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Chapter 16: Problem 97

A \(50.0-\mathrm{mL}\) sample of \(0.0413 \mathrm{MAgNO}_{3}(a q)\) is added to \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaIO}_{3}(a q)\). Calculate the \(\left[\mathrm{Ag}^{+}\right]\) at equilibrium in the resulting solution. \(\left[K_{\text {sp }}\right.\) for \(\mathrm{AgIO}_{3}(s)=3.17 \times 10^{-8}\).]

Short Answer

Expert verified
The equilibrium concentration of Ag⁺ ions in the resulting solution is \(1.08 \times 10^{-9} \thinspace M\).

Step by step solution

01

Calculate initial moles of ions

First, we calculate the initial moles of ions in the solutions using the given concentrations and volumes. For AgNO₃ and NaIO₃ solutions: Moles of Ag⁺ ions: \(moles_{Ag} = concentration × volume = (0.0413 \thinspace M)(0.050 \thinspace L) = 0.002065 \thinspace mol\) Moles of IO₃⁻ ions: \(moles_{IO_{3}} =concentration × volume = (0.100 \thinspace M)(0.050 \thinspace L) = 0.005 \thinspace mol\)
02

Determine limiting reactants and moles at equilibrium

As some of the Ag⁺ ions will combine with IO₃⁻ ions to form AgIO₃, we have to find out the limiting reactant in this reaction. Since they react in a 1:1 ratio, the limiting reactant will be the one with the lowest amount of moles: Limiting reactant: Ag⁺ ions Now, we can find the moles of ions left at equilibrium: Moles of Ag⁺ ions at equilibrium: \(0.002065 \thinspace mol - 0.002065 \thinspace mol = 0 \thinspace mol\) Moles of IO₃⁻ ions at equilibrium: \(0.005 \thinspace mol - 0.002065 \thinspace mol = 0.002935 \thinspace mol\) Moles of AgIO₃ precipitate formed: \(0.002065 \thinspace mol\)
03

Calculate equilibrium concentrations of ions

Next, we will find the equilibrium concentrations of ions in the mixture. To do this, we need to find the total volume of the mixture: Total volume of the mixture: \(0.050 \thinspace L + 0.050 \thinspace L = 0.100 \thinspace L\) Now, we can calculate the equilibrium concentrations of IO₃⁻ ions in the mixture: Concentration of IO₃⁻ ions at equilibrium: \( \dfrac{0.002935 \thinspace mol}{0.100 \thinspace L} =0.02935 \thinspace M\)
04

Set up and solve for the equilibrium concentration of Ag⁺ ions

The Ksp expression for the equilibrium of AgIO₃ is given as follows: \(K_{sp} = [Ag^{+}][IO_{3}^{-}]\) Using the known Ksp value for AgIO₃ and the calculated equilibrium concentration of IO₃⁻ ion, we can find the equilibrium concentration of Ag⁺ ion: \(3.17 \times 10^{-8} = [Ag^{+}](0.02935)\) To find the equilibrium concentration of Ag⁺ ions, simply divide the Ksp value by the equilibrium concentration of IO₃⁻ ions: \([Ag^{+}] = \dfrac{3.17 \times 10^{-8}}{0.02935} = 1.08 \times 10^{-9} \thinspace M\)
05

Final answer

The equilibrium concentration of Ag⁺ ions in the resulting solution is \(1.08 \times 10^{-9} \thinspace M\).

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Most popular questions from this chapter

\(\mathrm{Ag}_{2} \mathrm{~S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{~S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\), \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\), and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{~mL}\) of \(3.00 \mathrm{M} \mathrm{NH}_{3}\) with \(500.0 \mathrm{~mL}\) of \(2.00 \times 10^{-3} M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). The stepwise equilib- ria are \(\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q)\) \(K_{1}=1.86 \times 10^{4}\) \(\mathrm{CuNH}_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q)\) \(K_{2}=3.88 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)\) \(K_{3}=1.00 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q)\) \(K_{4}=1.55 \times 10^{2}\)

Consider a solution made by mixing \(500.0 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} \cdot \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\begin{aligned} \mathrm{AgNH}_{3}{ }^{+} \text {and } \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+: \\ \mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}{ }^{+}(a q) & K_{1}=2.1 \times 10^{3} \\ \mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) & K_{2}=8.2 \times 10^{3} \end{aligned}\) Determine the concentration of all species in solution.

On a hot day, a \(200.0-\mathrm{mL}\) sample of a saturated solution of \(\mathrm{Pb} \mathrm{I}_{2}\) was allowed to evaporate until dry. If \(240 \mathrm{mg}\) of solid \(\mathrm{PbI}_{2}\) was collected after evaporation was complete, calculate the \(K_{\text {sp }}\) value for \(\mathrm{Pb} \mathrm{I}_{2}\) on this hot day.

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: $$ \begin{aligned} \mathrm{Cu}^{+}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \operatorname{CuX}(a q) & K_{1}=1.0 \times 10^{2} \\ \mathrm{CuX}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{2}-(a q) & K_{2}=1.0 \times 10^{4} \\ \mathrm{CuX}_{2}^{-}(a q)+\mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}{ }^{2-}(a q) & K_{3}=1.0 \times 10^{3} \end{aligned} $$ with an overall reaction \(\mathrm{Cu}^{+}(a q)+3 \mathrm{X}^{-}(a q) \rightleftharpoons \mathrm{CuX}_{3}^{2-}(a q) \quad K=1.0 \times 10^{9}\) Calculate the following concentrations at equilibrium. a. \(\mathrm{CuX}_{3}^{2-}\) b. \(\mathrm{CuX}_{2}^{-}\) c. \(\mathrm{Cu}^{+}\)
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