Problem 102 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Silver ions are slowly added in a solution with \(\left[\mathrm{Br}^{-}\right]=\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{CO}_{3}^{2-}\right]=\left[\mathrm{AsO}_{4}^{3-}\right]\) \(=0.1 \mathrm{M}\). Which compound will precipitate first? (a) \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5 \times 10^{-13}\right)\) (b) \(\mathrm{AgCl}\left(K_{\mathrm{sp}}=1.8 \times 10^{-10}\right)\) (c) \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{\mathrm{sp}}=8.1 \times 10^{-12}\right)\) (d) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1 \times 10^{-22}\right)\)

Step-by-Step Solution

Answer

AgBr will precipitate first because it has the lowest K_{sp} value, which means it will reach a state of saturation and begin to precipitate at a lower concentration of silver ions compared to the other compounds.

Step 1: Identify the Relevant Equations

Write down the solubility product (K_{sp}) expressions for each of the silver compounds. For AgBr: K_{sp} = [Ag^+][Br^-], for AgCl: K_{sp} = [Ag^+][Cl^-], for Ag_2CO_3: K_{sp} = [Ag^+]^2[CO_3^{2-}], and for Ag_3PO_4: K_{sp} = [Ag^+]^3[PO_4^{3-}].

Step 2: Calculate the Ionization Product

For each silver compound, substitute the given concentration of the anion into the K_{sp} expression to find the ionization product (Q) that will be reached when silver ions are added.

Step 3: Compare Ionization Product with K_{sp}

The compound that will precipitate first is the one whose ionization product (Q) exceeds its K_{sp} by the largest margin when silver ions are added.

Step 4: Identify the First Precipitate

Calculate Q for each compound by assuming that the concentration of silver ions ([Ag^+]) needed to reach the K_{sp} value is just about to be exceeded. The compound with the smallest [Ag^+] that results in Q ≥ K_{sp} will precipitate first.

Key Concepts

Precipitation Chemistry

In precipitation chemistry, solubility rules play a crucial role in predicting whether a solid will form when solutions are mixed. When two aqueous solutions containing ions are combined, insoluble compounds can form and precipitate out of solution. Precipitation is governed by the solubility of the compound in water, which is quantitatively expressed by the solubility product constant ((K_{sp})). This constant is unique for every solid compound and depends on temperature. To determine which compound precipitates first, we must compare the ion concentrations in the solution with the respective (K_{sp}) values. The one with the ion product ((Q)) that exceeds the (K_{sp}) by the largest amount will precipitate first. This practical application of precipitation chemistry is vital in fields ranging from water treatment to pharmaceuticals.

Ionization Product

Ionization product ((Q)) is a term that reflects the product of molar concentrations of the ions involved in a precipitate at any given moment. It's essentially a snapshot of where the reaction stands before reaching equilibrium. To calculate (Q), you use the same formula as for the solubility product ((K_{sp})), but with the current ion concentrations. If (Q) < (K_{sp}), the solution is unsaturated and no precipitation occurs. However, if (Q) > (K_{sp}), the solution is supersaturated, and precipitation is expected. Hence, comparing (Q) with (K_{sp}) helps in predicting whether a compound will precipitate from a solution at any point in time.

Equilibrium Constant

The equilibrium constant ((K)) is a fundamental concept in chemical equilibrium, establishing the ratio of product concentrations to reactant concentrations at equilibrium. In the context of solubility, the solubility product constant ((K_{sp})) acts as an equilibrium constant for the dissolution of sparingly soluble ionic compounds. It signifies a saturated solution where the rate of dissolution equals the rate of precipitation. Each ionic compound has a specific (K_{sp}) value, which facilitates the prediction of solubility and consequently the formation of a precipitate in a reaction. When a solution's ion concentrations change, the reaction may shift to restore equilibrium, thereby forming more solid or dissolving some of it back into ions.

Silver Halides Solubility

Silver halides, such as silver chloride (AgCl) and silver bromide (AgBr), are known for their low solubility in water, making them a classic example in solubility product discussions. The solubility product constants for these compounds are quite small, reflecting their limited solubility. For instance, (AgCl) has a (K_{sp}) of 1.8 x 10^-10, whereas (AgBr) has an even lower (K_{sp}) of 5 x 10^-13. This characteristic is essential in photographic processes, where the reduction of the silver ion by light leads to the formation of microscopic silver particles, creating an image. Understanding the solubility of silver halides also allows chemists to manipulate conditions to selectively precipitate one silver halide over another, as demonstrated in the exercise on determining which silver compound precipitates first.