Problem 53 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) If the molar solubility of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2} \mathrm{~g} \mathrm{SrF}_{2}\) dissolves per \(100 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C} .\) Calculate the solubility product for \(\mathrm{SrF}_{2} .(\mathbf{c})\) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

Step-by-Step Solution

Answer

(a) The \(K_{sp}\) of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(3.86 \times 10^{-11}\). (b) The \(K_{sp}\) of \(\mathrm{SrF}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(5.32 \times 10^{-10}\). (c) The molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) is \(1.14 \times 10^{-3}\, \mathrm{mol/L}\).

Step 1: (a) Finding the \(K_{sp}\) of \(\mathrm{CaF}_{2}\)

We are given the molar solubility of \(\mathrm{CaF}_{2}\): \(1.24 \times 10^{-3} \, \mathrm{mol/L}\). First, we write the dissolution equation: \(\mathrm{CaF}_{2} \, (s) \rightleftharpoons \mathrm{Ca^{2+}} \, (aq) + 2 \, \mathrm{F^{-}} \, (aq)\) For each mole of \(\mathrm{CaF}_{2}\) that dissolves, we get one mole of \(\mathrm{Ca^{2+}}\) and two moles of \(\mathrm{F^{-}}\). So, the concentrations at equilibrium are: \( [\mathrm{Ca^{2+}}] = 1.24 \times 10^{-3} \, \mathrm{mol/L} \\ [\mathrm{F^{-}}] = 2 \times 1.24 \times 10^{-3} \, \mathrm{mol/L} \) Now we can find the \(K_{sp}\): \( K_{sp} = [\mathrm{Ca^{2+}}] \times [\mathrm{F^{-}}]^2 \\ K_{sp} = (1.24 \times 10^{-3}) \times (2 \times 1.24 \times 10^{-3})^2 \\ K_{sp} = 3.86 \times 10^{-11} \) So, the \(K_{sp}\) of \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(3.86 \times 10^{-11}\).

Step 2: (b) Finding the \(K_{sp}\) of \(\mathrm{SrF}_{2}\)

We are given the grams of \(\mathrm{SrF}_{2}\) dissolved in \(100 \, \mathrm{mL}\) of solution: \(1.1 \times 10^{-2} \, \mathrm{g}\). First, we need to find the molar solubility. The molar mass of \(\mathrm{SrF}_{2}\) is: \(M = 87.62 \, \mathrm{ (Sr)} + 2 \times 19 \, \mathrm{ (F)} = 125.62\, \mathrm{g/mol}\) Now we find the molar solubility: \( \frac{1.1 \times 10^{-2} \, \mathrm{g}}{125.62\, \mathrm{g/mol} \times 0.1\, \mathrm{L}} = 8.75 \times 10^{-4}\, \mathrm{mol/L} \) Now we write the dissolution equation of \(\mathrm{SrF}_{2}\): \( \mathrm{SrF}_{2} \, (s) \rightleftharpoons \mathrm{Sr^{2+}} \, (aq) + 2\, \mathrm{F^{-}} \, (aq) \) Proceeding like before, we have: \( [\mathrm{Sr^{2+}}] = 8.75 \times 10^{-4} \, \mathrm{mol/L} \\ [\mathrm{F^{-}}] = 2 \times 8.75 \times 10^{-4} \, \mathrm{mol/L} \) Now we can find the \(K_{sp}\): \( K_{sp} = [\mathrm{Sr^{2+}}] \times [\mathrm{F^{-}}]^2 \\ K_{sp} = (8.75 \times 10^{-4}) \times (2 \times 8.75 \times 10^{-4})^2 \\ K_{sp} = 5.32 \times 10^{-10} \) So, the \(K_{sp}\) of \(\mathrm{SrF}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(5.32 \times 10^{-10}\).

