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Chapter 14: Problem 98

At \(25^{\circ} \mathrm{C}\), the solubility of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), in water is \(6.1 \mathrm{mg} / \mathrm{L}\). (a) What are the equilibrium molar concentrations of \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) ?\) (b) Calculate \(K_{\mathrm{sp}}\) for calcium oxalate.

Short Answer

Expert verified
The equilibrium molar concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions are both \(4.76 \times 10^{-5}\,\mathrm{mol/L}\), and the solubility product of calcium oxalate, \(K_{sp}\), is \(2.27 \times 10^{-9}\) at \(25^{\circ}\mathrm{C}\).

Step by step solution

01

Convert solubility to molar concentration for calcium oxalate

First, we need to convert the solubility of calcium oxalate from mg/L to moles/L (molar concentration). To do this, we'll divide by the molar mass of calcium oxalate: Molar mass of \(\mathrm{CaC}_{2}\mathrm{O}_{4} = 40.08\,(\mathrm{Ca}) + 24.01 \times 2\,(\mathrm{C}) + 16.00 \times 4\,(\mathrm{O}) = 128.1 \,\mathrm{g/mol}\) Now, convert the solubility into molar concentration: \(C_{\mathrm{CaC}_{2}\mathrm{O}_{4}} = \frac{6.1\,\mathrm{mg/L}}{128.1\,\mathrm{g/mol}} \times \frac{1\, \mathrm{g}}{1000\, \mathrm{mg}} = 4.76 \times 10^{-5}\,\mathrm{mol/L}\)
02

Find the equilibrium molar concentrations of ions

Given that the dissolution of calcium oxalate can be represented as: \(\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \) Since the molar ratio of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(1:1\), we'll have the same equilibrium molar concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) as the molar concentration of calcium oxalate: \(C_{\mathrm{Ca}^{2+}} = C_{\mathrm{C}_{2} \mathrm{O}_{4}^{2-}} = 4.76 \times 10^{-5}\,\mathrm{mol/L}\)
03

Calculate \(K_{sp}\) for calcium oxalate

Now we can use the molar concentrations and the formula for \(K_{sp}\) to find the solubility product of calcium oxalate: \(K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]\) \(K_{sp} = (4.76 \times 10^{-5})^2 = 2.27 \times 10^{-9}\) The solubility product of calcium oxalate is \(K_{sp} = 2.27 \times 10^{-9}\) at \(25^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

Write the equilibrium constant expression for (a) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{PO}_{4}^{3-}(a q)\) (c) \(\mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \rightleftarrows \mathrm{PbI}_{2}(s)\) (d) \(\mathrm{Ca}^{2+}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) \(\stackrel{\mathrm{CaCO}_{3}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)}\)

Will \(K_{\text {eq }}\) for an endothermic reaction increase or decrease when the reaction mixture is (a) heated and (b) cooled? Explain your answer.

Write the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+\) \(\mathrm{CO}_{2}(g)\)

At \(25^{\circ} \mathrm{C}\), the solubility in water of the moderately soluble salt silver acetate, \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is \(10.6 \mathrm{~g} / \mathrm{L}\). (a) Write the chemical equation for the dissolving of silver acetate in water. (b) Write the \(K_{\mathrm{sp}}\) expression for silver acetate. (c) Calculate the value of \(K_{\mathrm{sp}}\) (show your work).

One way to calculate the value of a reaction's equilibrium constant is to perform the reaction, let it come to equilibrium, measure the concentration of all the reactants and products, and then plug those concentrations into the equilibrium constant expression and calculate its value. (a) A student performs the reaction \(\mathrm{A}(a q)+2 \mathrm{~B}(a q) \rightleftarrows \mathrm{C}(a q)\) starting with \(2.0 \mathrm{M} \mathrm{A}\) and \(4.0 \mathrm{M} \mathrm{B}\). He finds that at equilibrium, the concentrations of \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) are \(0.020 \mathrm{M}, 0.040 \mathrm{M}\), and \(1.98\) \(\mathrm{M}\), respectively. What is the value of this reaction's equilibrium constant (write your answer using scientific notation)? (b) Next, the student repeats the experiment, but this time he starts with \(3.0 \mathrm{M} \mathrm{A}\) and 5.0 M B. What value will he get for \(K_{\text {eq }}\) when he measures the equilibrium concentrations and plugs them into the equilibrium constant expression? (Hint: Think about why equilibrium constant is called a constant.)
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