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Chapter 16: Problem 77

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+\) is $1.7 \times 10^{7} . K_{\mathrm{sp}}\( for \)\mathrm{AgCl}\( is \)1.6 \times 10^{-10} .$ Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0M\) \(\mathrm{NH}_{3}.\)

Short Answer

Expert verified
The molar solubility of AgCl in 1.0M NH3 solution is approximately \(1.2 \times 10^{-4} M\).

Step by step solution

01

Write the balanced chemical equations

Write the balanced chemical equation for the formation of the complex ion Ag(NH3)2+ from Ag+ and NH3, and also the dissolution of AgCl. Formation of Ag(NH3)2+ complex ion: \[Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+\] Dissolution of AgCl: \[AgCl(s) \rightleftharpoons Ag^+ + Cl^-\]
02

Write the expressions for Kf and Ksp

Write the expressions for the formation constant (Kf) and the solubility product constant (Ksp) for the given reactions. For the formation of Ag(NH3)2+ complex ion: \[K_f = [\ce{Ag(NH3)2^+}] / ([\ce{Ag^+}] [\ce{NH3}]^2)\] For the dissolution of AgCl: \[K_{sp} = [\ce{Ag^+}] [\ce{Cl^-}]\]
03

Set up and solve the equation relating Ksp and molar solubility

Let the molar solubility of \(\ce{AgCl}\) in \(\ce{NH3}\) be x. From the balanced chemical equation of AgCl dissolution, we can set up the equilibrium concentrations: \[[\ce{Ag^+}] = [\ce{Cl^-}] = x\] Plug these equilibrium concentrations into the Ksp expression: \[K_{sp} = x^2\] Now, solve for x: \[x = \sqrt{K_{sp}}\]
04

Set up and solve the equation relating Kf and equilibrium concentrations

Since the initial concentration of NH3 is 1.0M, after reaction with Ag+, the equilibrium concentration of NH3 will be: \[[\ce{NH3}] = 1.0 - 2x\] Since Ag+ reacts with NH3 to form Ag(NH3)2+, equilibrium concentrations of Ag+ and Ag(NH3)2+ are related as: \[[\ce{Ag(NH3)2^+}] = x\] Now, plug these equilibrium concentrations into the Kf expression: \[K_f = x / (x (1.0 - 2x)^2)\] Solve for x: \[x = \sqrt[3]{\cfrac{K_f}{(1.0 - 2x)^2}}\]
05

Combine the equations and solve for molar solubility

Combine the equations obtained in Step 3 and Step 4 and substitute the given values for Kf and Ksp: \[x = \sqrt[3]{\cfrac{1.7 \times 10^{7}}{(\ 1.0 - 2 \sqrt{1.6 \times 10^{-10}}\ )^2}}\] Now, solve for x, which is the molar solubility of AgCl in 1.0M NH3: \[x \approx 1.2 \times 10^{-4} M\] The molar solubility of AgCl in 1.0M NH3 solution is approximately \(1.2 \times 10^{-4} M\).

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Most popular questions from this chapter

A solution is formed by mixing \(50.0 \mathrm{mL}\) of \(10.0 \mathrm{M}\) \(\mathrm{NaX}\) with \(50.0 \mathrm{mL}\) of $2.0 \times 10^{-3} \mathrm{M} \mathrm{CuNO}_{3} .\( Assume that \)\mathrm{Cu}^{+}$ forms complex ions with \(\mathrm{X}^{-}\) as follows: $$\mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}(a q) \quad K_{1}=1.0 \times 10^{2}$$ $$\operatorname{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}^{-}(a q) \qquad K_{2}=1.0 \times 10^{4}$$ $$\operatorname{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K_{3}=1.0 \times 10^{3}$$ with an overall reaction $$\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \qquad K=1.0 \times 10^{9}$$ Calculate the following concentrations at equilibrium a. \(\mathrm{CuX}_{3}^{2-} \quad\) b. \(\mathrm{CuX}_{2}^{-} \quad\) c. \(\mathrm{Cu}^{+}\)

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of $\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}$ b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

When 100.0 \(\mathrm{mL}\) of 2.00 $\mathrm{M} \mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{3}\( is added to 100.0 \)\mathrm{mL}$ of \(3.00 \mathrm{M} \mathrm{KIO}_{3},\) a precipitate of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) forms. Calculate the equilibrium concentrations of \(\mathrm{Ce}^{3+}\) and \(\mathrm{IO}_{3}^{-}\) in this solution. $\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}=3.2 \times 10^{-10} .\right]$

Calculate the solubility of solid $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\( in a \)0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}$ solution.

A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)
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