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Chapter 18: Problem 54

Calculate \(\left[\mathrm{Cu}^{2+}\right]\) in a \(0.10 \mathrm{M} \mathrm{CuSO}_{4}(\) aq) solution that is also \(6.0 \mathrm{M}\) in free \(\mathrm{NH}_{3}\). \(\mathrm{Cu}^{2+}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}(\mathrm{aq})\) \(K_{\mathrm{f}}=1.1 \times 10^{13}\)

Short Answer

Expert verified
\([\mathrm{Cu}^{2+}]\) in the solution is approximately \(1.2 \times 10^{-19} M\).

Step by step solution

01

Initial Concentrations

Set up initial concentrations. From the initial molarity of \(\mathrm{CuSO}_{4}\), it is known that the initial concentration of \(\mathrm{Cu}^{2+}\) is 0.10 M. Also, the initial molar concentration of \(\mathrm{NH}_{3}\) is given as 6.0 M.
02

Establish change based on reaction

Discuss the change in concentrations that occur after the reaction. Based on the stoichiometry of the reaction, for each mole of \(\mathrm{Cu}^{2+}\) that reacts, four moles of \(\mathrm{NH}_{3}\) are consumed. Therefore, assuming all the \(\mathrm{Cu}^{2+}\) reacts, the change in concentration for \(\mathrm{NH}_{3}\) will be 4 times the initial concentration for \(\mathrm{Cu}^{2+}\). The \(\mathrm{NH}_{3}\) will decrease by (4)(0.10 M) = 0.40 M, giving a final concentration of 6.0 M - 0.40 M = 5.6 M.
03

Formation constant

Using the formation constant, set up an expression for the equilibrium. The right side of the equilibrium is predominant according to the high formation constant, so the concentration of the complex ion can be approximated by the initial concentration of \(\mathrm{Cu}^{2+}\), this is \(0.10 M\). Also, the concentration of \(\mathrm{NH}_{3}\) has already been calculated, so we presume \([\mathrm{Cu}^{2+}]\) is very small and can be considered as \(x\). Plugging all values into the equilibrium constant equation \(K_{f}=\frac{[\mathrm{Cu}(NH_{3})_{4}]^{2+}}{[\mathrm{Cu}^{2+}][NH_{3}]^{4}}\), this simplifies to \(1.1 \times 10^{13} = \frac{0.10}{x(5.6)^4}\).
04

Calculate concentration

Rearrange and solve for \([\mathrm{Cu}^{2+}]\). Solving the previous equation will find the concentration of \([\mathrm{Cu}^{2+}]\) in the solution. The answer is approximately \(1.2 \times 10^{-19} M\).

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Most popular questions from this chapter

The electrolysis of \(\mathrm{MgCl}_{2}(\mathrm{aq})\) can be represented as \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) The electrolysis of a 315 mL sample of \(0.185 \mathrm{M} \mathrm{MgCl}_{2}\) is continued until \(0.652 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) has been collected. Will \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate when electrolysis is carried to this point? [Hint: Notice that \(\left[\mathrm{Mg}^{2+}\right]\) remains constant throughout the electrolysis, but \(\left.\left[\mathrm{OH}^{-}\right] \text {increases. }\right]\)

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-11}\right)\) in (a) pure water; (b) \(0.0862 \mathrm{M}\) \(\mathrm{MgCl}_{2} ;\) (c) \(0.0355 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\).

To precipitate as \(\mathrm{Ag}_{2} \mathrm{S}(\mathrm{s}),\) all the \(\mathrm{Ag}^{+}\) present in \(338 \mathrm{mL}\) of a saturated solution of \(\mathrm{AgBrO}_{3}\) requires \(30.4 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) measured at \(23^{\circ} \mathrm{C}\) and \(748 \mathrm{mm} \mathrm{Hg} .\) What is \(K_{\mathrm{sp}}\) for \(\mathrm{AgBrO}_{3} ?\)

Which of the following would be most effective, and which would be least effective, in reducing the concentration of the complex ion \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) in a solution: \(\mathrm{HCl}, \mathrm{NH}_{3},\) or \(\mathrm{NH}_{4} \mathrm{Cl} ?\) Explain your choices.

Which of the following has the highest molar solubility? (a) \(\mathrm{MgF}_{2}, K_{\mathrm{sp}}=3.7 \times 10^{-8}\) \(\mathrm{MgCO}_{3}\), \(K_{\mathrm{sp}}=3.5 \times 10^{-8} ;(\mathrm{c}) \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-25}\); (d) \(\mathrm{Li}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=3.2 \times 10^{-9}\).
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