Question

A solution contains \(0.018\) mole each of \(\mathrm{I}^{-}, \mathrm{Br}^{-}\), and \(\mathrm{Cl}^{-}\). When the solution is mixed with 200. \(\mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\) what mass of \(\mathrm{AgCl}(s)\) precipitates out, and what is \(\left[\mathrm{Ag}^{+}\right] ?\) Assume no volume change. $$ \begin{aligned} \text { AgI: } K_{\text {sp }} &=1.5 \times 10^{-16} \\ \mathrm{AgBr:} K_{\text {sp }} &=5.0 \times 10^{-13} \\ \mathrm{AgCl}: K_{\text {sp }} &=1.6 \times 10^{-10} \end{aligned} $$

Short answer

When the solution is mixed with \(200 \mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\), \(1.72 \mathrm{g}\) of \(\mathrm{AgCl}(s)\) precipitates out, and the remaining concentration of \(\mathrm{Ag}^{+}\) in the solution is \(0.06 \mathrm{M}\).

Step-by-step solution

Calculate the initial moles of Ag+ ions

First, we need to find out how many moles of \(\mathrm{Ag}^{+}\) ions are present in the \(200 \mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\) solution. To do this, we'll multiply the volume (in liters) by the molarity: Initial moles of \(\mathrm{Ag}^{+}\) = Volume (L) × Molarity Initial moles of \(\mathrm{Ag}^{+}\) = \((200 \times 10^{-3}\) L\() \times 0.24\) M Initial moles of \(\mathrm{Ag}^{+}\) = \(0.048\) moles Initially, we have \(0.048\) moles of \(\mathrm{Ag}^{+}\) ions.

Determine moles of Ag+ ions that react with each anion

Since we have an equal amount of each anion (\(\mathrm{I}^{-}, \mathrm{Br}^{-}, \mathrm{Cl}^{-}\)) and the formation of their respective silver compounds (\(\mathrm{AgI}, \mathrm{AgBr}, \mathrm{AgCl}\)) is ordered by their solubility product constants (\(K_{sp}\)), we can conclude that all the \(0.018\) moles of \(\mathrm{I}^{-}\) ions will react first with \(\mathrm{Ag}^{+}\) ions to form \(\mathrm{AgI}\), followed by all the \(0.018\) moles of \(\mathrm{Br}^{-}\) ions reacting with the remaining \(\mathrm{Ag}^{+}\) ions to form \(\mathrm{AgBr}\). Moles of \(\mathrm{Ag}^{+}\) ions reacting with \(\mathrm{I}^{-}\) = \(0.018\) moles Moles of \(\mathrm{Ag}^{+}\) ions reacting with \(\mathrm{Br}^{-}\) = \(0.018\) moles

Find the remaining moles of Ag+ ions

Now, we calculate the remaining moles of \(\mathrm{Ag}^{+}\) ions after the reactions with \(\mathrm{I}^{-}\) and \(\mathrm{Br}^{-}\): Remaining moles of \(\mathrm{Ag}^{+}\) = Initial moles of \(\mathrm{Ag}^{+}\) - Moles of \(\mathrm{Ag}^{+}\) ions reacting with \(\mathrm{I}^{-}\) - Moles of \(\mathrm{Ag}^{+}\) ions reacting with \(\mathrm{Br}^{-}\) Remaining moles of \(\mathrm{Ag}^{+}\) = \(0.048 - 0.018 - 0.018\) Remaining moles of \(\mathrm{Ag}^{+}\) = \(0.012\) moles

Compute the concentration of Ag+ ions

To find the final concentration of \(\mathrm{Ag}^{+}\) ions, divide the remaining moles of \(\mathrm{Ag}^{+}\) by the total volume (assuming no volume change): \(\left[\mathrm{Ag}^{+}\right] = \frac{0.012 \text{ moles}}{0.2 \text{ L}}\) \(\left[\mathrm{Ag}^{+}\right] = 0.06 \mathrm{M}\)

Calculate the mass of precipitated AgCl(s)

Since all the remaining moles of \(\mathrm{Ag}^{+}\) ions will react with \(\mathrm{Cl}^{-}\) ions forming \(\mathrm{AgCl}\), we can calculate the mass of precipitated \(\mathrm{AgCl}\) by multiplying the moles of remaining \(\mathrm{Ag}^{+}\) by the molar mass of \(\mathrm{AgCl}\) (143.32 g/mol): Mass of precipitated \(\mathrm{AgCl}(s)\) = \(0.012\) moles × \(143.32 \frac{\text{g}}{\text{mol}}\) Mass of precipitated \(\mathrm{AgCl}(s)\) = \(1.72 \text{g}\) In summary, when the solution is mixed with \(200 \mathrm{mL}\) of \(0.24 \mathrm{M} \mathrm{AgNO}_{3}\), \(1.72 \mathrm{g}\) of \(\mathrm{AgCl}(s)\) precipitates out, and the remaining concentration of \(\mathrm{Ag}^{+}\) in the solution is \(0.06 \mathrm{M}\).

Key concepts

Understanding Solubility Product Constants

In the realm of chemistry, specifically when dealing with precipitation reactions, the solubility product constant (\( K_{sp} \)) plays a pivotal role. This constant gives us valuable insights into the solubility of ionic compounds in water.

The solubility product constant is a unique value for each sparingly soluble compound, essentially capturing the point at which a solution becomes saturated with a particular ionic compound, leading to the formation of a precipitate. In the context of our exercise, various silver halides, such as AgI, AgBr, and AgCl, each have their own specific solubility product constants. These constants aid in the determination of which compound will precipitate first from a solution containing multiple potential precipitates.

In simpler terms, the lower the value of the constant, the less soluble the compound is. Therefore, AgI, with the smallest K_{sp}, precipitates before AgBr and AgCl. This sequence is instrumental in determining the composition of the final solution and the mass of precipitate formed, as demonstrated in the original exercise.

Molar Mass Calculations in Precipitation Reactions

Molar mass calculations are foundational in stoichiometry, which involves quantifying elements and compounds during chemical reactions. The molar mass is the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol).

For a compound like AgCl, which consists of one silver ion (Ag+) and one chloride ion (Cl-), the molar mass is the sum of the molar masses of these two ions. Since Ag has an approximate molar mass of 107.87 g/mol and Cl has one of 35.45 g/mol, the molar mass of AgCl would be roughly 143.32 g/mol. This is crucial in determining how much of a precipitate will form, which can be calculated by multiplying the number of moles of the compound by its molar mass, as shown in our exercise where the resulting precipitate mass of AgCl was calculated after a precipitation reaction.

Ion Concentration Determination

Determining ion concentration is a significant part of chemistry, especially when it comes to reactions in solution. It involves finding the number of moles of an ion per liter of solution, which is expressed as molarity (M). This concept is applied, for instance, when calculating the concentration of silver ions remaining in a solution after a precipitation reaction.

In the exercise, the concentration of Ag+ ions is calculated by dividing the remaining moles of Ag+ after the reaction by the total volume of the solution. Importantly, one must take into account the stoichiometry of the reaction and the initial conditions, such as volumes and molarity, which impact the final concentration. Changes in the concentrations of ions can influence the direction and extent of chemical equilibria, as well as the potential for further precipitation reactions.