Question

A solution contains \(1.0 \times 10^{-5} \mathrm{MAg}^{+}\) and \(2.0 \times 10^{-6} \mathrm{M} \mathrm{CN}^{-}\). Will \(\mathrm{AgCN}(s)\) precipitate? \(\left(K_{\text {sp }}\right.\) for \(\mathrm{AgCN}(s)\) is \(2.2 \times 10^{-12}\).)

Short answer

Yes, AgCN will precipitate since the reaction quotient (Q = \(2.0 \times 10^{-11}\)) is greater than the solubility product constant (Ksp = \(2.2 \times 10^{-12}\)).

Step-by-step solution

Write the dissociation equation for AgCN

Write the dissociation equation for the reaction of AgCN precipitating: \[ \mathrm{AgCN}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{CN}^{-}(aq) \]

Calculate the Reaction Quotient (Q)

Calculate Q using the given concentrations of Ag⁺ and CN⁻ ions: \[ Q = [\mathrm{Ag}^{+}][\mathrm{CN}^{-}] \] Given concentrations: Ag⁺ = \(1.0 \times 10^{-5} \mathrm{M}\) CN⁻ = \(2.0 \times 10^{-6} \mathrm{M}\) Calculate Q: \[ Q = (1.0 \times 10^{-5})(2.0 \times 10^{-6}) = 2.0 \times 10^{-11} \]

Compare Q with Ksp

Compare the calculated Q value with the given Ksp value for AgCN: Q = \(2.0 \times 10^{-11}\) Ksp = \(2.2 \times 10^{-12}\) Since Q > Ksp, it means the reaction will shift to the left to reach equilibrium, causing AgCN to precipitate.

Key concepts

Chemical Equilibrium

Understanding the balance of a chemical reaction is essential when predicting how much product will form. Chemical equilibrium is the state of a reaction where the rates of the forward and reverse reactions are equal, meaning there is no net change in the concentration of reactants and products. It is a dynamic equilibrium because molecules continue to react, but the overall concentrations remain constant over time.

When a solution contains ions that can form a precipitate, as seen in the exercise, equilibrium will play a significant role in determining if and when precipitation will occur. If more product (precipitate) starts to form than is being dissolved, the reaction will shift towards the reactants to re-establish equilibrium.

Reaction Quotient (Q)

To predict the direction in which a reaction at nonequilibrium conditions will proceed, chemists use the reaction quotient (Q). Q is calculated using the same expression as the equilibrium constant (K), but with initial concentrations instead of equilibrium concentrations.

If Q is less than the solubility product constant (Ksp), the reaction will proceed forward towards forming more precipitate. Conversely, if Q is greater than Ksp, the reaction will favor the formation of reactants to reduce the concentrations of ions in the solution. In the given exercise, Q's value being greater than Ksp indicates the solution is supersaturated, leading to precipitation.

Precipitation Reactions

Precipitation reactions involve the formation of a solid from the reaction of ions in solution. When the product of the ionic concentrations exceeds the solubility product constant (Ksp), a precipitate will form, as the solid is less soluble than needed to maintain those ion concentrations in solution.

In the context of the exercise, the presence of Ag⁺ and CN⁻ ions in a solution can lead to the formation of AgCN solid. The solubility equilibrium for AgCN involves dynamically balancing the solid form with its dissolved ions. The driving factor behind precipitation is reaching a state where the solution can no longer hold all the dissolved ions, and excess begins to settle out as a solid.

Dissociation Equation

The dissociation equation represents the breakdown of a solid ionic compound into its constituent ions in a solution. For AgCN, the dissociation equation is:

\[ AgCN(s) \rightleftharpoons Ag^{+}(aq) + CN^{-}(aq) \]
This equation is pivotal in understanding solubility equilibria. It displays how AgCN dissociates into Ag⁺ and CN⁻ ions in water. To predict whether a precipitate forms, knowing the concentrations of these ions, as given in the exercise, allows the calculation of the reaction quotient, Q. The comparison of Q with the Ksp then predicts precipitation.