Chapter 16: Problem 12
Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
Chapter 16: Problem 12
Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.
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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} \mathrm{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 \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$
Calculate the mass of manganese hydroxide that dissolves to form 1300 mL of a saturated manganese hydroxide solution. For $\mathrm{Mn}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=2.0 \times 10^{-13}.$
Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\) $\left(K_{\mathrm{sp}}=4 \times 10^{-38}\right)$ in each of the following. a. water b. a solution buffered at pH \(=5.0\) c. a solution buffered at pH\(=11.0\)
Write equations for the step wise formation of each of the following complex ions. a. \(N i(C N)_{4}^{2-}\) b. \(V\left(C_{2} O_{4}\right)_{3}^{3-}\)
A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and $1.0 \times 10^{-3} M$ \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and $K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .$
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