Step 3: (c) Finding the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\)

We are given the \(K_{sp}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\): \(6.0 \times 10^{-10}\) and need to find its molar solubility. We write the dissolution equation: \( \mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2} \, (s) \rightleftharpoons \mathrm{Ba^{2+}} \, (aq) + 2 \, \mathrm{IO}_{3}^{-} \, (aq) \) Let \(s\) be the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\). We get one mole of \(\mathrm{Ba^{2+}}\) and two moles of \(\mathrm{IO}_{3}^{-}\). Thus, at equilibrium, we have: \( [\mathrm{Ba^{2+}}] = s \, \mathrm{mol/L} \\ [\mathrm{IO}_{3}^{-}] = 2s \, \mathrm{mol/L} \) Now substitute these values into the \(K_{sp}\) expression: \( K_{sp} = [\mathrm{Ba^{2+}}] \times [\mathrm{IO}_{3}^{-}]^2 \\ 6.0 \times 10^{-10} = (s) \times (2s)^2 \\ 6.0 \times 10^{-10} = 4s^3 \) Now solve for \(s\): \(s^3 = \frac{6.0 \times 10^{-10}}{4} = 1.5 \times 10^{-10}\) \(s = \sqrt[3]{1.5 \times 10^{-10}} = 1.14 \times 10^{-3}\, \mathrm{mol/L}\) So, the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) is \(1.14 \times 10^{-3}\, \mathrm{mol/L}\).

Key Concepts

Molar Solubility

Molar solubility is an important concept in chemistry. It refers to the number of moles of a solute that can dissolve per liter of solution until the solution reaches saturation. At saturation, the dissolved solute is in equilibrium with the undissolved solute.
This concept is key when working with sparingly soluble salts, such as calcium fluoride (CaF\(_2\)) or strontium fluoride (SrF\(_2\)).
Understanding molar solubility helps us predict how a salt will behave in a solution:

• A higher molar solubility indicates that more solute can dissolve, leading to a higher concentration of ions in solution.
• If we know the molar solubility, we can deduce information about the solution's ionic concentration at equilibrium.

This is crucial for calculating other important parameters such as the solubility product constant, which we will discuss shortly.

Dissolution Equation

The dissolution equation illustrates how a solid salt dissociates into its constituent ions in solution. These equations are essential as they provide the stoichiometric relationships between the solute and the ions in solution.
For instance, in the case of calcium fluoride, the dissolution equation is:\[\mathrm{CaF}_{2} \, (s) \rightleftharpoons \mathrm{Ca^{2+}} \, (aq) + 2 \, \mathrm{F^{-}} \, (aq)\]This equation tells us that one mole of CaF\(_2\) yields one mole of Ca\(^{2+}\) ions and two moles of F\(^-\) ions upon dissolution.
It's important to note the stoichiometry:

• The coefficients in the balanced equation show the ratio of the ions produced.
• For each mole of solid dissolved, twice as many fluoride ions are produced compared to calcium ions.

This same method applies when writing dissolution equations for other salts like SrF\(_2\) and Ba(IO\(_3\))\(_2\), helping us to understand and predict the composition of equilibrium solutions.

Solubility Constant Calculations

The solubility product constant, or \(K_{sp}\), is a measure of the extent to which a salt can dissolve in water. It is derived from the concentrations of the ions at equilibrium, as shown in the dissolution equation.
Calculations involving \(K_{sp}\) usually start by identifying the ion concentrations in a saturated solution. For example, for calcium fluoride:\[K_{sp} = [\mathrm{Ca^{2+}}] \times [\mathrm{F^{-}}]^2\]Using the molar solubility of CaF\(_2\) as \(1.24 \times 10^{-3} \, \mathrm{mol/L}\), we calculate:

• [Ca\(^{2+}\)] = \(1.24 \times 10^{-3} \, \mathrm{mol/L}\)
• [F\(^{-}\)] = \(2 \times 1.24 \times 10^{-3} \, \mathrm{mol/L}\)

The \(K_{sp}\) value is then found by plugging these concentrations into the formula.
Understanding \(K_{sp}\) calculations is vital for predicting whether a precipitate will form when two solutions are mixed, determining reaction conditions, and tailoring chemical processes.