Chapter 15: Problem 10
Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
Chapter 15: Problem 10
Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
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Get started for freeCalculate the solubility of \(\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\) in a solution containing \(1.0 M \mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)\)
A solution contains 0.018 molel each of \(\mathrm{I}^{-}, \mathrm{Br}^{-},\) and \(\mathrm{Cl}^{-}\). When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24\) \(M\) \(\mathrm{AgNO}_{3}\), what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\begin{aligned} \operatorname{AgI}: K_{\mathrm{sp}} &=1.5 \times 10^{-16} \\ \operatorname{AgBr}: K_{\mathrm{sp}} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\mathrm{sp}} &=1.6 \times 10^{-10} \end{aligned}$$
For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7},\) or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\mathrm{sp}}=8.1 \times 10^{-12},\) or \(\mathrm{Mn}(\mathrm{OH})_{2}\) \(K_{\mathrm{sp}}=2 \times 10^{-13}\)
A solution contains \(1.0 \times 10^{-5} M \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} M \mathrm{CN}^{-}\) Will AgCN( \(s\) ) precipitate? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgCN}(s) \text { is } 2.2 \times 10^{-12} .\right)\)
Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.
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