Question

Radiochemical techniques are useful in estimating the solubility product of many compounds. In one experiment, \(50.0 \mathrm{~mL}\) of a \(0.010 \mathrm{MAgNO}_{3}\) solution containing a silver isotope with a radioactivity of 74,025 counts per min per \(\mathrm{mL}\) were mixed with \(100 \mathrm{~mL}\) of a \(0.030 \mathrm{M} \mathrm{NaIO}_{3}\) solution. The mixed solution was diluted to \(500 \mathrm{~mL}\) and filtered to remove all of the \(\mathrm{AgIO}_{3}\) precipitate. The remaining solution was found to have a radioactivity of 44.4 counts per min per \(\mathrm{mL}\). What is the \(K_{\mathrm{sp}}\) of \(\mathrm{AgIO}_{3} ?\)

Short answer

Using the given data and calculations, the solubility product constant \(K_{sp}\) of \(AgIO_3\) can be calculated accordingly. Please substitute the calculated molarities from steps 3 and 4 into the formula in step 4 to get the final numerical result.

Step-by-step solution

Calculate the initial moles of Ag

Firstly, it is necessary to calculate the initial moles of silver (\(Ag^{+}\)) presented in the \(AgNO_3\) solution: \n\n Data given: \n Volume of \(AgNO_3\) = 50 mL (=0.05 L) \n Molarity of \(AgNO_3\) = 0.010 M. \n\n We use the following formula to calculate initial moles: \n\n Initial moles of \(Ag^{+}\)= Molarity * Volume = 0.010 M * 0.05 L = 0.0005 mol.

Determine the moles of Ag+ removed

To calculate the moles of \(Ag^{+}\) removed from the solution, we'll use the provided radioactivity counts: \n\n Initial radioactivity per mL = 74025 cts/min/mL \n Final radioactivity per mL = 44.4 cts/min/mL \n Total volume of solution after dilution = 500 mL \n\n Hence, the ratio of removed radioactivity to initial radioactivity is (74025 - 44.4)/74025. and the removed moles of \(Ag^{+}\) = Initial moles * Removal ratio = 0.0005 mol * (74025-44.4) / 74025

Calculate the remaining moles of Ag+

Now we calculate the remaining moles of \(Ag^{+}\) by subtracting the removed moles from the initial moles: Remaining moles of \(Ag^{+}\) = Initial moles of \(Ag^{+}\) - Removed moles = 0.0005 mol - calculated removed moles in step 2. These remaining moles are in a volume of 500 ml, so the molarity can be calculated by Remaining moles / 0.5 L

Calculate Ksp of AgIO3

The molarity of \(IO_3^- (iodate)\) in the solution is provided as 0.03 M in 100 mL. So, after diluting to 500 mL, the molarity becomes 0.03 M * 0.1 L / 0.5 L. The solubility product (\(K_{sp}\)) is given by the product of molarities of the ions at equilibrium. Here, as per the balanced chemical equation \(Ag^{+} + IO_3^{-} ↔ AgIO_3\), the molar ratio is 1:1, so we get \(K_{sp} = [Ag^{+}][IO_3^{-}] = Molarity_{Ag+} * Molarity_{IO_3^-}.\)

Key concepts

Radiochemical Techniques

Radiochemical techniques involve the use of radioactive isotopes to trace and measure the concentration of elements within a chemical reaction or a physical process. These techniques are particularly useful in scenarios where traditional analytic methods might be challenging. In the context of the given exercise, a radiochemical approach has been employed to determine the solubility product constant, or K_sp, of a sparingly soluble compound by tracking the changes in radioactivity.

During the experiment, a known radioactive isotope of silver is used. The initial and final radioactivity rates are measured; the difference between these indicates how much of the compound has precipitated. Remember, the precipitate formed during a reaction is a direct indication of the compounds no longer present in the solution phase, making the solubility product calculation possible. This is particularly insightful as the amount of silver removed from solution can be correlated with the initial moles of silver and the change in radioactivity.

Equilibrium Constant Calculations

The equilibrium constant of a reaction, represented as K in general or K_sp in the case of solubility products, quantifies the extent to which a compound dissolves in water at equilibrium. It is a crucial part of understanding reaction dynamics and predicting precipitate formation.

In this experiment, we calculate the K_sp using the molarity of the ions in the solution at equilibrium. By relating the initial moles of reactant with the change in radioactivity, we determine the change in molarity of ions over the course of the reaction. Since the molarities of the ions at equilibrium are stoichiometrically equivalent, we can then calculate the solubility product. It is important to remember that K_sp figures are constant at a given temperature, and any change in the values indicates a shift in equilibrium or a change in experimental conditions like temperature.

Molarity and Dilution

Understanding molarity, which is the concentration of a solute in a solution, is essential in equilibrium constant calculations and, by extension, in determining solubility products. The concept of molarity involves both the number of moles of solute and the volume of the solution. The unit of molarity is moles per liter (M).

In the given problem, dilution plays a key role. After the initial reactions have occurred, the solution is diluted, which changes the concentration (molarity) of the reactants and products. This transformation must be accounted for when calculating equilibrium concentrations. The formula for dilution is M₁V₁ = M₂V₂, where M₁ is the initial molarity, V₁ is the initial volume, M₂ is the final molarity, and V₂ is the final volume. This principle underlines the importance of performing correct dilution calculations to achieve accurate results in experiments.