Calculating the Molar Solubility The solubility of a substance is the quantity that dissolves to form a saturated solution and can be represented by the solubility product expression. The solubility product expression is equivalent to the product of the concentrations after each is raised to a power that is equal to its coefficient in the balanced equilibrium equation. For example, see the following reaction involving calcium phosphate and its corresponding solubility product expression: Ca3(PO4)2(s) = 3Ca2+(aq) + 2PO4(aq) Ksp = [Ca2+]^3 * [PO4]^2 The molar solubility of a slightly soluble salt is the number of moles of solute that dissolve per liter of solution and has the units M. For the example above, every mole of solute that dissolves produces 3 moles of Ca2+ and 2 moles of PO4, so the molar solubility is equivalent to [Ca2+]^3 / [PO4]^2. If MgCrO4 were added to water to form 1 L of solution, which species’ concentration will be equal to the molar solubility? Check all that apply: CrO4, Mg2+, MgCrO4
The Correct Answer and Explanation is:
In this scenario, we are dealing with the solubility product expression (Ksp) and molar solubility of a salt, MgCrO4. The general form of the solubility product expression for an ionic compound is:Salt (solid)⇌Cation (aqueous)+Anion (aqueous)\text{Salt (solid)} \rightleftharpoons \text{Cation (aqueous)} + \text{Anion (aqueous)}Salt (solid)⇌Cation (aqueous)+Anion (aqueous)
Step-by-Step Explanation
For magnesium chromate (MgCrO4), the dissociation in water would be:MgCrO4 (s)⇌Mg2+(aq)+CrO42−(aq)\text{MgCrO4 (s)} \rightleftharpoons \text{Mg}^{2+} (aq) + \text{CrO4}^{2-} (aq)MgCrO4 (s)⇌Mg2+(aq)+CrO42−(aq)
The solubility product expression (Ksp) for this reaction would be:Ksp=[Mg2+][CrO42−]\text{Ksp} = [\text{Mg}^{2+}] [\text{CrO4}^{2-}]Ksp=[Mg2+][CrO42−]
Where:
- [Mg2+][ \text{Mg}^{2+} ][Mg2+] is the concentration of magnesium ions in solution.
- [CrO42−][ \text{CrO4}^{2-} ][CrO42−] is the concentration of chromate ions in solution.
Molar Solubility
The molar solubility is defined as the number of moles of solute (MgCrO4) that dissolve in 1 liter of solution to form a saturated solution. Let’s assume the molar solubility of MgCrO4 is s, which means that s moles of MgCrO4 dissolve in 1 liter of solution.
For each mole of MgCrO4 that dissolves:
- 1 mole of Mg2+\text{Mg}^{2+}Mg2+ is produced, so the concentration of Mg2+\text{Mg}^{2+}Mg2+ will be s.
- 1 mole of CrO42−\text{CrO4}^{2-}CrO42− is produced, so the concentration of CrO42−\text{CrO4}^{2-}CrO42− will also be s.
Therefore, the concentrations of the species in the solution are:
- [Mg2+]=s[ \text{Mg}^{2+} ] = s[Mg2+]=s
- [CrO42−]=s[ \text{CrO4}^{2-} ] = s[CrO42−]=s
Which species’ concentration will be equal to the molar solubility?
The molar solubility s refers to the number of moles of MgCrO4 that dissolve to give the concentrations of the ions. Therefore, the concentration of both Mg²⁺ and CrO₄²⁻ ions will be equal to the molar solubility s.
Answer:
- The concentration of Mg²⁺ will be equal to the molar solubility.
- The concentration of CrO₄²⁻ will be equal to the molar solubility.
- The concentration of MgCrO4 itself, being a solid, does not directly correspond to the molar solubility in the solution. Thus, MgCrO4 will not have a concentration equal to the molar solubility.
Conclusion:
- The correct species whose concentration will be equal to the molar solubility are Mg²⁺ and CrO₄²⁻.
