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Chapter 16: Problem 26

The concentration of \(\mathrm{Ag}^{+}\) in a solution saturated with \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s)\) is \(2.2 \times 10^{-4}\) M. Calculate \(K_{\text {sp }}\) for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\).

Short Answer

Expert verified
The solubility product constant, \(K_{sp}\), for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is approximately \(5.36\times10^{-11}\).

Step by step solution

01

Write the dissolution reaction

The dissolution of \(\mathrm{Ag}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) in water can be represented as: \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \rightleftharpoons 2\mathrm{Ag}^{+}(aq) + \mathrm{C}_{2}\mathrm{O}_{4}^{2-}(aq)\)
02

Write the expression for \(K_{sp}\)

Based on the balanced equation, we can write the solubility product constant (\(K_{sp}\)) expression as: \(K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]\)
03

Determine the concentration of \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\)

Given that the concentration of \(\mathrm{Ag}^{+}\) is \(2.2 \times 10^{-4}\) M, we can find the concentration of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions using the stoichiometry of the dissolution reaction, which is a 2:1 ratio in concentrations of \(\mathrm{Ag}^{+}\) to \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\): \([\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = \dfrac{1}{2}[\mathrm{Ag}^{+}] = \dfrac{1}{2}(2.2\times10^{-4}\,\text{M}) = 1.1\times10^{-4}\,\text{M}\)
04

Calculate \(K_{sp}\)

Now that we have the concentrations of the ions, we can plug them into the \(K_{sp}\) expression: \(K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]=(2.2 \times 10^{-4})^{2}(1.1\times10^{-4})=5.356\times10^{-11}\) So, the solubility product constant, \(K_{sp}\), for \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is approximately \(5.36\times10^{-11}\).

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Most popular questions from this chapter

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?

Approximately \(0.14 \mathrm{~g}\) nickel(II) hydroxide, \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\), dissolves per liter of water at \(20^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{Pb} \mathrm{I}_{2}\right.\) ) is \(1.4 \times 10^{-8}\). Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 \mathrm{M}\) NaI

\(\mathrm{Ag}_{2} \mathrm{~S}(s)\) has a larger molar solubility than \(\mathrm{CuS}\) even though \(\mathrm{Ag}_{2} \mathrm{~S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.
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