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Chapter 18: Problem 53

In a solution that is \(0.0500 \mathrm{M}\) in \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\) and \(0.80 \mathrm{M}\) in free \(\mathrm{CN}^{-}\), the concentration of \(\mathrm{Cu}^{+}\) is \(6.1 \times 10^{-32} \mathrm{M}\) Calculate \(K_{\mathrm{f}}\) of \(\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}\). \(\mathrm{Cu}^{+}(\mathrm{aq})+4 \mathrm{CN}^{-}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}(\mathrm{aq}) \quad K_{\mathrm{f}}=?\)

Short Answer

Expert verified
The formation constant \(K_f\) of \([Cu(CN)_4]^{3-}\) is approximately \(6.41 \times 10^{36}\).

Step by step solution

01

Understand the equilibrium concept

The equilibrium constant \(K\) defines the ratio of the concentrations of the products to the reactants at equilibrium. In the reaction \(Cu^+ + 4CN^- \rightarrow [Cu(CN)_4]^{3-}\), the formation constant \(K_f\) is thus given by: \[K_f = \frac{([Cu(CN)_4]^{3-})}{[Cu^+][CN^-]^4}\]
02

Substitute given concentration values

Substitute the given concentration values of \([Cu(CN)_4]^{3-}\) = \(0.0500 M\), \([Cu^+]\) = \(6.1 \times 10^{-32} M\), and \([CN^-]\) = \(0.80 M\) into the equation for \(K_f\). So: \[K_f = \frac{(0.0500)}{(6.1 \times 10^{-32})(0.80)^4}\]
03

Solve for \(K_f\)

Carrying out the calculation we get: \[K_f \approx 6.41 \times 10^{36}\]

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Most popular questions from this chapter

All but two of the following solutions yield a precipitate when the solution is also made \(2.00 \mathrm{M}\) in \(\mathrm{NH}_{3}\). Those two are (a) \(\mathrm{MgCl}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{FeCl}_{3}(\mathrm{aq})\); (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq}) ;(\mathrm{d}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\); (e) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(\mathrm{aq})\).

Which of the following solids are likely to be more soluble in acidic solution and which in basic solution? Which are likely to have a solubility that is independent of pH? Explain. (a) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} ;\) (b) \(\mathrm{MgCO}_{3} ;\) (c) \(\mathrm{CdS}\); (d) \(\mathrm{KCl} ;\) (e) \(\mathrm{NaNO}_{3} ;\) (f) \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Which of the following solids is (are) more soluble in an acidic solution than in pure water: \(\mathrm{KCl}\), \(\mathrm{MgCO}_{3}\), \(\mathrm{FeS}, \mathrm{Ca}(\mathrm{OH})_{2,}\) or \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} ?\) Explain.

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)

The addition of \(\mathrm{HCl}(\mathrm{aq})\) to a solution containing several different cations produces a white precipitate. The filtrate is removed and treated with \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})\) in 0.3 M HCl. No precipitate forms. Which of the following conclusions is (are) valid? Explain. (a) \(\mathrm{Ag}^{+}\) or \(\mathrm{Hg}_{2}^{2+}\) (or both) is probably present. (b) \(\mathrm{Mg}^{2+}\) is probably not present. (c) \(\mathrm{Pb}^{2+}\) is probably not present. (d) \(\mathrm{Fe}^{2+}\) is probably not present.
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