Problem 73
Question
A solution contains \(2.0 \times 10^{-4} \mathrm{MAg}^{+}\)and \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Pb}^{2+}\). If \(\mathrm{NaI}\) is added, will \(\mathrm{AgI}^{\mathrm{I}}\left(K_{4 p}=8.3 \times 10^{-17}\right)\) or \(\mathrm{PbI}_{2}\) \(\left(K_{\text {sp }}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(I^{-}\)needed to begin precipitation.
Step-by-Step Solution
Verified Answer
AgI will precipitate first, and the concentration of I⁻ needed to begin precipitation is \(4.15 \times 10^{-13}\ \mathrm{M}\).
1Step 1: Write the solubility product expressions for AgI and PbI₂
The solubility product expressions for the ionic compounds AgI and PbI₂ are given by:
\(K_{4p}(AgI) = [Ag^+][I^-] \)
\(K_{sp}(PbI_2) = [Pb^{2+}][I^-]^2\)
Where [Ag⁺], [I⁻], and [Pb²⁺] are the molar concentrations of the respective ions in the solution.
2Step 2: Substitute the given values and solve for [I⁻]
In order to determine the concentration of I⁻ needed to begin precipitation, substitute the known values of K₄ₚ, Kₛₚ, [Ag⁺], and [Pb²⁺] into the solubility product expressions and solve for [I⁻]:
For AgI:
\(K_{4p} = [Ag^+][I^-]\)
\(8.3 \times 10^{-17} = (2.0 \times 10^{-4})[I^-]\)
\([I^-] = \frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} = 4.15 \times 10^{-13}\ \mathrm{M}\)
For PbI₂:
\(K_{sp} = [Pb^{2+}][I^-]^2\)
\(7.9 \times 10^{-9} = (1.5 \times 10^{-3})[I^-]^2\)
\([I^-] = \sqrt{\frac{7.9 \times 10^{-9}}{1.5 \times 10^{-3}}} = 7.25 \times 10^{-4}\ \mathrm{M}\)
3Step 3: Compare the [I⁻] values and determine which compound will precipitate first
Now that we have calculated the concentrations of I⁻ required to begin precipitation for AgI and PbI₂, we can compare these values to determine which compound will precipitate first:
[I⁻] needed for AgI precipitation: \(4.15 \times 10^{-13}\ \mathrm{M}\)
[I⁻] needed for PbI₂ precipitation: \(7.25 \times 10^{-4}\ \mathrm{M}\)
Since the concentration of I⁻ required to begin precipitation of AgI is much smaller than that required for PbI₂, AgI will precipitate first.
4Step 4: Specify the concentration of I⁻ needed to begin precipitation
The concentration of I⁻ needed to begin precipitation in this case is the one required for AgI precipitation, as it precipitates first:
[I⁻] needed for precipitation: \(4.15 \times 10^{-13}\ \mathrm{M}\)
Key Concepts
Precipitation ReactionsIonic EquilibriumKsp (Solubility Product Constant)
Precipitation Reactions
When a solution reaches a point where a particular ionic compound exceeds its solubility limit, it forms a precipitate – a solid that separates from the solution. Such reactions, known as precipitation reactions, are pivotal in the field of chemistry as they are instrumental in the separation and analysis of compounds.
In our exercise, when NaI is added to a solution containing Ag+ and Pb2+ ions, an equilibrium competition begins to assess which compound, AgI or PbI2, will precipitate first. The outcome is fundamentally determined by the respective solubility product constants for these compounds. To predict the precipitate, we calculate the concentration of I− ions required to initiate the formation of each solid, which depends on the initial ion concentrations and the solubility product constants.
In our exercise, when NaI is added to a solution containing Ag+ and Pb2+ ions, an equilibrium competition begins to assess which compound, AgI or PbI2, will precipitate first. The outcome is fundamentally determined by the respective solubility product constants for these compounds. To predict the precipitate, we calculate the concentration of I− ions required to initiate the formation of each solid, which depends on the initial ion concentrations and the solubility product constants.
Ionic Equilibrium
Ionic equilibrium unfolds when there is a balance between the forward and reverse processes of ion formation and deposition in solution. It's a state of dynamic balance where the rate of the forward reaction – dissolving solid into ions – equals the rate of the reverse reaction – reformation of the solid from its ions. This balance defines the solubility of a substance in a solution at specific conditions.
For the solubility equilibrium of sparingly soluble ionic compounds like AgI and PbI2, the product of the molar concentrations of the ions at equilibrium is a constant, known as the solubility product constant, at a given temperature. The solubility product is crucial in solving problems where we need to understand when a precipitate will start to form as exemplified in the step-by-step solution provided for our exercise.
For the solubility equilibrium of sparingly soluble ionic compounds like AgI and PbI2, the product of the molar concentrations of the ions at equilibrium is a constant, known as the solubility product constant, at a given temperature. The solubility product is crucial in solving problems where we need to understand when a precipitate will start to form as exemplified in the step-by-step solution provided for our exercise.
Ksp (Solubility Product Constant)
The Ksp, or Solubility Product Constant, quantifies the solubility of ionic compounds in a solution. It is defined for a saturated solution in equilibrium and varies with the temperature. A key aspect to note is that Ksp values are unique for different substances at a given temperature.
In the textbook exercise, the calculation involving Ksp serves as the underpinning for determining which compound – AgI or PbI2 – will precipitate first. The Ksp values allow us to compute the concentrations at which the respective ions will start to combine and form a solid precipitate. Intuitively, lower Ksp values generally suggest lower solubility and an increased likelihood of precipitation under a set of concentration conditions, explaining why AgI, with a lower Ksp, precipitates before PbI2 in the given solution.
In the textbook exercise, the calculation involving Ksp serves as the underpinning for determining which compound – AgI or PbI2 – will precipitate first. The Ksp values allow us to compute the concentrations at which the respective ions will start to combine and form a solid precipitate. Intuitively, lower Ksp values generally suggest lower solubility and an increased likelihood of precipitation under a set of concentration conditions, explaining why AgI, with a lower Ksp, precipitates before PbI2 in the given solution.
Other exercises in this chapter
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