Chapter 17: Problem 53
The solubility of an ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3}\) (molar mass \(=288 \mathrm{~g}\) ) is \(3.6 \times 10^{-17} \mathrm{~g} / \mathrm{L}\). What is \(K_{\mathrm{sp}}\) for this compound?
Short Answer
Expert verified
The \( K_{sp} \) for \( \mathrm{M}_{2} \mathrm{X}_{3} \) is \( 3.91 \times 10^{-93} \).
Step by step solution
Over 22 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product
A solubility product, denoted as \(K_{sp}\), is an important concept in chemistry that describes the solubility of an ionic compound in water. It provides a numerical value for the equilibrium constant in dissolution reactions. This value helps predict how much of a compound can dissolve in water, forming ions.
For any ionic compound, the solubility product expression involves the product of the ion concentrations each raised to the power of their stoichiometric coefficients. It essentially represents the point at which the rates of dissolution and precipitation of a compound are balanced.
In the given exercise, we calculate \(K_{sp}\) to understand how much of \( \mathrm{M}_{2} \mathrm{X}_{3}\) can dissolve in water before reaching saturation. Since \(K_{sp}\) numbers are typically very small, they indicate that only a limited amount of the compound can dissolve before reaching an equilibrium state.
For any ionic compound, the solubility product expression involves the product of the ion concentrations each raised to the power of their stoichiometric coefficients. It essentially represents the point at which the rates of dissolution and precipitation of a compound are balanced.
In the given exercise, we calculate \(K_{sp}\) to understand how much of \( \mathrm{M}_{2} \mathrm{X}_{3}\) can dissolve in water before reaching saturation. Since \(K_{sp}\) numbers are typically very small, they indicate that only a limited amount of the compound can dissolve before reaching an equilibrium state.
Ionic Compound Dissolution
When ionic compounds dissolve in water, they separate into individual ions in a process known as dissolution. The extent and behavior of this dissolution depend on the compound’s nature and the solution conditions.
For the compound \(\mathrm{M}_{2} \mathrm{X}_{3}\), it dissolves according to the reaction: \(\mathrm{M}_{2} \mathrm{X}_{3} (s) \leftrightarrow 2 \mathrm{M}^{3+} (aq) + 3 \mathrm{X}^{2-} (aq)\). Here, two ions of \(\mathrm{M}^{3+}\) and three ions of \(\mathrm{X}^{2-}\) are formed for every molecule of \(\mathrm{M}_{2} \mathrm{X}_{3}\) that dissolves.
Understanding this reaction is crucial since the ions formed will dictate the concentrations needed for the \(K_{sp}\) expression. Additionally, the dissolution of ionic compounds involves overcoming lattice energy, which is the energy needed to separate the ions in the crystalline solid form.
For the compound \(\mathrm{M}_{2} \mathrm{X}_{3}\), it dissolves according to the reaction: \(\mathrm{M}_{2} \mathrm{X}_{3} (s) \leftrightarrow 2 \mathrm{M}^{3+} (aq) + 3 \mathrm{X}^{2-} (aq)\). Here, two ions of \(\mathrm{M}^{3+}\) and three ions of \(\mathrm{X}^{2-}\) are formed for every molecule of \(\mathrm{M}_{2} \mathrm{X}_{3}\) that dissolves.
Understanding this reaction is crucial since the ions formed will dictate the concentrations needed for the \(K_{sp}\) expression. Additionally, the dissolution of ionic compounds involves overcoming lattice energy, which is the energy needed to separate the ions in the crystalline solid form.
Molar Solubility
Molar solubility is a measure of the number of moles of solute that can dissolve in a liter of solution until the saturation point is reached. In our context, it's essential to convert the given solubility from grams per liter to moles per liter.
For the compound \(\mathrm{M}_{2} \mathrm{X}_{3}\), the conversion step helps us express its solubility in terms of molarity, which is then used to calculate ion concentrations. In this exercise, the molar solubility is calculated as \(1.25 \times 10^{-19} \mathrm{~mol/L}\), showing the amount of \(\mathrm{M}_{2} \mathrm{X}_{3}\) that can dissolve before saturation is reached.
Recognizing molar solubility is crucial because it sets the foundation for understanding how much of each ion is present in the solution, thus feeding directly into the calculation of the solubility product.
For the compound \(\mathrm{M}_{2} \mathrm{X}_{3}\), the conversion step helps us express its solubility in terms of molarity, which is then used to calculate ion concentrations. In this exercise, the molar solubility is calculated as \(1.25 \times 10^{-19} \mathrm{~mol/L}\), showing the amount of \(\mathrm{M}_{2} \mathrm{X}_{3}\) that can dissolve before saturation is reached.
Recognizing molar solubility is crucial because it sets the foundation for understanding how much of each ion is present in the solution, thus feeding directly into the calculation of the solubility product.
Ion Concentration Calculations
Once you know the molar solubility, you can calculate the concentration of each ion in the solution. This involves multiplying the molar solubility by the number of each ion produced per formula unit of the compound.
For \(\mathrm{M}_{2} \mathrm{X}_{3}\), the dissolution produces 2 \(\mathrm{M}^{3+}\) ions and 3 \(\mathrm{X}^{2-}\) ions. Calculations give us \([\mathrm{M}^{3+}] = 2.5 \times 10^{-19} \mathrm{~M}\) and \([\mathrm{X}^{2-}] = 3.75 \times 10^{-19} \mathrm{~M}\).
These concentrations are crucial for computing the \(K_{sp}\) because they indicate the proportion of ions in the solution. By carefully calculating these values, you ensure that the \(K_{sp}\) reflects the true solubility equilibrium of the ionic compound.
For \(\mathrm{M}_{2} \mathrm{X}_{3}\), the dissolution produces 2 \(\mathrm{M}^{3+}\) ions and 3 \(\mathrm{X}^{2-}\) ions. Calculations give us \([\mathrm{M}^{3+}] = 2.5 \times 10^{-19} \mathrm{~M}\) and \([\mathrm{X}^{2-}] = 3.75 \times 10^{-19} \mathrm{~M}\).
These concentrations are crucial for computing the \(K_{sp}\) because they indicate the proportion of ions in the solution. By carefully calculating these values, you ensure that the \(K_{sp}\) reflects the true solubility equilibrium of the ionic compound.
