(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
The solubility product constants (Ksp) at the given temperatures are: (a) \(K_{sp} (CaF_2) = 3.87 \times 10^{-10}\), (b) \(K_{sp} (SrF_2) = 3.72 \times 10^{-9}\), and (c) the molar solubility of \(Ba(IO_3)_2\) at \(25^{\circ} \mathrm{C}\) is \(5.93 \times 10^{-4} \mathrm{mol / L}\).

Step by step solution

01

Write the balanced solubility equilibrium equation for the compound

For calcium fluoride (CaF2), the balanced solubility equilibrium equation is: \[CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}\]
02

Set up the expression for the solubility product constant (Ksp)

The solubility product constant expression for this equilibrium is given by: \[K_{sp} = [Ca^{2+}][F^{-}]^2\]
03

Calculate Ksp using molar solubility

Given that the molar solubility of CaF2 is \(1.24 \times 10^{-3}\mathrm{mol / L}\), let's substitute it into the Ksp expression. Let x = molar solubility of CaF2 = \(1.24 \times 10^{-3}\mathrm{mol / L}\) Then, \[[Ca^{2+}] = x = 1.24 \times 10^{-3}M\] \[[F^{-}] = 2x = 2(1.24 \times 10^{-3})M\] Now, substitute these concentrations into the Ksp expression: \[K_{sp} = (1.24 \times 10^{-3})(2(1.24 \times 10^{-3}))^{2}\]
04

Solve for Ksp

Calculate the Ksp value: \[K_{sp} = 3.87 \times 10^{-10}\] #b) Find Ksp for SrF2#
05

Calculate the molar solubility

We are given the mass of SrF2 which dissolves per 100 mL of the solution at 25˚C: \(1.1 \times 10^{-2}\mathrm{g / 100 mL} = 1.1 \times 10^{-2}\mathrm{g / 0.1 L}\) Since molar solubility is in mol/L, we'll convert the amount of SrF2 to moles. Given the molar mass of SrF2 = 125.62 g/mol, \[\frac{1.1 \times 10^{-2}\,\mathrm{g}}{0.1\,\mathrm{L}} \times \frac{1\,\mathrm{mol}}{125.62\,\mathrm{g}} = x_{SrF2} = 8.76 \times 10^{-4}\mathrm{mol / L }\]
06

Calculate Ksp using molar solubility

Set up the balanced solubility equilibrium equation for SrF2: \[SrF_{2(s)} \rightleftharpoons Sr^{2+}_{(aq)} + 2F^{-}_{(aq)}\] And the Ksp expression is: \[K_{sp} = [Sr^{2+}][F^{-}]^2\] Using the molar solubility (x) we calculated in Step 1: \[[Sr^{2+}] = x = 8.76 \times 10^{-4}M\] \[[F^{-}] = 2x = 2(8.76 \times 10^{-4})M\] Now substitute these concentrations into the Ksp expression: \[K_{sp} = (8.76 \times 10^{-4})(2(8.76 \times 10^{-4}))^2\]
07

Solve for Ksp

Calculate the Ksp value: \[K_{sp} = 3.72 \times 10^{-9}\] #c) Find the molar solubility of Ba(IO3)2#
08

Write the balanced solubility equilibrium equation

For barium iodate (Ba(IO3)2), the balanced solubility equilibrium equation is: \[Ba(\mathrm{IO}_{3})_{2(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2IO_{3}^{-}_{(aq)}\]
09

Set up the expression for the solubility product constant (Ksp)

The solubility product constant expression for this equilibrium is given by: \[K_{sp} = [Ba^{2+}][IO_{3}^{-}]^2\]
10

Calculate molar solubility using Ksp

We are given that the Ksp of Ba(IO3)2 at 25˚C is \(6.0 \times 10^{-10}\). Let x = molar solubility of Ba(IO3)2 Then, \[[Ba^{2+}] = x\] \[[IO_{3}^{-}] = 2x\] Substitute these concentrations into the Ksp expression: \[6.0 \times 10^{-10} = x(2x)^2\]
11

Solve for the molar solubility

Isolate x by solving the equation from Step 3: \[x =\sqrt[3]{\frac{6.0 \times 10^{-10}}{4}}\] Calculate the molar solubility: \[x = 5.93 \times 10^{-4} \mathrm{mol / L}\]

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Most popular questions from this chapter

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Calculate the \(\mathrm{pH}\) at the equivalence point for titrating 0.200 \(\mathrm{M}\) solutions of each of the following bases with 0.200 \(M \mathrm{HBr} :(\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\)

A sample of 0.1687 gof an unknown monoprotic acid was dissolved in 25.0 mL. of water and titrated with 0.1150 \(\mathrm{M}\) NaOH. The acid required 15.5 \(\mathrm{mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After 7.25 mL of base had been added in the titration, the pH was found to be \(2.85 .\) What is the \(K_{a}\) for the unknown acid?

A 20.0 -mL sample of 0.200 \(\mathrm{M}\) HBr solution is titrated with 0.200 \(\mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) of the solution after the following volumes of base have been added: (a) 15.0 \(\mathrm{mL}\) \((\mathbf{b}) 19.9 \mathrm{mL},(\mathbf{c}) 20.0 \mathrm{mL},(\mathbf{d}) 20.1 \mathrm{mL},(\mathbf{e}) 35.0 \mathrm{mL}\)

Which of the following solutions is a buffer? (a) 0.10\(M\) \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(0.10 \mathrm{MCH}_{3} \mathrm{CONa},(\mathbf{b}) 0.10 \mathrm{MCH}_{3} \mathrm{COOH}\) (c) 0.10 \(\mathrm{M} \mathrm{HCl}\) and \(0.10 \mathrm{M} \mathrm{NaCl},(\mathbf{d})\) both a and \(\mathrm{c},(\mathbf{e})\) all of a, \(\mathrm{b},\) and \(\mathrm{c} .\)

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