Question

Will precipitation occur in the following cases? (a) \(0.10 \mathrm{mg}\) NaCl is added to \(1.0 \mathrm{L}\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}(\mathrm{aq})\). (b) One drop \((0.05 \mathrm{mL})\) of \(0.10 \mathrm{M} \mathrm{KBr}\) is added to 250 mL of a saturated solution of AgCl. (c) One drop \((0.05 \mathrm{mL})\) of \(0.0150 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) is added to \(3.0 \mathrm{L}\) of a solution with \(2.0 \mathrm{mg} \mathrm{Mg}^{2+}\) per liter.

Short answer

For (a) and (b), based on the calculated ion product which must be compared with the corresponding Ksp, determine if precipitation occurs. For (c), again based on the calculated ion product and the given Ksp for Mg(OH)2, determine if precipitation will occur.

Step-by-step solution

Analyze the first scenario

The task is to determine if \(0.10 \mathrm{mg}\) NaCl in \(1.0 \mathrm{L}\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}(\mathrm{aq})\) will precipitate. Find the concentration of the Cl- ions from the NaCl, then calculate the ion product [Ag+][Cl-]. Compare it to the reported Ksp value of AgCl which is \(1.8 × 10^{-10}\). If the ion product is greater than the Ksp, precipitation will occur.

Analyze the second scenario

For the second scenario, one drop or \(0.05 \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{KBr}\) is added to 250 mL of a saturated solution of AgCl. Calculate the molarity of Br- in the solution, then, determine the new ion product of AgCl with this additional Br-. Again compare the ion product with the Ksp value of AgCl.

Analyze the third scenario

In the third scenario, to check if precipitation occurs with \(0.05 \mathrm{mL}\) of $0.0150 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\( is added to \)3.0 \mathrm{L}$ of a solution with \(2.0 \mathrm{mg} \mathrm{Mg}^{2+}\) per liter, determine the concentration of OH- ions and the concentration of Mg2+ ions. Rule out the possibility of forming NaMg(OH)3 because Na+ is a spectator ion. The relevant salt for precipitation here is Mg(OH)2. Calculate the ion product [Mg2+][OH-]^2 and compare it to the Ksp for Mg(OH)2, which is \(1.5 × 10^{-11}\). If the ion product is larger than the Ksp, a precipitate will form.

Key concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, commonly represented as Ksp, is crucial when analyzing chemical precipitation reactions. It provides a quantitative measure of a sparingly soluble compound's solubility in water. Specifically, the Ksp value is the product of the molar concentrations of the ions that make up a solid compound, each raised to the power of their stoichiometric coefficients from the balanced dissolution reaction.

The Ksp is, thus, an equilibrium constant and only applies to the solubility of ionic solids. To make it simple, think of Ksp as a threshold: If the ion product (calculated by multiplying together the ion concentrations in a solution) exceeds this threshold value, the solid compound will precipitate out of solution. Conversely, if the ion product is less than the Ksp, the compound remains soluble.

Let's use the exercise's example: AgCl has a Ksp of 1.8 × 10^-10. This means that the product of the concentrations of silver ions (Ag+) and chloride ions (Cl-) in a saturated solution at equilibrium cannot exceed this value without resulting in the formation of a solid precipitate of AgCl. Ksp values are specific for each compound at a given temperature and are key factors in predicting whether precipitation will occur in a given scenario.

Ion Concentration Calculation

The calculation of ion concentration is an essential step in predicting whether a chemical precipitation reaction will occur in a solution. Here's how to approach it: First, understand the concepts of molarity and dilution. Molarity represents the number of moles of solute per liter of solution. When a compound dissociates in water, it releases a definitive number of ions depending on its formula.

For example, to calculate the concentration of Cl- ions from the dissolved NaCl in the exercise, you would first convert the mass of NaCl to moles (using its molecular weight) and then divide by the total volume of the solution. Remember that each mole of NaCl yields one mole of Cl- ions when it dissociates.

In the exercise, we are presented with various amounts and volumes to consider when calculating ion concentrations. In such calculations, always account for the volume of both the solute and the solvent as this can affect the final concentration. Moreover, in situations where dilution occurs (as with the drop of KBr added to AgCl solution), the volume change and resulting concentration must be recalculated. Accurate ion concentration calculation is critical when comparing to the Ksp to determine the potential for precipitation.

Precipitate Formation Analysis

Analysis of precipitate formation involves understanding the conditions under which a compound will come out of solution as a solid. After computing the ionic concentrations, as previously discussed, the next step is to compare them to the Ksp to predict precipitation. A simplified way to think about this process is to imagine the ions in solution as guests at a party—the Ksp value is the size of the venue. If too many guests (ions) arrive, they can't all fit, so some have to leave, resulting in precipitation.

In the provided exercise scenarios, different ionic compounds are added to various solutions, and the task is to predict whether this addition will result in the exceedance of the Ksp value. For instance, when a drop of NaOH is introduced to a Mg2+ solution, one must consider the formation of Mg(OH)2. To predict whether Mg(OH)2 will precipitate, the concentration of Mg2+ and OH- ions is multiplied together (with the OH- ion concentration squared, as two hydroxide ions are involved in the reaction) to get the ion product. Then, this ion product is compared with the Ksp of Mg(OH)2.

In a study session, students can comprehend this analysis by noting that precipitation is favored when the ion product exceeds the Ksp. If a student strives for a complete understanding of scenarios involving precipitation, staying attentive to such comparisons becomes crucial. Moreover, considering all potential precipitates and disregarding spectator ions (as with Na+ in the NaOH example) is essential for accurate analysis.