Problem 67 - Chemistry: The Central Science

Question

A solution contains \(20 \times 10^{-4} \mathrm{M} \mathrm{Ag}^{+}\) 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 \(\mathrm{PbI}_{2}\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipi- tate first? Specify the concentration of \(\mathrm{I}^{-}\) needed to begin precipitation.

Step-by-Step Solution

Answer

In conclusion, AgI will precipitate first when NaI is added to the solution, and the concentration of I- ions needed to begin precipitation is \(4.15 \times 10^{-14} M\).

Step 1: Write down the given information

We are given: - Concentration of Ag+ ions: \(20 \times 10^{-4} M\) - Concentration of Pb2+ ions: \(1.5 \times 10^{-3} M\) - Solubility product constants: \(K_{sp}(AgI) = 8.3 \times 10^{-17}\), and \(K_{sp}(PbI_2) = 7.9 \times 10^{-9}\)

Step 2: Calculate the Q value for AgI and PbI2

For a compound to precipitate, the ion product (Q) in the solution must be greater than the solubility product (Ksp). Let's find the expression of Q for AgI and PbI2. For the reaction: \(Ag^{+} + I^{-} \rightleftharpoons AgI\), the \(Q=[Ag^+][I^-]\). Similarly, for the reaction: \(Pb^{2+} + 2I^{-} \rightleftharpoons PbI_2\), the \(Q=[Pb^{2+}][I^-]^2\).

Step 3: Find the I- concentration for each compound to reach Ksp

We will now find the I- concentration to reach Ksp for each compound. For AgI: \[Q = K_{sp}\] \([Ag^+][I^-] = 8.3 \times 10^{-17}\) \([I^-] = \frac{8.3 \times 10^{-17}}{20 \times 10^{-4}} = 4.15 \times 10^{-14} M\) For PbI2: \([Pb^{2+}][I^-]^2 = 7.9 \times 10^{-9}\) \([I^-]^2 = \frac{7.9 \times 10^{-9}}{1.5 \times 10^{-3}}\] \([I^-] = \sqrt{5.27 \times 10^{-6}} = 2.3 \times 10^{-3} M\)

Step 4: Determine which compound precipitates first

As we can see, the AgI requires a lower concentration of I- ions (\(4.15 \times 10^{-14} M\)) to reach its Ksp compared to PbI2 (\(2.3 \times 10^{-3} M\)). Therefore, AgI will precipitate first when NaI is added to the solution.

Step 5: Specify the concentration of I- needed to begin precipitation

The concentration of I- ions needed to initiate precipitation is the concentration required for AgI to reach its Ksp, as AgI precipitates first. Thus, the required I- concentration is \(4.15 \times 10^{-14} M\). In conclusion, AgI will precipitate first when NaI is added to the solution, and the concentration of I- ions needed to begin precipitation is \(4.15 \times 10^{-14} M\).

Key Concepts

Solubility Calculations

Understanding solubility calculations is crucial for predicting whether a substance will dissolve in a solution. Solubility is a measure of how much of a particular solute can be dissolved in a solvent at a given temperature. When we speak of solubility in the context of ionic compounds, we refer to the maximum amount of substance that can dissolve before the solution becomes saturated and precipitation begins.

The solubility of a substance is often represented by the solubility product constant (K_sp), which is specific to each ionic compound at a particular temperature. The K_sp is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation. For instance, the solubility product for silver iodide (AgI) is calculated as follows: \[K_{sp} = [Ag^+][I^-]\]

In our example, to prevent the calculation from becoming complex with simultaneous equilibrium systems, we can assume that only one compound begins to precipitate at a time. This allows us to tackle the solubility calculation for each potentially precipitating compound independently.

Precipitation of Ionic Compounds

Moving onto the concept of precipitation of ionic compounds, precipitation occurs when the concentration of ionic species in a solution exceeds the solubility of the compound they form. This causes the compound to come out of solution, forming a solid. In our exercise, we are comparing two potential precipitates, AgI and PbI₂, and assessing which will form first upon the addition of iodide ions (I^-) to the solution.

By calculating the concentration of iodide ions needed to reach the solubility product constant for both AgI and PbI₂, we can predict the order of precipitation. A lower iodide concentration required to reach K_sp for AgI suggests it will precipitate before PbI₂. This is a result of the different K_sp values, which imply that AgI is less soluble than PbI₂. Hence, as soon as the I^- concentration in our exercise reaches \(4.15 \times 10^{-14} M\), AgI begins to precipitate.

Chemical Equilibrium

Finally, let's discuss chemical equilibrium, the state where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. It is essential to note that this does not mean the reactants and products are in equal concentrations but rather that their concentrations have stabilized at a particular ratio which remains unchanged.

The solubility product constant (K_sp) is actually an equilibrium constant for the dissolving process of a salt. When the solution is saturated and additional solute is added, the system will reach a new equilibrium by forming a precipitate, which keeps the ion product (Q) in balance with the K_sp.

In the case of our exercise, as sodium iodide (NaI) is added, iodide ions (I^-) increase in the solution. When the concentration of I^- ions reaches a point where K_sp is exceeded for AgI, equilibrium shifts to form a solid precipitate of AgI. Thus, understanding chemical equilibrium can help predict the conditions under which a precipitate will form in a saturated solution.