Problem 107 - General Chemistry: Principles and Modern Applications

Question

Will AgI(s) precipitate from a solution with \(\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]=0.012 \mathrm{M}, \left[\mathrm{CN}^{-}\right]=1.05 \mathrm{M}, \) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\) For \( \mathrm{AgI}, K_{\mathrm{sp}}=8.5 \times 10^{-17} ; =\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, K_{\mathrm{f}}=5.6 \times 10^{18}\).

Step-by-Step Solution

Answer

The precipitation of AgI from the solution will depend on the comparison of the calculated Q value with the given Ksp value. If Q > Ksp, then AgI will precipitate.

Step 1: Calculate the concentration of Ag+ ions

The expression for the formation constant (Kf) for the complex ion is given by:\nKf = [\(\mathrm{Ag+}\)]/[\(\mathrm{Ag(CN)_{2}^{-}}\)][\(\mathrm{CN-}\)]².\nWe are given that Kf = \(5.6 \times 10^{18}\), [\(\mathrm{Ag(CN)_{2}^{-}}\)] = 0.012M and [\(\mathrm{CN-}\)] = 1.05M. Solving the equation for [\(\mathrm{Ag+}\)], we obtain:\n[\(\mathrm{Ag+}\)] = Kf × [\(\mathrm{Ag(CN)_{2}^{-}}\)] × [\(\mathrm{CN-}\)]²\nSubstituting the given values into this equation, we can calculate [\(\mathrm{Ag+}\)]

Step 2: Compute the ionic product

To decide if AgI will precipitate, we have to calculate the ionic product for the AgI and compare it with the given solubility product (Ksp). The ionic product (Q) is given by Q = [\(\mathrm{Ag+}\)] × [\(\mathrm{I-}\)]. Using the calculated [\(\mathrm{Ag+}\)] from Step 1 and the given [\(\mathrm{I-}\)] = 2.0M, we can compute Q.

Step 3: Compare ionic product with solubility product

The final step is to compare Q with Ksp. If Q > Ksp, then AgI will precipitate. So, we compare the calculated Q from Step 2 with the given Ksp = \(8.5 \times 10^{-17}\)

Key Concepts

Solubility Product (Ksp)

The solubility product, often abbreviated as Ksp, is a constant that helps determine whether a salt will dissolve in water. It's specific to each compound at a given temperature. When you work with Ksp, you're dealing with a type of equilibrium related to the solubility of a compound. This equilibrium is where the solid and its ions in solution are balanced.

For a salt like AgI (silver iodide), which is considered sparingly soluble, Ksp is the product of the concentrations of its ions raised to the power of their stoichiometric coefficients. In the case of AgI, it dissolves to form one Ag⁺ ion and one I⁻ ion, so the expression for the solubility product is:

• \[ K_{ ext{sp}} = [ ext{Ag}^+] imes [ ext{I}^-] \]

If the product of the concentrations of these ions exceeds Ksp, the solution becomes supersaturated, and the excess ions will precipitate as a solid.
In our problem, determining if AgI will precipitate involves comparing the Ksp of AgI with the ionic product, which indicates the actual concentrations of Ag⁺ and I⁻ in the solution.

Complex Ion Formation

Complex ions are fascinating entities that result from the union of a metal ion with one or more ligands, which are ions or molecules that can donate an electron pair. This bond formation creates a complex ion with unique chemical behavior.

In our scenario, we focus on \( \left[\text{Ag(CN)}_2^-\right] \), which is a complex ion of silver (Ag⁺) bound to two cyanide ions (CN⁻).

• The formation constant, often represented as \( K_{ ext{f}} \), quantifies the stability of a complex ion in solution.
• A larger \( K_{ ext{f}} \) value, like \( 5.6 \times 10^{18} \), signifies a very stable complex ion.

The given exercise uses \( K_{ ext{f}} \) to find the concentration of free \( ext{Ag}^+ \) ions in the solution.
The chemical equilibrium for the formation of this ion is:
\[ K_{ ext{f}} = \frac{[\text{Ag}^+][\text{CN}^-]^2}{[\text{Ag(CN)}_2^-]} \]
Once you've solved for \( [\text{Ag}^+] \), you can then proceed to calculate whether the concentration of \( ext{Ag}^+ \) and \( ext{I}^- \) exceeds the solubility product (Ksp), which indicates if precipitation occurs.

Ionic Product

The ionic product (Q) is a crucial figure that tells us about the saturation level of a solution regarding a particular salt. It's calculated similarly to the solubility product (Ksp) but uses the actual concentrations available in the liquid.

When you compute the ionic product for a compound like AgI, it helps determine whether precipitation will occur:

• If Q < Ksp, the solution is unsaturated, and more solute can dissolve.
• If Q = Ksp, the solution is at equilibrium, and no more solute can dissolve without causing precipitation.
• If Q > Ksp, the solution is supersaturated, and the excess solute will start to precipitate out.

To find the ionic product in the problem, multiply \( [\text{Ag}^+] \) by \( [\text{I}^-] \). The outcome will tell us if the Ag⁺ and I⁻ ions in the solution exceed the capacity indicated by the Ksp of AgI.
Using this comparison, you can conclude whether AgI precipitates in your solution.