Question

From the solubility data given, calculate the solubility products for the following compounds: (a) \(\mathrm{SrF}_{2}\), \(7.3 \times 10^{-2} \mathrm{~g} / \mathrm{L},\) (b) \(\mathrm{Ag}_{2} \mathrm{PO}_{4}, 6.7 \times 10^{-3} \mathrm{~g} / \mathrm{L}\)

Short answer

\(K_{sp}\) for \(\mathrm{SrF}_2\) is \(7.83 \times 10^{-10}\), and \(K_{sp}\) for \(\mathrm{Ag}_2\mathrm{PO}_4\) is \(1.64 \times 10^{-12}\).

Step-by-step solution

Identify the Compounds and Their Dissolution Process

The compounds given are \(\mathrm{SrF}_2\) and \(\mathrm{Ag}_2\mathrm{PO}_4\). Start by writing their dissolution equations. \(\mathrm{SrF}_2 \rightarrow \mathrm{Sr}^{2+} + 2\mathrm{F}^{-}\) and \(\mathrm{Ag}_2\mathrm{PO}_4 \rightarrow 2 \mathrm{Ag}^+ + \mathrm{PO}_4^{3-}\)

Calculate Moles Per Liter from Solubility

Convert the solubility from grams per liter to moles per liter for each compound. \(\mathrm{SrF}_2\) has a molar mass of \(125.6\, \mathrm{g/mol}\): \[ \text{Molarity} = \frac{7.3 \times 10^{-2} \, \mathrm{g/L}}{125.6 \, \mathrm{g/mol}} = 5.81 \times 10^{-4} \, \mathrm{mol/L} \]. For \(\mathrm{Ag}_2\mathrm{PO}_4\), with molar mass \(418.58\, \mathrm{g/mol}\): \[ \text{Molarity} = \frac{6.7 \times 10^{-3} \, \mathrm{g/L}}{418.58 \, \mathrm{g/mol}} = 1.6 \times 10^{-5} \, \mathrm{mol/L} \].

Write the Expression for Ksp

The \(K_{sp}\) for each reaction is based on the concentrations of the ions at equilibrium. For \(\mathrm{SrF}_2\), \(K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{F}^-]^2\), and for \(\mathrm{Ag}_2\mathrm{PO}_4\), \(K_{sp} = [\mathrm{Ag}^+]^2[\mathrm{PO}_4^{3-}]\).

Substitute the Molarity Values into Ksp Expressions

For \(\mathrm{SrF}_2\), substitute \([\mathrm{Sr}^{2+}] = 5.81 \times 10^{-4} \, \mathrm{mol/L}\) and \([\mathrm{F}^-] = 2 \times 5.81 \times 10^{-4} \, \mathrm{mol/L} = 1.162 \times 10^{-3} \, \mathrm{mol/L}\)\ into the equation: \(K_{sp} = (5.81 \times 10^{-4})(1.162 \times 10^{-3})^2\). Calculate to get \(K_{sp} = 7.83 \times 10^{-10}\). For \(\mathrm{Ag}_2\mathrm{PO}_4\), substitute \([\mathrm{Ag}^+] = 2 \times 1.6 \times 10^{-5} \, \mathrm{mol/L} = 3.2 \times 10^{-5} \, \mathrm{mol/L} \) and \([\mathrm{PO}_4^{3-}] = 1.6 \times 10^{-5} \, \mathrm{mol/L}\)\ into the equation: \(K_{sp} = (3.2 \times 10^{-5})^2 (1.6 \times 10^{-5})\). Calculate to get \(K_{sp} = 1.64 \times 10^{-12}\).

Solution Verification

Double check calculations from Step 4 to ensure there are no arithmetic mistakes in the \(K_{sp}\) calculation for both \(\mathrm{SrF}_2\) and \(\mathrm{Ag}_2\mathrm{PO}_4\). Confirm values are within the expected range for these types of compounds.

Key concepts

Chemical Equilibrium

Understanding **chemical equilibrium** is key to grasping how solubility products work. In the context of dissolving ionic compounds like \(\mathrm{SrF}_2\) and \(\mathrm{Ag}_2\mathrm{PO}_4\), when these substances dissolve in water, they reach a point where the rate of dissolving equals the rate of recrystallizing.
- This state is called equilibrium.
- At this point, the concentrations of ions in solution remain constant.
The equilibrium expression for a dissolution reaction provides the **solubility product constant** \(K_{sp}\). This constant reflects the saturated solution's ion concentrations at equilibrium. Think of \(K_{sp}\) as a way to describe how much of the compound can dissolve in water. Calculating \(K_{sp}\) involves substituting the ion concentrations back into the equilibrium expression. For example, \[K_{sp} = [\mathrm{Sr}^{2+}][\mathrm{F}^-]^2\]for \(\mathrm{SrF}_2\) showcases the balance of products of the dissolved ions.

Ionic Compound Dissolution

When an **ionic compound dissolves**, it separates into its constituent ions. This process is called dissociation. Taking \(\mathrm{SrF}_2\) and \(\mathrm{Ag}_2\mathrm{PO}_4\) as examples, their dissolution reactions are written as:- \(\mathrm{SrF}_2 \rightarrow \mathrm{Sr}^{2+} + 2\mathrm{F}^-\)- \(\mathrm{Ag}_2\mathrm{PO}_4 \rightarrow 2\mathrm{Ag}^+ + \mathrm{PO}_4^{3-}\)
Understanding how many ions each compound produces is critical because it affects the concentration of ions in the solution. For \(\mathrm{SrF}_2\), one mole dissolves to produce: - 1 mole of \(\mathrm{Sr}^{2+}\) - 2 moles of \(\mathrm{F}^-\)This ratio is vital for calculating the **ion concentrations** needed to find the solubility product \(K_{sp}\). Similarly, the dissolution of \(\mathrm{Ag}_2\mathrm{PO}_4\) yields:- 2 moles of \(\mathrm{Ag}^+\)- 1 mole of \(\mathrm{PO}_4^{3-}\)
The more ions an ionic compound breaks into, the more complex its solubility product equation becomes.

Molarity Conversion

Converting **solubility from grams per liter to moles per liter** is a fundamental step to calculate \(K_{sp}\). This conversion is necessary because equilibrium calculations require concentrations in molarity (mol/L). Here's how to do it:1. **Find the molar mass** of the compound. This is the sum of the atomic masses of the elements in the compound.2. **Use the formula**: \[ \text{Molarity} = \frac{\text{Solubility in g/L}}{\text{Molar Mass in g/mol}} \]
For instance, with \(\mathrm{SrF}_2\), you have:- Solubility = \(7.3 \times 10^{-2}\, \mathrm{g/L}\)- Molar mass = \(125.6 \, \mathrm{g/mol}\)This converts to:\[\text{Molarity} = \frac{7.3 \times 10^{-2}}{125.6} = 5.81 \times 10^{-4} \, \mathrm{mol/L}\]Perform similar steps for \(\mathrm{Ag}_2\mathrm{PO}_4\). These calculations let you determine how many moles of ions are present in one liter of saturated solution, helping find the solubility product.