Problem 59 - Chemistry and Chemical Reactivity

Question

You have a solution that has a lead(II) ion concentration of 0.0012 M. If enough soluble chloride-containing salt is added so that the \(\mathrm{Cl}^{-}\) concentration is \(0.010 \mathrm{M},\) will PbCl \(_{2}\) precipitate?

Step-by-Step Solution

Answer

PbCl sert{\(_2\)} will not precipitate because Q < Ksp.

Step 1: Determine the Solubility Product Constant

Lead(II) chloride, PbCl\(_2\), is a slightly soluble salt. The solubility product constant (Ksp) for PbCl\(_2\) is needed to determine if precipitation will occur. The Ksp for PbCl\(_2\) is 1.7 x 10^-5.

Step 2: Write the Dissolution Equation

The dissolution of PbCl\(_2\) in water can be represented by the equation:\[\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2 \text{Cl}^{-}(aq)\]This equation shows that for every mole of PbCl\(_2\) that dissolves, one mole of Pbsert{\(^{2+}\)} is produced and two moles of Clert{\(-\)} are produced.

Step 3: Calculate the Reaction Quotient, Q

The reaction quotient \(Q\) can be calculated using the concentrations of the ions:\[Q = [\text{Pb}^{2+}] [\text{Cl}^{-}]^2\]Given, \([\text{Pb}^{2+}] = 0.0012\, M\) and \([\text{Cl}^{-}] = 0.010\, M\):\[Q = (0.0012) \times (0.010)^2 = 0.0012 \times 0.0001 = 1.2 \times 10^{-7}\]

Step 4: Compare Q with Ksp

To determine if precipitation will occur, compare \(Q\) to \(Ksp\):- If \(Q > Ksp\), precipitation occurs.- If \(Q < Ksp\), no precipitation occurs.- If \(Q = Ksp\), the solution is at equilibrium, and no more precipitation occurs.Since \(Q = 1.2 \times 10^{-7}\) and \(Ksp = 1.7 \times 10^{-5}\), \(Q < Ksp\).

Key Concepts

Precipitation Reaction

In a precipitation reaction, two soluble salts in a solution mix to form an insoluble salt, known as a precipitate.
This process plays a critical role in various fields, from chemistry labs to wastewater treatment.
Here's how to identify if a precipitation reaction will occur:

• Determine the dissolved ions in the solution.
• Use the solubility rules to predict if any combination forms an insoluble compound.
• Calculate the reaction quotient, \(Q\), to compare with the solubility product constant, \(K_{sp}\). If \(Q > K_{sp}\), precipitation occurs.

Understanding these steps helps in predicting when and why precipitates form. It's all about the balance between solubility and reaction conditions.

Lead(II) Chloride

Lead(II) chloride, \(\text{PbCl}_2\), is a compound known for its limited solubility in water. This property makes it interesting for studying precipitation reactions.
PbCl\(_2\) breaks down in water into lead ions \((\text{Pb}^{2+})\) and chloride ions \((\text{Cl}^-)\).
Here's the breakdown:

• 1 mole of \(\text{PbCl}_2\) forms 1 mole of \(\text{Pb}^{2+}\)
• 2 moles of \(\text{Cl}^-\)

The limited solubility is governed by its solubility product constant \(K_{sp}\), which is \(1.7 \times 10^{-5}\) for PbCl\(_2\). Understanding this helps predict if and when \(\text{PbCl}_2\) will precipitate in a reaction.

Reaction Quotient

The reaction quotient, \(Q\), helps predict whether a precipitate will form in a solution at a given moment. This is essential for understanding equilibrium conditions in chemical reactions.
Here's how to calculate \(Q:\)

• Start with the concentrations of the ions present in the solution.
• For \(\text{PbCl}_2\), \(Q\) is given by \[Q = [\text{Pb}^{2+}] [\text{Cl}^-]^2\]
• Plug in the concentrations: \([\text{Pb}^{2+}] = 0.0012 \; \text{M}\) and \([\text{Cl}^-] = 0.010 \; \text{M}\)
• Calculate: \(Q = (0.0012) \times (0.010)^2 = 1.2 \times 10^{-7}\)

Then, compare \(Q\) to the \(K_{sp}\):- If \(Q > K_{sp}\), precipitation occurs.- If \(Q < K_{sp}\), no precipitation occurs.- If \(Q = K_{sp}\), the solution is at equilibrium.
In this case, \(Q < K_{sp}\), meaning no more \(\text{PbCl}_2\) will precipitate.