Chapter 16: Problem 14
Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.
Chapter 16: Problem 14
Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.
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Get started for freeUse the following data to calculate the \(K_{\text {sp }}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\).
A mixture contains \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Mn}^{2+}\) and is saturated with \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\). Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\text {sp }}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{se}}\) for \(\mathrm{MnS}=2.3 \times 10^{-13} .\)
Silver chloride dissolves readily in \(2 M \mathrm{NH}_{3}\) but is quite insoluble in \(2 M \mathrm{NH}_{4} \mathrm{NO}_{4} .\) Explain.
A solution is prepared by adding \(0.10\) mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owendl }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}{ }^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$ 5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}} $$ for the overall reaction $$ \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) $$
The overall formation constant for \(\mathrm{HgI}_{4}{ }^{2-}\) is \(1.0 \times 10^{30} .\) That is, $$ 1.0 \times 10^{30}=\frac{\left[\mathrm{HgI}_{4}{ }^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}} $$ What is the concentration of \(\mathrm{Hg}^{2+}\) in \(500.0 \mathrm{~mL}\) of a solution that was originally \(0.010 \mathrm{M} \mathrm{Hg}^{2+}\) and \(0.78 \mathrm{M} \mathrm{I}^{-}\) ? The reaction is $$ \mathrm{Hg}^{2+}(a q)+4 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{HgI}_{4}^{2-}(a q) $$
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