex] |
| Replace your solute species with x’s.
Since [latex]\begin{equation} [\mathrm{Pb}^{2+}]=[\mathrm{C}_2 \mathrm{O}_4^{2-}] \end{equation}[/latex] [latex]\begin{equation} 8.5 \times 10^{-9}=(\mathrm{x})(x)=\mathrm{x}^2 \end{equation}[/latex] |
| Solve for x algebraically
[latex]\begin{aligned} 8.5 \times 10^{-9} & =x^2 \\ \sqrt{8.5 \times 10^{-9}} & =\sqrt{x^2} \end{aligned}[/latex] [latex]\begin{equation} \text { 9. } 2 \times 10^{-5}=\mathrm{x}=[\mathrm{Pb}^{2+}]=[\mathrm{C}_2 \mathrm{O}_4^{2-}] \end{equation}[/latex] Since the concentration of each ion is 9.2 x 10-5 M, the solubility of PbC2O4 (aq) is 9.2 x 10-5 M. |
Check Your Work
Since this is a slightly soluble salt, we expect the solubility 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 Ksp value, you can calculate the molar concentrations for each species involved.
Does your answer make chemical sense?
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PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (2).
- Question 7.E.24(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.24(b) is modified from question 4 (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. Blackstock, L.; Brewer, S.; Jensen, A. In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.
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).
