Problem 31 - Chemistry: Principles and Reactions

Question

A solution is \(0.035 M\) in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(0.035 \mathrm{M}\) in \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\). Solid \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is added without changing the volume of the solution. (a) Which salt, \(\mathrm{PbSO}_{4}\) or \(\mathrm{PbCrO}_{4}\), will precipitate first? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) when the salt in (a) first begins to precipitate?

Step-by-Step Solution

Answer

Additionally, what is the concentration of Pb2+ when the salt in part (a) first begins to precipitate? Answer: PbSO4 will precipitate first, and the concentration of Pb2+ when PbSO4 first begins to precipitate is 4.57 x 10^{-7} M.

Step 1: Find the Ksp values of PbSO4 and PbCrO4

Look up the solubility product constants (Ksp) of PbSO4 and PbCrO4 in a textbook or online reference. The Ksp values are: PbSO4: Ksp = 1.6 x 10^{-8} PbCrO4: Ksp = 1.2 x 10^{-12}

Step 2: Calculate Q for both salts and compare Q to Ksp

Ion product (Q) is calculated as follows: Q = [Pb2+][SO42-] for PbSO4 Q = [Pb2+][CrO42-] for PbCrO4 At the beginning, before any lead salt has precipitated, the concentrations of sulfate and chromate ions are equal to 0.035 M as given in the problem. Thus, we can write: Q(PbSO4) = [Pb2+](0.035) Q(PbCrO4) = [Pb2+](0.035) Divide each Q by their respective Ksp values: Q(PbSO4)/Ksp(PbSO4) = [Pb2+](0.035) / (1.6 x 10^{-8}) Q(PbCrO4)/Ksp(PbCrO4) = [Pb2+](0.035) / (1.2 x 10^{-12}) The salt for which the ratio Q/Ksp is smaller will precipitate first.

Step 3: Determine which salt precipitates first

Comparing the Q/Ksp values for both salts, we find that: [Pb2+](0.035) / (1.6 x 10^{-8}) < [Pb2+](0.035) / (1.2 x 10^{-12}) So PbSO4 will precipitate first. Answer to part (a): PbSO4 will precipitate first.

Step 4: Calculate the concentration of Pb2+ when PbSO4 begins to precipitate

Since PbSO4 is determined to precipitate first, we will use its Ksp value to find the concentration of Pb2+ when precipitation starts: Ksp(PbSO4) = 1.6 x 10^{-8} = [Pb2+][SO42-] Given that [SO42-] = 0.035 M, we can solve for [Pb2+]: [Pb2+] = Ksp(PbSO4) / [SO42-] [Pb2+] = (1.6 x 10^{-8}) / 0.035 [Pb2+] = 4.57 x 10^{-7} M Answer to part (b): The concentration of Pb2+ when PbSO4 first begins to precipitate is 4.57 x 10^{-7} M.

Key Concepts

Solubility Product Constant

The solubility product constant, often denoted as \( K_{sp} \), is a crucial concept for understanding how and when a salt will precipitate from a solution. It offers insight into the maximum amount of solute that can dissolve in a solution before it begins to form a solid. The smaller the \( K_{sp} \), the less soluble the compound is.

For any given compound \( AB \), the solubility product is expressed as:\[ K_{sp} = [A^+][B^-] \]

This constant is typically found in tables, as it varies for different compounds. Understanding \( K_{sp} \) allows chemists to predict whether a precipitate will form under certain conditions, which is crucial for chemical reactions in labs and industries.

Ion Product

The ion product \( (Q) \) is a snapshot of the concentrations of the ions in a solution at any given time. Similar to \( K_{sp} \), the ion product helps us determine whether a solution is unsaturated, saturated, or supersaturated.

Here's how it works:

• If \( Q < K_{sp} \), the solution is unsaturated and no precipitate will form.
• If \( Q = K_{sp} \), the solution is at equilibrium, and it is saturated.
• If \( Q > K_{sp} \), the solution is supersaturated, and precipitation will occur.

The calculation for \( Q \) is analogous to \( K_{sp} \), using the current concentrations of the ions in the solution. This comparison between \( Q \) and \( K_{sp} \) is essential in predicting the formation of a precipitate.

Lead(II) Sulfate

Lead(II) sulfate, or \( \text{PbSO}_{4} \), is a white solid that is relatively insoluble in water. Its solubility product constant (\( K_{sp} \)) is \( 1.6 \times 10^{-8} \). This low \( K_{sp} \) value means it readily precipitates out of solution under certain conditions.

This compound is significant in understanding the precipitation process in solutions containing lead. In many instances, such as the given exercise, \( \text{PbSO}_{4} \) precipitates before other lead compounds due to its specific \( K_{sp} \). Understanding which conditions lead to its formation can help in various applications such as water treatment and battery manufacturing.

Lead(II) Chromate

Lead(II) chromate, known as \( \text{PbCrO}_{4} \), is a bright yellow compound with a \( K_{sp} \) value of \( 1.2 \times 10^{-12} \), indicating it is even less soluble than lead(II) sulfate. This makes it an important compound in pigments and paints.

When calculating the precipitation potential in a solution containing both sulfate and chromate ions, \( \text{PbCrO}_{4} \) typically precipitates after \( \text{PbSO}_{4} \) due to its smaller \( K_{sp} \). Knowing its properties helps chemists determine the order of precipitation in mixed ionic solutions, thus playing a crucial role in experimental chemistry settings.