Chapter 15: 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 15: 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 freeIn the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} M\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \quad K_{\mathrm{overall}}=?$$
What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{L}\) of a \(1.0-M \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=\) \(4.0 \times 10^{-11}\) and \(K_{\mathrm{a}}\) for \(\mathrm{HF}=7.2 \times 10^{-4} .\) Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).
The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: $$\begin{aligned} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}^{2-}(a q) \\ +\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) & K=? \end{aligned}$$ If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{mol} / \mathrm{L},\) calculate the equilibrium constant for the preceding reaction.
\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.
a. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in water, ignoring the basic properties of \(\left.\mathrm{F}^{-} . \text {(For } \operatorname{Sr} \mathrm{F}_{2}, K_{\mathrm{sp}}=7.9 \times 10^{-10} .\right)\). b. Would the measured molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) be greater than or less than the value calculated in part a? Explain. c. Calculate the molar solubility of \(\operatorname{Sr} \mathrm{F}_{2}\) in a solution buffered at \(\mathrm{pH}=2.00 .\left(K_{\mathrm{a}} \text { for } \mathrm{HF} \text { is } 7.2 \times 10^{-4} .\right)\).
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