Question

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).

Short answer

The order of precipitation as a source of Pb²⁺ is added gradually to the solution is: PbF₂ → PbS → Pb₃(PO₄)₂.

Step-by-step solution

Write the solubility equilibrium expressions for each salt

The solubility equilibrium expressions for PbF₂, PbS, and Pb₃(PO₄)₂ based on their Ksp values: 1. PbF₂ ⇌ Pb²⁺ + 2F⁻ 2. PbS ⇌ Pb²⁺ + S²⁻ 3. Pb₃(PO₄)₂ ⇌ 3Pb²⁺ + 2PO₄³⁻

Calculate the concentration of Pb²⁺ required to reach Ksp for each salt

To find the concentration of Pb²⁺ required for each salt to reach its Ksp value, set up equations using the solubility equilibrium expressions and solve for the concentration of Pb²⁺. 1. For PbF₂: Ksp(PbF₂) = [Pb²⁺][F⁻]² 4×10⁻⁸ = (x)(2×10⁻⁴)² Solve for x to find the concentration of Pb²⁺ required to reach Ksp for PbF₂. 2. For PbS: Ksp(PbS) = [Pb²⁺][S²⁻] 7×10⁻²⁹ = (x)(1×10⁻⁴) Solve for x to find the concentration of Pb²⁺ required to reach Ksp for PbS. 3. For Pb₃(PO₄)₂: Ksp(Pb₃(PO₄)₂) = [Pb²⁺]³[PO₄³⁻]² 1×10⁻⁵⁴ = (x)³(1×10⁻⁴)² Solve for x to find the concentration of Pb²⁺ required to reach Ksp for Pb₃(PO₄)₂.

Determine the order of precipitation

After solving for the concentration of Pb²⁺ required to reach Ksp for each salt, compare the values to determine the order of precipitation. The salt with the lowest required concentration of Pb²⁺ will precipitate first, followed by the salt with the next lowest concentration, and so on. PbF₂ requires less Pb²⁺ to reach its Ksp value compared to PbS and Pb₃(PO₄)₂. So, PbF₂ will precipitate first. Next, compare the values for PbS and Pb₃(PO₄)₂. PbS requires less Pb²⁺ to reach its Ksp value compared to Pb₃(PO₄)₂. Therefore, PbS will precipitate second, and Pb₃(PO₄)₂ will precipitate last. The order of precipitation is: PbF₂ → PbS → Pb₃(PO₄)₂.

Key concepts

Ksp values

Understanding the solubility product constant, commonly referred to as Ksp, is crucial when delving into the solubility of ionic compounds in aqueous solutions. The Ksp value represents the extent to which a compound can dissolve in water and is a measure of the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation.

These values allow chemists to determine whether a precipitate will form when two solutions are mixed. A precipitate is a solid that forms when the product of the ion concentrations exceeds the Ksp. If the product of the concentrations (called the reaction quotient, Q) is lower than the Ksp, no precipitation will occur and the ions will remain dissolved. When the Q equals the Ksp, the solution is at equilibrium, and the rate at which the solid dissolves is equal to the rate at which it precipitates.

In the described exercise, knowledge of Ksp values for PbF₂ (\(4 \times 10^{-8}\)), PbS (\(7 \times 10^{-29}\)), and Pb₃(PO₄)₂ (\(1 \times 10^{-54}\)) is used to foresee the order of precipitation. The lower the Ksp, the less soluble the compound is, and consequently, the more likely it is to precipitate out of solution at lower concentrations of the ions.

Precipitation Reaction

The concept of a precipitation reaction revolves around the formation of an insoluble solid from the reaction of two soluble compounds in aqueous solution. Precipitation reactions are a subset of chemical reactions, where ionic compounds exchange ions and produce at least one insoluble product, the precipitate.

Using the rules of solubility and the solubility product constant (Ksp), it's possible to predict which compound would precipitate first in a given situation. One example is adding a source of Pb²⁺ ions to a solution containing fluoride, sulfate, and phosphate ions. The order of precipitation will depend on the relative solubility of the potential precipitates that can form, which is directly linked to the Ksp values of their corresponding salts.

The exercise demonstrates this by showing that as Pb²⁺ is gradually added, the solid that forms first is the one associated with the highest solubility product (lowest Ksp), since it needs a smaller concentration of Pb²⁺ to exceed its solubility threshold and begin to precipitate.

Concentration Calculation

The process of concentration calculation is fundamental in predicting the occurrence of precipitation reactions. It involves calculating the molarity, or molar concentration, of ions in solution, which is crucial when comparing to Ksp values to assess saturation levels and the potential for precipitate formation.

In the context of the exercise provided, the calculation required solving for the concentration of Pb²⁺ ions that would lead to the saturated solution conditions for each salt. This involves algebraically manipulating the solubility equilibrium expressions and the provided initial ion concentrations. Once computed, these concentrations reveal the threshold at which adding more Pb²⁺ ions will result in exceeding the Ksp value and precipitating a corresponding salt.

For example, to find the concentration of Pb²⁺ needed for PbF₂ to reach its Ksp, we set up the equation \(4 \times 10^{-8} = (x)(2 \times 10^{-4})^2\) and solve for x. The calculated concentrations allow us to compare the solubility of each lead salt and determine the order in which they will precipitate from the solution.