Chapter 15: Problem 2
Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.
Chapter 15: Problem 2
Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.
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Get started for freeWrite equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{Ni}(\mathrm{CN})_{4}^{2-}\) b. \(\mathrm{V}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}\)
A solution contains \(2.0 \times 10^{-3} M \mathrm{Ce}^{3+}\) and \(1.0 \times 10^{-2} M\) IO \(_{3}^{3-}\) Will \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} \text { is } 3.2 \times 10^{-10} .\right]\)
The solubility of the ionic compound \(\mathrm{M}_{2} \mathrm{X}_{3},\) having a molar mass of \(288 \mathrm{g} / \mathrm{mol},\) is \(3.60 \times 10^{-7} \mathrm{g} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) of the compound.
a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\).
When aqueous KI is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{Hg} \mathrm{I}_{4}^{2-} .\) )
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