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 freeCalculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{ED} \mathrm{TA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{EDTA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+}\) is Consider a solution with \(0.010 \mathrm{~mol} \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to \(1.0 \mathrm{~L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing \(0.050 \mathrm{M} \mathrm{Na}_{4} \mathrm{EDTA} .\) Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?
A solution is \(1 \times 10^{-4} \mathrm{M}\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{~S}\), and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\text {sp }}\) values are \(K_{\text {sp }}\left(\mathrm{PbF}_{2}\right)=\) \(4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29}\), and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}\)
A solution contains \(0.018\) mol each of \(\mathrm{I}^{-}, \mathrm{Br}^{-}\), and \(\mathrm{Cl}^{-}\). When the solution is mixed with \(200 . \mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\), what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$\begin{aligned}\text { AgI: } K_{\text {sp }} &=1.5 \times 10^{-16} \\\\\text { AgBr: } K_{\text {sp }} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\text {sp }} &=1.6 \times 10^{-10}\end{aligned}$$
Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-} .\) In terms of solubility, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the \(\mathrm{pH}\) dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\), as a function of \(\left[\mathrm{H}^{+}\right]\), obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is \(40.0\) and \(K_{\mathrm{se}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the \(\mathrm{pH}\) range \(4-12\).
A \(50.0-\mathrm{mL}\) sample of \(0.00200 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{NaIO}_{3-}\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgIO}_{3}\) is \(3.0 \times 10^{-8}\).)
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