Problem 52 - Chemistry: Principles and Reactions

Question

A 35-mL sample of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(35 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) NaI labeled with I-131. The following reaction occurs. $$\mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{AgI}(s)$$ The filtrate is found to have an activity of \(2.50 \times 10^{3}\) counts per minute per milliliter. The \(0.050 \mathrm{M}\) NaI solution had an activity of \(1.25 \times 10^{10}\) counts per minute per milliliter. Calculate \(K_{\text {sp }}\) for AgI.

Step-by-Step Solution

Answer

Based on the provided solution, calculate the \(K_{sp}\) for AgI using the equilibrium concentrations of Ag\(^+\) and I\(^-\) ions found in the mixed solution.

Step 1: Calculate the initial amount of I\(^-\) ions

Before mixing, only NaI has I\(^-\) ions in it. Since the concentration is \(0.050\,\mathrm{M}\) and the volume is \(35\,\mathrm{mL}\), we can calculate the initial moles of I\(^-\) ions: $$\text{moles of I}^{-} = (0.050\,\mathrm{M})(0.035\,\mathrm{L}) = 1.75\times10^{-3}\,\mathrm{mol}$$

Step 2: Calculate the remaining amount of I\(^-\) ions

The filtrate has an activity of \(2.50\times10^3\,\mathrm{counts/min/mL}\), and the original NaI solution had an activity of \(1.25\times10^{10}\,\mathrm{counts/min/mL}\). The ratio of activities gives us the ratio of remaining I\(^-\) ions: $$\frac{\mathrm{remaining\, I}^{-}\, \mathrm{ions}}{\mathrm{initial\, I}^{-}\, \mathrm{ions}} = \frac{2.50\times10^3\,\mathrm{counts/min/mL}}{1.25\times10^{10}\,\mathrm{counts/min/mL}}$$ $$\mathrm{remaining\, I}^{-}\, \mathrm{ions} = 1.75\times10^{-3}\,\mathrm{mol} \times \frac{2.50\times10^3\,\mathrm{counts/min/mL}}{1.25\times10^{10}\,\mathrm{counts/min/mL}} = 3.50\times10^{-8}\,\mathrm{mol}$$

Step 3: Calculate the remaining amount of Ag\(^+\) ions

Since the reaction has a 1:1 stoichiometry, the remaining amount of Ag\(^+\) ions will be equal to the remaining amount of I\(^-\) ions: $$\mathrm{remaining\, Ag}^{+}\, \mathrm{ions} = 3.50\times10^{-8}\,\mathrm{mol}$$

Step 4: Calculate the equilibrium concentrations of Ag\(^+\) and I\(^-\) ions

Now we need to find the concentrations of Ag\(^+\) and I\(^-\) ions in the 70 mL mixture: $$[\mathrm{Ag}^+] = \frac{3.50\times10^{-8}\,\mathrm{mol}}{0.070\,\mathrm{L}} = 5.00\times10^{-7}\,\mathrm{M}$$ $$[\mathrm{I}^-] = \frac{3.50\times10^{-8}\,\mathrm{mol}}{0.070\,\mathrm{L}} = 5.00\times10^{-7}\,\mathrm{M}$$

Step 5: Calculate \(K_{sp}\) for AgI

Finally, we can calculate the solubility product constant (\(K_{sp}\)) for AgI using the equilibrium concentrations of Ag\(^+\) and I\(^-\) ions: $$K_{sp} = [\mathrm{Ag}^+][\mathrm{I}^-] = (5.00\times10^{-7})(5.00\times10^{-7})=2.50\times10^{-13}$$ So, the \(K_{sp}\) for AgI is \(2.50\times10^{-13}\).

Key Concepts

Chemical Equilibrium

In the world of chemistry, chemical equilibrium refers to a condition where the rates of the forward and reverse reactions are equal, resulting in no net change in the amounts of products and reactants. It is a dynamic state, meaning that the reactions are still occurring, but at identical rates, so the system's macroscopic properties remain constant. For the reaction \( \mathrm{Ag}^{+}(aq) + \mathrm{I}^{-}(aq) \longrightarrow \mathrm{AgI}(s) \), chemical equilibrium is achieved when the rate at which \( \mathrm{Ag}^{+} \) ions combine with \( \mathrm{I}^{-} \) ions to form solid \( \mathrm{AgI} \) equals the rate at which \( \mathrm{AgI} \) breaks apart into its ions.

Stoichiometry

With regard to stoichiometry, it is essentially the mathematics behind the science of chemistry. It involves calculating the quantities of reactants and products involved in a chemical reaction. In our particular example, the stoichiometry of the reaction dictates a 1:1 molar ratio between silver ions \( \mathrm{Ag}^{+} \) and iodide ions \( \mathrm{I}^{-} \). This means that for every mole of \( \mathrm{Ag}^{+} \) that reacts, one mole of \( \mathrm{I}^{-} \) is required, and vice versa. Understanding this relationship is crucial when calculating concentrations of ions at equilibrium and determining the solubility product constant, \( K_{sp} \).

Molarity and Concentration

In the field of chemistry, molarity and concentration are terms frequently used to describe the strength of a solution. Molarity, denoted as \( M \) and measured in moles per liter (mol/L), is the number of moles of solute divided by the volume of the solution in liters. Concentrations are crucial when dealing with reactions in solution as they directly affect the reaction rates and the position of equilibrium. By calculating the concentration of the remaining \( \mathrm{Ag}^{+} \) and \( \mathrm{I}^{-} \) ions post-reaction, we were able to determine the solubility product constant for silver iodide (AgI).

Radioactive Decay and Activity

The phenomenon of radioactive decay is associated with unstable isotopes, which release energy by emitting radiation to become more stable. This radioactive emission is what we refer to as activity, typically measured in counts per minute (cpm) or becquerels (Bq). In solving our exercise, we analyzed the decay of I-131, an isotope of iodine. The activity of a radioactive sample can provide insight into the quantity of radioactive substance present. By comparing the activities of the initial and remaining \( \mathrm{I}^{-} \) ions, we were able to deduce the amount of iodine that precipitated out as \( \mathrm{AgI} \) and calculate the solubility product constant.