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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鈦� 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 AgNO鈧� and NaIO鈧� solutions: Moles of Ag鈦� ions: \(moles_{Ag} = concentration 脳 volume = (0.0413 \thinspace M)(0.050 \thinspace L) = 0.002065 \thinspace mol\) Moles of IO鈧冣伝 ions: \(moles_{IO_{3}} =concentration 脳 volume = (0.100 \thinspace M)(0.050 \thinspace L) = 0.005 \thinspace mol\)
02

Determine limiting reactants and moles at equilibrium

As some of the Ag鈦� ions will combine with IO鈧冣伝 ions to form AgIO鈧�, 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鈦� ions Now, we can find the moles of ions left at equilibrium: Moles of Ag鈦� ions at equilibrium: \(0.002065 \thinspace mol - 0.002065 \thinspace mol = 0 \thinspace mol\) Moles of IO鈧冣伝 ions at equilibrium: \(0.005 \thinspace mol - 0.002065 \thinspace mol = 0.002935 \thinspace mol\) Moles of AgIO鈧� 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 IO鈧冣伝 ions in the mixture: Concentration of IO鈧冣伝 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鈦� ions

The Ksp expression for the equilibrium of AgIO鈧� is given as follows: \(K_{sp} = [Ag^{+}][IO_{3}^{-}]\) Using the known Ksp value for AgIO鈧� and the calculated equilibrium concentration of IO鈧冣伝 ion, we can find the equilibrium concentration of Ag鈦� ion: \(3.17 \times 10^{-8} = [Ag^{+}](0.02935)\) To find the equilibrium concentration of Ag鈦� ions, simply divide the Ksp value by the equilibrium concentration of IO鈧冣伝 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鈦� ions in the resulting solution is \(1.08 \times 10^{-9} \thinspace M\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

chemical equilibrium
Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. In the context of solubility, the establishment of an equilibrium state involves the dissolving solid and its constituent ions in a saturated solution. When AgNO鈧� and NaIO鈧� are mixed, initially there is precipitation until a dynamic equilibrium is achieved. Here, the undissolved AgIO鈧� solid coexists with its ions in solution. At equilibrium, the concentrations of the dissolved ions remain constant, governed by the solubility product constant, or Ksp. This reflects the balance of partial dissolution and precipitation processes.
limiting reactant
In a chemical reaction, the limiting reactant is the substance that is totally consumed first, limiting the amount of products formed. When combining AgNO鈧� and NaIO鈧�, we must first calculate the initial moles of Ag鈦� and IO鈧冣伝 ions present. With a 1:1 stoichiometric ratio for forming AgIO鈧�, whichever reactant has fewer moles will be the limiting reactant. Here, Ag鈦� ions are scarce compared to IO鈧冣伝, confirming Ag鈦� as the limiting reactant. Consequently, all Ag鈦� ions are consumed to form the precipitate, while excess IO鈧冣伝 ions remain in the solution, influencing their equilibrium concentration.
equilibrium concentration
Equilibrium concentration refers to the concentration of reactants or products in a reaction mixture when a chemical equilibrium is reached. After the precipitation of AgIO鈧�, the remaining concentration of ions in the solution must be recalculated. Using the total volume of the mixture, the concentration of IO鈧冣伝 is found after considering the amount that has reacted. Similarly, the Ag鈦� concentration at equilibrium can be indirectly calculated using the Ksp expression, which relates the product of the ion concentrations to the Ksp value, ensuring that the system remains at equilibrium.
molarity
Molarity is a measure of concentration, expressed as moles of solute per liter of solution (mol/L). To solve problems involving aqueous reactions, such as finding the equilibrium concentration of ions, it is essential to determine molarity accurately. When AgNO鈧� and NaIO鈧� are mixed, their volumes sum to create a new total solution volume. This volume is pivotal for finding molarity at any stage of the reaction. By dividing the remaining or excess moles of ions by the total volume post-reaction, molarity gives insight into how concentrated each ion is in the solution post-equilibrium.

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Most popular questions from this chapter

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 \(1.0 \times 10^{-3} \mathrm{M}\), respec- tively, in a \(1.5-M \mathrm{NH}_{3}\) solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). \(\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K_{\text {overall }}=?\)

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. \(\mathbf{a}\). water b. \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}(a q)\), \(\mathrm{Ba}^{2+}(a q)\), and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{BaBr}_{2}\). (For \(\left.\mathrm{BaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)

Use the following data to calculate the \(K_{\text {sp }}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\).

When aqueous \(\mathrm{KI}\) is added gradually to mercury(II) nitrate, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. ( Hint: \(\mathrm{Hg}^{2+}\) reacts with \(\mathrm{I}^{-}\) to form \(\mathrm{HgI}_{4}{ }^{2-} .\) )
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