Question

A solution is \(1 \times 10^{-4} \mathrm{M}\) 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 \(K_{\text {sp }}\) values are \(K_{\text {sp }}\left(\mathrm{PbF}_{2}\right)=\) \(4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29}\), 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 when a source of Pb²⁺ is added to the given solution would be: PbS → Pb₃(PO₄)₂ → PbF₂. This is based on the Pb²⁺ concentrations required for precipitation, which follow the order PbS < Pb₃(PO₄)₂ < PbF₂.

Step-by-step solution

Identify the ionic equations for the three compounds

In this situation, the solution initially contains NaF, Na2S, and Na3PO4; and, a source of Pb²⁺ ions will be added. The potentially formed compounds are: PbF2, PbS, and Pb3(PO4)2. Write the ionic equations for the formation of these compounds: 1. Pb²⁺ (aq) + 2F⁻ (aq) → PbF₂ (s) 2. Pb²⁺ (aq) + S²⁻ (aq) → PbS (s) 3. 3Pb²⁺ (aq) + 2PO₄³⁻ (aq) → Pb₃(PO₄)₂ (s)

Calculate the solubility of each compound

Using the Ksp values and the ionic equations, we can calculate the solubility of each compound in terms of Pb²⁺ concentration: 1. PbF₂: Ksp = [Pb²⁺][F⁻]² = \(4 \times 10^{-8}\) ⇒ [Pb²⁺] = \(1 / (4 \times 10^{-8})\) since \(2 [F^{-}] = [Pb^{2+}]\) 2. PbS: Ksp = [Pb²⁺][S²⁻] = \(7 \times 10^{-29}\) ⇒ [Pb²⁺] = \(7 \times 10^{-29}\) since \([S^{2-}] = [Pb^{2+}]\) 3. Pb₃(PO₄)₂: Ksp = [Pb²⁺]³[PO₄³⁻]² = \(1 \times 10^{-54}\) ⇒ [Pb²⁺] = \(\sqrt[3]{1 / (1 \times 10^{-54})}\) = \(1 \times 10^{-18}\) since \((2/3) [PO_4^{3-}] = [Pb^{2+}]\)

Determine the order of precipitation

Compare the values of [Pb²⁺] for all three compounds: 1. PbF₂: [Pb²⁺] = \(1 / (4 \times 10^{-8})\) 2. PbS: [Pb²⁺] = \(7 \times 10^{-29}\) 3. Pb₃(PO₄)₂: [Pb²⁺] = \(1 \times 10^{-18}\) Since the Pb²⁺ concentrations required for precipitation follow the order PbS < Pb₃(PO₄)₂ < PbF₂, the order of precipitation would be: PbS → Pb₃(PO₄)₂ → PbF₂

Key concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, commonly abbreviated as Ksp, is essential when exploring the solubility of ionic compounds in aqueous solutions. It is a unique value for every sparingly soluble compound and represents the extent to which a compound can dissociate into its ions in solution. This constant is crucial for predicting whether a precipitation reaction will occur under specified conditions.

Let's take a closer look at what Ksp tells us. It is calculated as the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced chemical equation. For instance, the Ksp for a generic compound AB₂ in water would be expressed as Ksp = [A⁺] x [B⁻]², where [A⁺] and [B⁻] are the concentrations of the ions. In very simple terms, a lower Ksp value indicates a compound that is less soluble in water.

When lead (Pb²⁺) is added to the solution containing NaF, Na₂S, and Na₃PO₄, it's necessary to compare the Ksp values of the possible lead compounds to predict the order of precipitation. PbF₂, PbS, and Pb₃(PO₄)₂ will form and precipitate out based on the solubility limits defined by their Ksp values.

Ionic Compounds

Ionic compounds, such as salts, are composed of positive and negative ions held together by strong electrostatic forces known as ionic bonds. In the case of our exercise, we are dealing with compounds of lead (Pb²⁺) combined with fluoride (F⁻), sulfide (S²⁻), and phosphate (PO₄³⁻) ions.

Ionic compounds typically dissolve in water by dissociating into their individual ions. The extent of their solubility varies and can be broadly predicted by the rules of solubility. However, Ksp provides a more precise and quantifiable approach, especially for compounds that are only sparingly soluble. Knowing the stoichiometry of these compounds, as well as their Ksp values, helps in predicting and explaining their behavior when they interact in a solution.

For example, in a given solubility equilibrium, high concentrations of certain ions can shift the balance, favoring the formation of a precipitate of the ionic compound. In our exercise, adding a source of Pb²⁺ ions to a solution initially containing NaF, Na₂S, and Na₃PO₄ will result in the formation of ionic compounds based on the solubility product constants.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form one or more insoluble products, called precipitates. This type of reaction is common in chemistry and can be described using the ion product (Q), which is calculated in the same manner as the Ksp. However, Q represents the actual concentrations in the solution at a given moment, as opposed to the equilibrium concentrations described by the Ksp.

A precipitation reaction will take place if the ion product exceeds the Ksp for any given ionic compound. This is a straightforward way to predict the occurrence of precipitation — by comparing Q with Ksp. If Q > Ksp, precipitation is favored; if Q < Ksp, the compounds will remain in solution.

In our exercise, as Pb²⁺ is gradually introduced to the solution, the ion product for PbS reaches its Ksp value first, followed by Pb₃(PO₄)₂, and lastly PbF₂. The first to reach this threshold will precipitate out of the solution, indicating the order in which the compounds will form precipitates: PbS precipitates first, followed by Pb₃(PO₄)₂, and finally PbF₂, in line with their respective solubility limits.