(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 \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

Short Answer

Expert verified
(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 by step solution

01

(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 \(3.86 \times 10^{-11}\).
02

(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 \(5.32 \times 10^{-10}\).
03

(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 \(1.14 \times 10^{-3}\, \mathrm{mol/L}\).

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Most popular questions from this chapter

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of \(K_{b}\) for \(B\) ?

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