Question

A solution contains \(2.0 \times 10^{-4} \mathrm{MAg}^{+}\) and \(1.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{Pb}^{2+}\). If NaI is added, will AgI \(\left(K_{s p}=8.3 \times 10^{-17}\right)\) or \(\mathrm{PbI}_{2}\) \(\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(1^{-}\) needed to begin precipitation.

Short answer

AgI will precipitate first when NaI is added to the solution. The concentration of I⁻ needed to begin precipitation is \(4.15 \times 10^{-13}\) M.

Step-by-step solution

Calculate the reaction quotient (Q) for each compound.

The reaction quotient, Q, can be calculated using the ion concentrations in solution. For each compound, Q is given by: AgI: \(Q_{1}= \mathrm{[Ag^{+}][I^{-}]} \) PbI₂: \(Q_{2}=\mathrm{[Pb^{2+}][I^{-}]}^2 \) Initially, the I⁻ concentration is 0.

Compare the Q values to the Ksp values to determine which compound will precipitate first.

From the given information, we know: Ag⁺ concentration = \(2.0 \times 10^{-4}\) M Pb²⁺ concentration: = \(1.5 × 10^{-3}\) M We will compare the Ksp values for AgI and PbI₂. The lower Ksp indicates the relative ease of precipitation. Ksp for AgI: \(8.3 \times 10^{-17}\) Ksp for PbI₂: \(7.9 \times 10^{-9}\) Since Ksp(AgI) < Ksp(PbI₂), AgI will precipitate first.

Calculate the concentration of I⁻ needed for the first compound to precipitate.

Now, we need to find the concentration of I⁻ needed for AgI to precipitate first. To do this, we set Q equal to the Ksp for AgI and solve for the I⁻ concentration: \[Q_{1} = K_{s p} (\mathrm{AgI})\] \[\mathrm{[Ag^{+}][I^{-}]}= 8.3 \times 10^{-17} \] \[\Rightarrow [I^{-}]=\frac{8.3 \times 10^{-17}}{\mathrm{[Ag^{ + }]}}\] \[ [I^{-}]=\frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} \] Now, calculating the concentration of I⁻: \[ [I^{-}] = 4.15 \times 10^{-13} \mathrm{M} \] So, the concentration of I⁻ needed to begin precipitation is \(4.15 \times 10^{-13}\) M. AgI will precipitate first when NaI is added to the solution.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, denoted as \( K_{sp} \), is a crucial concept in understanding solubility and precipitation reactions. It describes the equilibrium state between a solid and its ions in solution. Essentially, \( K_{sp} \) is a measure of how much of a compound can dissolve in water. For a generic salt \( AB \), which dissociates into \( A^+ \) and \( B^- \) ions, the \( K_{sp} \) expression is written as: \[ K_{sp} = [A^+][B^-] \] When a solution reaches the point where no more solute can dissolve, it’s at equilibrium, and the product of the ion concentrations equals \( K_{sp} \).

• If the product of the ion concentrations in a solution exceeds \( K_{sp} \), the solution is supersaturated, and a precipitate will form.
• If the ion product is less than \( K_{sp} \), the solution is unsaturated, and no precipitate will form.

In our example, the \( K_{sp} \) values for AgI and PbI₂ suggest that AgI will precipitate first because its \( K_{sp} \) is much lower, indicating it's less soluble.

Precipitation Reactions

Precipitation reactions occur when dissolved ionic species in a solution combine to form an insoluble solid, or precipitate. Understanding when a precipitation reaction happens involves comparing the current state of the solution to the solubility product constant (\( K_{sp} \)). Here’s the basic process:

• Initially, ions are freely moving in the solution. When the specific ion concentrations reach or exceed a certain level, they combine to form a solid.
• For example, if iodine ions (\( I^- \)) are added to a solution containing silver ions (\( Ag^+ \)), they'll form AgI if the product \( [Ag^+][I^-] \) equals or surpasses the \( K_{sp} \).

In the case of AgI and PbI₂, since AgI has a lower \( K_{sp} \), it means that less iodine is needed for it to start precipitating compared to PbI₂. Thus, adding \( NaI \) to the solution will cause AgI to appear first as a solid.

Reaction Quotient (Q)

The reaction quotient \( Q \) is a valuable tool in predicting whether precipitation will occur in a solution. It helps to determine the current state of a reaction relative to its equilibrium by using the initial concentrations of the reacting species. The formula for \( Q \) mirrors the equilibrium expression, which, for AgI for example, would be: \[ Q = [Ag^+][I^-] \] By comparing \( Q \) to \( K_{sp} \), you can determine the tendency for a reaction to proceed.

• If \( Q < K_{sp} \), the solution is below saturation, meaning no precipitate will form yet as the solution can still dissolve more ions.
• If \( Q = K_{sp} \), the system is at equilibrium, and the solution is just saturated.
• If \( Q > K_{sp} \), the solution is supersaturated, hence a precipitate will form.

In our exercise, we calculated \( Q \) for both AgI and PbI₂ by setting it equal to the \( K_{sp} \) and solving for \( [I^-] \) to find the ion concentration needed to begin precipitation. This, in turn, allows us to predict which compound will precipitate from the solution first.