Question

A solution contains three anions with the following concentrations: \(0.20 \mathrm{MCrO}_{4}^{2-}, 0.10 \mathrm{MCO}_{3}^{2-},\) and \(0.010 \mathrm{MCl}^{-}\). If a dilute \(\mathrm{AgNO}_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{4 p}=1.2 \times 10^{-12}\right), \mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{4 p}=8.1 \times 10^{-12}\right)\) or \(\mathrm{AgCl}\left(K_{\mathrm{sp}}=1.8 \times 10^{-10}\right) ?\)

Short answer

AgCl precipitates first.

Step-by-step solution

Understanding Solubility Product

Each potential precipitate has a solubility product constant, or \(K_{sp}\), which indicates its solubility in water. Lower \(K_{sp}\) values typically mean the compound is less soluble and more likely to precipitate first.

Writing the Solubility Product Expressions

For each compound, set up the expression for \(K_{sp}\): 1. \( ext{Ag}_2 ext{CrO}_4\): \(K_{sp} = [ ext{Ag}^+]^2[ ext{CrO}_4^{2-}]\)2. \( ext{Ag}_2 ext{CO}_3\): \(K_{sp} = [ ext{Ag}^+]^2[ ext{CO}_3^{2-}]\)3. \( ext{AgCl}\): \(K_{sp} = [ ext{Ag}^+][ ext{Cl}^-]\)

Calculating Required [Ag+] for Precipitation

Determine the concentration of \([ ext{Ag}^+]\) needed to start precipitation for each anion:1. For \( ext{Ag}_2 ext{CrO}_4\): \([Ag^+] = \sqrt{\frac{K_{sp}}{[ ext{CrO}_4^{2-}]}} = \sqrt{\frac{1.2 \times 10^{-12}}{0.20}}\)2. For \( ext{Ag}_2 ext{CO}_3\): \([Ag^+] = \sqrt{\frac{K_{sp}}{[ ext{CO}_3^{2-}]}} = \sqrt{\frac{8.1 \times 10^{-12}}{0.10}}\)3. For \( ext{AgCl}\): \([Ag^+] = \frac{K_{sp}}{[ ext{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.010}\)

Comparing [Ag+] Values to Determine First Precipitate

Calculate each required \([Ag^+]\):1. \(\text{Ag}_2\text{CrO}_4\): \([Ag^+] = \sqrt{6 \times 10^{-12}} \approx 2.45 \times 10^{-6}\)2. \(\text{Ag}_2\text{CO}_3\): \([Ag^+] = \sqrt{8.1 \times 10^{-11}} \approx 9.0 \times 10^{-6}\)3. \(\text{AgCl}\): \([Ag^+] = 1.8 \times 10^{-8}\)Since the lowest \([Ag^+]\) required for precipitation is for \( ext{AgCl}\), it will precipitate first.

Key concepts

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid, known as a precipitate. This happens due to the formation of a compound that has a low solubility in water. When a solution of \(\text{AgNO}_3\) is added to a solution containing different anions, the silver ions (\(\text{Ag}^+\)) react with available anions to form potential precipitates such as \(\text{Ag}_2\text{CrO}_4\), \(\text{Ag}_2\text{CO}_3\), or \(\text{AgCl}\). Which compound forms first depends on their respective solubility products.
These reactions are often predictable. By measuring the concentration of ions and comparing the product of the ion concentrations to known \(K_{sp}\) values, it's possible to predict whether a precipitate will form. Lower \(K_{sp}\) values suggest compounds are less soluble and more likely to precipitate.

Chemical Equilibrium

Chemical equilibrium in a reaction is reached when the rates of the forward and backward reactions are equal. In the context of precipitation and dissolution, this is when the rate at which the compound dissolves is equal to the rate at which it precipitates out of solution.

For the compounds involved in our precipitation example, the compounds are described by their \(K_{sp}\), which is a type of equilibrium constant. This constant is specific to a given temperature and reflects the balance between the dissolution and precipitation of the compound in a saturated solution. Once equilibrium is reached, no more net change in the concentration of any species is observed, although both the forward and reverse reactions continue to occur.

Ksp (Solubility Product Constant)

The solubility product constant, \(K_{sp}\), is a crucial concept in determining the solubility of a compound. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. The \(K_{sp}\) expression for a salt is derived from the concentration of the ions in the solution at saturation.

Each compound has a unique \(K_{sp}\) value. Smaller \(K_{sp}\) values indicate a lower solubility. In our example of the silver salt precipitates, \(\text{AgCl}\) with a \(K_{sp}\) of \(1.8 \times 10^{-10}\) has a higher solubility than \(\text{Ag}_2\text{CrO}_4\), which has a \(K_{sp}\) of \(1.2 \times 10^{-12}\). The smallest \(K_{sp}\) usually results in the fastest formation of a precipitate. When the ionic product exceeds the solubility product, precipitation occurs, thus establishing which compound precipitates first.