Chapter 17: Problem 52
The solubility of an ionic compound MX (molar mass = \(346 \mathrm{~g}\) ) is \(4.63 \times 10^{-3} \mathrm{~g} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for this compound?
Short Answer
Expert verified
The solubility product constant, \(K_{sp}\), for the compound is \(1.80 \times 10^{-10}\).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ionic Compound
Ionic compounds are substances made up of charged particles called ions. These ions are held together by strong electrostatic forces in a solid lattice structure.
The compound MX in our exercise is an example of an ionic compound. It consists of cations (positively charged ions) and anions (negatively charged ions).
The compound MX in our exercise is an example of an ionic compound. It consists of cations (positively charged ions) and anions (negatively charged ions).
- The cation in this exercise is M⁺.
- The anion is X⁻.
Molar Solubility
Molar solubility represents the number of moles of a compound that can dissolve to form a liter of solution. It helps us understand how much of the compound can saturate a given volume of solvent.
In the exercise, the solubility of the compound MX is given in grams per liter, and it needs to be converted into molar solubility units (moles per liter).
To do this conversion, use the formula:\[ \text{Molar Solubility} = \frac{\text{Solubility in g/L}}{\text{Molar Mass}} \]This conversion is essential because it allows us to relate the physical amount of substance that dissolves with the chemical stoichiometry in the solution.
For MX, with the given values:\[ \frac{4.63 \times 10^{-3} \text{ g/L}}{346 \text{ g/mol}} = 1.34 \times 10^{-5} \text{ mol/L} \]Thus, the molar solubility of MX is \(1.34 \times 10^{-5} \text{ mol/L}\). This value is crucial as it determines the ion concentrations at equilibrium.
In the exercise, the solubility of the compound MX is given in grams per liter, and it needs to be converted into molar solubility units (moles per liter).
To do this conversion, use the formula:\[ \text{Molar Solubility} = \frac{\text{Solubility in g/L}}{\text{Molar Mass}} \]This conversion is essential because it allows us to relate the physical amount of substance that dissolves with the chemical stoichiometry in the solution.
For MX, with the given values:\[ \frac{4.63 \times 10^{-3} \text{ g/L}}{346 \text{ g/mol}} = 1.34 \times 10^{-5} \text{ mol/L} \]Thus, the molar solubility of MX is \(1.34 \times 10^{-5} \text{ mol/L}\). This value is crucial as it determines the ion concentrations at equilibrium.
Solubility Equilibrium
Solubility equilibrium occurs when a solid compound dissolves in a solvent until no more can dissolve, reaching a state of balance between dissolution and precipitation. For ionic compounds like MX, this means balancing between the solid and its ions in solution.
The dissociation reaction of MX can be written as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]In this scenario, the process is reversible, allowing ions to go back to solid form and vice versa, maintaining a dynamic equilibrium.
At equilibrium, the concentration of the dissolved ions remains constant because the rate of dissolution equals the rate of precipitation.
Understanding this concept is key to grasping how solubility products work, as the concentrations of the individual ions are crucial inputs in the calculation of the solubility product constant.
The dissociation reaction of MX can be written as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]In this scenario, the process is reversible, allowing ions to go back to solid form and vice versa, maintaining a dynamic equilibrium.
At equilibrium, the concentration of the dissolved ions remains constant because the rate of dissolution equals the rate of precipitation.
Understanding this concept is key to grasping how solubility products work, as the concentrations of the individual ions are crucial inputs in the calculation of the solubility product constant.
Ksp Calculation
The solubility product constant, or \(K_{sp}\), is a value that indicates the degree to which a compound will dissolve in water. It is determined at equilibrium concentrations of the ions in the solution.
For the ionic compound MX, whose dissociation is given as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]The expression for \(K_{sp}\) is:\[ K_{sp} = [ \text{M}^{+}][ \text{X}^{-}] \]In our exercise, both ion concentrations are equal due to the 1:1 stoichiometry, and equal to the molar solubility \(1.34 \times 10^{-5} \text{ mol/L}\).
Substituting these concentrations into the \(K_{sp}\) formula, we compute:\[ K_{sp} = (1.34 \times 10^{-5})(1.34 \times 10^{-5}) = 1.80 \times 10^{-10} \]This value helps predict how much of the compound can dissolve and form ions in solution under specific conditions. It’s a fundamental concept in chemistry that aids in understanding solubility and reactions in aqueous solutions.
For the ionic compound MX, whose dissociation is given as:\[ \text{MX}_{(s)} \rightleftharpoons \text{M}^{+}_{(aq)} + \text{X}^{-}_{(aq)} \]The expression for \(K_{sp}\) is:\[ K_{sp} = [ \text{M}^{+}][ \text{X}^{-}] \]In our exercise, both ion concentrations are equal due to the 1:1 stoichiometry, and equal to the molar solubility \(1.34 \times 10^{-5} \text{ mol/L}\).
Substituting these concentrations into the \(K_{sp}\) formula, we compute:\[ K_{sp} = (1.34 \times 10^{-5})(1.34 \times 10^{-5}) = 1.80 \times 10^{-10} \]This value helps predict how much of the compound can dissolve and form ions in solution under specific conditions. It’s a fundamental concept in chemistry that aids in understanding solubility and reactions in aqueous solutions.
