Problem 89 - Chemistry A Molecular Approach

Question

Use the given molar solubilities in pure water to calculate \(K_{\mathrm{sp}}\) for each compound. a. \(\mathrm{MX} ;\) molar solubility \(=3.27 \times 10^{-11} \mathrm{M}\) b. \(\mathrm{PbF}_{2} ;\) molar solubility \(=5.63 \times 10^{-3} \mathrm{M}\) c. \(\mathrm{MgF}_{2} ;\) molar solubility \(=2.65 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

a. \(K_{\mathrm{sp}}\) for MX is \( 1.07 \times 10^{-21} \). b. \(K_{\mathrm{sp}}\) for \(\mathrm{PbF}_{2}\) is \( 7.15 \times 10^{-8} \). c. \(K_{\mathrm{sp}}\) for \(\mathrm{MgF}_{2}\) is \( 7.44 \times 10^{-12} \).

Step 1: Write the Dissociation Equation for MX

First, write the dissociation reaction of MX in water. MX dissolves to form M\textsuperscript{+} and X\textsuperscript{-} ions:\[\mathrm{MX(s)} \rightleftharpoons \mathrm{M\textsuperscript{+}(aq)} + \mathrm{X\textsuperscript{-}(aq)}\]

Step 2: Express the Solubility Product Constant for MX

Using the equilibrium concentrations of the ions from the dissociation of MX, we can express the solubility product constant, \(K_{\mathrm{sp}}\), as follows:\[K_{\mathrm{sp}} = [\mathrm{M\textsuperscript{+}}][\mathrm{X\textsuperscript{-}}]\]Since the stoichiometry is 1:1, the concentration of each ion is equal to the molar solubility of MX.

Step 3: Calculate the Solubility Product for MX

Plug the molar solubility of MX into the expression for \(K_{\mathrm{sp}}\) and calculate its value:\[K_{\mathrm{sp}} = (3.27 \times 10^{-11})(3.27 \times 10^{-11}) = 1.07 \times 10^{-21}\]

Step 4: Write the Dissociation Equation for \(\mathrm{PbF}_{2}\)

Write the dissociation reaction of \(\mathrm{PbF}_{2}\) in water:\[\mathrm{PbF}_{2}(s) \rightleftharpoons \mathrm{Pb\textsuperscript{2+}(aq)} + 2\mathrm{F\textsuperscript{-}(aq)}\]

Step 5: Express the Solubility Product Constant for \(\mathrm{PbF}_{2}\)

The expression for \(K_{\mathrm{sp}}\) of \(\mathrm{PbF}_{2}\) is based on the equilibrium concentrations:\[K_{\mathrm{sp}} = [\mathrm{Pb\textsuperscript{2+}}][\mathrm{F\textsuperscript{-}}]^2\]The concentration of \(\mathrm{F\textsuperscript{-}}\) ions is twice the molar solubility, as each unit of \(\mathrm{PbF}_{2}\) produces two fluoride ions.

Step 6: Calculate the Solubility Product for \(\mathrm{PbF}_{2}\)

Use the molar solubility of \(\mathrm{PbF}_{2}\) to calculate \(K_{\mathrm{sp}}\):\[K_{\mathrm{sp}} = (5.63 \times 10^{-3})(2 \times 5.63 \times 10^{-3})^2 = (5.63 \times 10^{-3})(11.26 \times 10^{-3})^2 = 7.15 \times 10^{-8}\]

Step 7: Write the Dissociation Equation for \(\mathrm{MgF}_{2}\)

The dissociation reaction of \(\mathrm{MgF}_{2}\) in water is:\[\mathrm{MgF}_{2}(s) \rightleftharpoons \mathrm{Mg\textsuperscript{2+}(aq)} + 2\mathrm{F\textsuperscript{-}(aq)}\]

Step 8: Express the Solubility Product Constant for \(\mathrm{MgF}_{2}\)

The expression for \(K_{\mathrm{sp}}\) of \(\mathrm{MgF}_{2}\) is:\[K_{\mathrm{sp}} = [\mathrm{Mg\textsuperscript{2+}}][\mathrm{F\textsuperscript{-}}]^2\]Here also, the concentration of \(\mathrm{F\textsuperscript{-}}\) ions is twice the molar solubility.

Step 9: Calculate the Solubility Product for \(\mathrm{MgF}_{2}\)

Using the molar solubility of \(\mathrm{MgF}_{2}\), calculate \(K_{\mathrm{sp}}\):\[K_{\mathrm{sp}} = (2.65 \times 10^{-4})(2 \times 2.65 \times 10^{-4})^2 = (2.65 \times 10^{-4})(5.30 \times 10^{-4})^2 = 7.44 \times 10^{-12}\]

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. In simple terms, it is the maximum amount of a substance that can be dissolved in a given volume of solvent at a specified temperature.

This concept is particularly important in the fields of chemistry and material science, as it helps predict the solubility of an ionic compound in water. It is expressed in units of molarity (M), which is moles per liter. For instance, if a compound has a molar solubility of 3.27 x 10^-11 M, it means that only a very tiny portion of that compound can dissolve in water, indicating a low level of solubility.

Dissociation Equation

The dissociation equation is a representation of the chemical reaction in which an ionic solid separates into its individual ions in a solution. It is the reversible process of a solid ionic compound dissolving in a solvent, such as water. For example, the dissociation equation for MX is:
\[\mathrm{MX(s)} \rightleftharpoons \mathrm{M^{+}(aq)} + \mathrm{X^{-}(aq)}\]
It illustrates that for each formula unit of MX that dissolves, one ion of M⁺ and one ion of X^- are released into the solution. This helps us understand the stoichiometry of the dissolution process and is crucial for calculating the solubility product constant, K_sp.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each species in a solution when the rate of its forward reaction (dissolving of the solid) equals the rate of its reverse reaction (precipitation out of solution). At this stage, the concentrations of the reactants and products remain constant over time.

Understanding equilibrium concentrations is vital in calculating the solubility product constant. In the case of ionic compounds in water, this concentration is generally equal to the molar solubility of the compound. However, for compounds like PbF₂ that dissociate into more than one ion per formula unit, the concentration of the ions will need to be adjusted accordingly in the K_sp equation.

Ksp Calculation

The solubility product constant (K_sp) is a special type of equilibrium constant that measures the solubility of sparingly soluble ionic compounds. It is calculated from the equilibrium concentrations of the ions in a saturated solution. Essentially, K_sp is the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced dissociation equation.

For example, the calculation of K_sp for a compound MX, where the molar solubility is 3.27 x 10^-11 M, would be:
\[K_{\mathrm{sp}} = (3.27 \times 10^{-11})(3.27 \times 10^{-11}) = 1.07 \times 10^{-21}\]
In this formula, we multiply the concentration of the M⁺ ion by the concentration of the X^- ion. The resulting K_sp value indicates the degree to which the compound can dissolve in water—lower K_sp values mean lower solubility.