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 \(K_{\text {sp }}\) 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 \(\mathrm{SrF}_{2}\). (c) The \(K_{\text {pp }}\) 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 \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

Short answer

The solubility products of \(\mathrm{CaF}_{2}\), \(\mathrm{SrF}_{2}\), and \(\mathrm{Ba(IO_3)_2}\) are found as follows: a) For \(\mathrm{CaF}_{2}\), the \(K_{sp} = 7.61 \times 10^{-10}\). b) For \(\mathrm{SrF}_{2}\), the \(K_{sp} = 2.68 \times 10^{-9}\). c) The molar solubility of \(\mathrm{Ba(IO_3)_2}\) at \(25^{\circ} \mathrm{C}\) is \(\approx 1.14 \times 10^{-3}\, \mathrm{mol/l}\).

Step-by-step solution

Write the balanced equation and the corresponding K_sp expression for CaF2

Given the formula of \(\mathrm{CaF}_{2}\), the balanced equation for its dissolution in water is: \[\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca^{2+}}(aq) + 2\mathrm{F^-}(aq)\] The solubility product expression for \(\mathrm{CaF}_{2}\) is: \[K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^{2}\]

Use the molar solubility to find concentrations of ions in solution

Given that the molar solubility of \(\mathrm{CaF}_2\) at \(35^{\circ} \mathrm{C}\) is \(1.24 \times 10^{-3} \mathrm{~mol/l}\), it means that for every liter of solution, there is \(1.24 \times 10^{-3}\, \mathrm{mol}\) of \(\mathrm{CaF}_2\) that has dissolved. From the balanced equation, we can see that 1 mole of \(\mathrm{CaF}_2\) produces 1 mole of \(\mathrm{Ca^{2+}}\) and 2 moles of \(\mathrm{F^-}\). Thus, the molar concentrations of the ions in the solution are: \[[\mathrm{Ca^{2+}}] = 1.24 \times 10^{-3} \mathrm{~mol/l}\] and \[[\mathrm{F^-}] = 2(1.24 \times 10^{-3}) \mathrm{~mol/l} = 2.48 \times 10^{-3} \mathrm{~mol/l}\]

Calculate K_sp using the ion concentrations

We can now plug the concentrations of ions into the \(K_{sp}\) expression and solve for the solubility product: \[K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^{2} = (1.24 \times 10^{-3})(2.48 \times 10^{-3})^2 = 7.61 \times 10^{-10}\] The solubility product for \(\mathrm{CaF}_{2}\) at \(35^{\circ} \mathrm{C}\) is \(7.61 \times 10^{-10}\). #b) Solubility and K_sp for SrF2#

Write the balanced equation for SrF2 and its solubility in water

The balanced equation for dissolution of \(\mathrm{SrF}_2\) in water is: \[\mathrm{SrF}_{2}(s) \rightleftharpoons \mathrm{Sr^{2+}}(aq) + 2\mathrm{F^-}(aq)\] The solubility product expression for \(\mathrm{SrF}_{2}\) is: \[K_{sp}=[\mathrm{Sr^{2+}}][\mathrm{F^-}]^{2}\]

Convert the solubility in grams to molar solubility

Given that \(1.1 \times 10^{-2}\, \mathrm{g}\) of \(\mathrm{SrF}_2\) dissolves per \(100\, \mathrm{mL}\) of solution, we must first find the molar concentration of \(\mathrm{SrF}_2\). Molar mass of \(\mathrm{SrF}_2\): \(= 87.62\, \mathrm{g/mol} (\mathrm{Sr}) + 2 \times 18.998\, \mathrm{g/mol} (\mathrm{F})\) \(= 125.62\, \mathrm{g/mol}\) Molar solubility of \(\mathrm{SrF}_2\): \(\frac{1.1 \times 10^{-2}\, \mathrm{g}}{125.62\, \mathrm{g/mol}} \times \frac{1}{0.100\, \mathrm{L}} \approx 8.76 \times 10^{-4}\, \mathrm{mol/l}\)

Use the molar solubility to find the concentrations of ions in solution

From the balanced equation, 1 mole of \(\mathrm{SrF}_2\) produces 1 mole of \(\mathrm{Sr^{2+}}\) and 2 moles of \(\mathrm{F^-}\). Thus, the molar concentrations of the ions in the solution are: \[[\mathrm{Sr^{2+}}] = 8.76 \times 10^{-4} \mathrm{~mol/l}\] and \[[\mathrm{F^-}] = 2(8.76 \times 10^{-4}) \mathrm{~mol/l} = 1.75 \times 10^{-3} \mathrm{~mol/l}\]

