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Chapter 15: Problem 64

In the presence of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) forms the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) If the equilibrium concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) are \(1.8 \times 10^{-17} M\) and \(1.0 \times 10^{-3} M,\) respectively, in a \(1.5-M \mathrm{NH}_{3}\) solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{\mathrm{overall}}=?$$

Short Answer

Expert verified
The overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is approximately \(1.097 \times 10^{14}\).

Step by step solution

01

List the given equilibrium concentrations

We are given that $$[\mathrm{Cu}^{2+}] = 1.8 \times 10^{-17} M$$ $$[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}] = 1.0 \times 10^{-3} M$$ And the concentration of the solution $$[\mathrm{NH}_{3}] = 1.5 M$$
02

Insert equilibrium concentrations into the K expression

Now that we have the equilibrium concentrations, we can plug them into our K expression: $$K_{\mathrm{overall}}=\frac{[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]}{[\mathrm{Cu}^{2+}][\mathrm{NH}_{3}]^4}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})[(1.5)^4]}$$
03

Calculate K

Now, we can calculate the K value for the complex ion: $$K_{\mathrm{overall}}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})[(1.5)^4]}=\frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})(5.0625)}$$ $$K_{\mathrm{overall}}=\frac{1.0 \times 10^{-3}}{9.1125 \times 10^{-17}}\approx 1.097 \times 10^{14}$$ Thus, the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) is approximately \(1.097 \times 10^{14}\).

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Most popular questions from this chapter

A solution contains \(1.0 \times 10^{-5} M \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} M \mathrm{CN}^{-}\) Will AgCN( \(s\) ) precipitate? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{AgCN}(s) \text { is } 2.2 \times 10^{-12} .\right)\)

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7}\) \(\operatorname{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{sp}}\) for this salt. Ignore any potential reactions of the ions with water.

Sodium tripolyphosphate \(\left(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\right)\) is used in many synthetic detergents. Its major effect is to soften the water by complexing \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions. It also increases the efficiency of surfactants, or wetting agents that lower a liquid's surface tension. The \(K\) value for the formation of \(\mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}\) is \(4.0 \times 10^{8} .\) The reaction is \(\mathrm{Mg}^{2+}(a q)+\mathrm{P}_{3} \mathrm{O}_{10}^{5-}(a q) \rightleftharpoons \mathrm{MgP}_{3} \mathrm{O}_{10}^{3-}(a q)\) Calculate the concentration of \(\mathrm{Mg}^{2+}\) in a solution that was originally \(50 .\) ppm \(\mathrm{Mg}^{2+}(50 . \mathrm{mg} / \mathrm{L} \text { of solution) after } 40 . \mathrm{g}\) \(\mathrm{Na}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\) is added to \(1.0 \mathrm{L}\) of the solution.

Calculate the solubility of solid \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1 \times 10^{-54}\right)\) in a \(0.10-M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{CoF}_{6}^{3-}\) b. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)
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