Calculate K_sp using the ion concentrations

We can now plug the concentrations of ions into the \(K_{sp}\) expression and solve for the solubility product: \[K_{sp}=[\mathrm{Sr^{2+}}][\mathrm{F^-}]^{2} = (8.76 \times 10^{-4})(1.75 \times 10^{-3})^2 = 2.68 \times 10^{-9}\] The solubility product of \(\mathrm{SrF}_{2}\) is \(2.68 \times 10^{-9}\). #c) Solubility and K_sp for Ba(IO3)2#

Write the balanced equation for Ba(IO3)2 and its solubility in water

The balanced equation for the dissolution of \(\mathrm{Ba(IO_3)_2}\) in water is: \[\mathrm{Ba(IO_3)_2}(s) \rightleftharpoons \mathrm{Ba^{2+}}(aq) + 2\mathrm{IO_3^-}(aq)\] The solubility product expression for \(\mathrm{Ba(IO_3)_2}\) is: \[K_{sp}=[\mathrm{Ba^{2+}}][\mathrm{IO_3^-}]^{2}\]

Calculate the molar solubility using the given K_sp value

We are given that the solubility product, \(K_{sp}\), for \(\mathrm{Ba(IO_3)_2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10}\). Let the molar solubility of \(\mathrm{Ba(IO_3)_2}\) be x. Then: \[[\mathrm{Ba^{2+}}] = x\] and \[[\mathrm{IO_3^-}] = 2x\] Now, we can plug these concentrations into the \(K_{sp}\) expression and solve for x: \[K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{IO_3^-}]^{2} = x(2x)^{2} = 4x^{3}\] \[x^{3} = \frac{K_{sp}}{4} = \frac{6.0 \times 10^{-10}}{4} = 1.5 \times 10^{-10}\] \[x = \sqrt[3]{1.5 \times 10^{-10}} \approx 1.14 \times 10^{-3}\, \mathrm{mol/l}\] The molar solubility of \(\mathrm{Ba(IO_3)_2}\) at \(25^{\circ} \mathrm{C}\) is \(\approx 1.14 \times 10^{-3}\, \mathrm{mol/l}\).

Key concepts

Ksp Calculation

Understanding the solubility product constant, commonly represented as Ksp, is central to predicting and explaining the solubility of ionic compounds in solution. The Ksp is a special kind of equilibrium constant that applies to the dissolution of sparingly soluble salts. When we talk about Ksp calculations, we are referring to the process of determining the extent to which a compound will dissolve in water.

Let's consider an example where the molar solubility of calcium fluoride, CaF2, is given. To calculate its Ksp value, one must write the balanced dissociation equation of the salt and construct the corresponding Ksp expression, which includes the concentrations of the dissolved ions raised to their stoichiometric coefficients in the balanced equation. Using the molar solubility provided, we can find the concentrations of the respective ions in the solution and then calculate the Ksp.

By grasping the mechanics behind Ksp, students can not only solve problems but also understand the factors affecting solubility, such as the common ion effect and the impact of pH changes.

Molar Solubility

Molar solubility is the number of moles of a solute that will dissolve per liter of solution to reach a state of dynamic equilibrium. In simple terms, it indicates how much of a substance can dissolve in a given quantity of solvent at a specific temperature. This concept is crucial because it helps us to quantify solubility and predict whether a precipitate will form under certain conditions. For instance, when given the solubility in grams like with SrF2, converting to molar solubility allows us to compare the solubility of different compounds unambiguously.

Molar solubility can also be tied back to Ksp. For example, if we know the Ksp value of barium iodate, Ba(IO3)2, we can determine its molar solubility by setting up an equation where the concentration of each ion is related to the molar solubility, then solving for this value. Molar solubility is a direct gauge of a compound's solubility and plays a pivotal role in many areas, including medicinal chemistry and environmental science.

Chemical Equilibrium

The concept of chemical equilibrium is pivotal in understanding how reactions progress and the conditions at which they balance out. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants or products over time. It's essential to understand that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their ratios remain constant.

When studying solubility equilibria, we are actually dealing with a specific type of equilibrium involving a solid and its dissolved ions. The equilibrium constant (Ksp) is an expression of this equilibrium state. By mastering chemical equilibrium, students can better understand the dissolution process of ionic compounds and predict the conditions under which they will precipitate out of the solution, which is incredibly useful in various fields, including biochemistry, pharmaceuticals, and materials science.

Dissolution of Ionic Compounds

Dissolution refers to the process by which ionic compounds dissociate into their constituent ions when they dissolve in a solvent like water. This process is influenced by the nature of the ionic compound and the solvent, temperature, and presence of other ions or compounds in the solution. Understanding the dissolution process is important for foreseeing the solubility behavior of a substance in different environments.

The dissolution of ionic compounds is not always complete, as seen with sparingly soluble salts such as CaF2 and SrF2 mentioned in our initial exercise. The extent to which these compounds dissolve is quantified by their solubility product constants. Students can apply this knowledge to predict whether a particular substance will dissolve in water and to what extent, which is crucial in industries like water treatment and mineral extraction where solubility plays a significant role in the efficiency of processes.