Problem 50 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

Since the calculated reaction quotient, Q, (\(2.0 \times 10^{-11}\)) is greater than the solubility product constant, Ksp, (\(2.2 \times 10^{-12}\)), AgCN will precipitate in the given solution.

Step 1: Write out the reaction and Ksp expression

First, write out the reaction for the dissociation of AgCN(s) and the Ksp expression. Reaction: \(AgCN(s) \rightleftharpoons Ag^+(aq) + CN^-(aq)\) Ksp expression: \(K_{sp} = [Ag^+][CN^-]\) The given \(K_{sp}\) value for AgCN is \(2.2 \times 10^{-12}\).

Step 2: Calculate the reaction quotient (Q)

Next, we can calculate the Q value using the given concentrations of Ag+ and CN- ions in the solution. Q = \([Ag^+]\)[\(CN^-\)\] Q = \( (1.0 \times 10^{-5})(2.0 \times 10^{-6})\) Now, calculate Q: Q = \( 2.0 \times 10^{-11}\)

Step 3: Compare Q and Ksp

Now, we will compare the calculated Q value to the given Ksp value to determine if AgCN will precipitate in the solution. Ksp = \(2.2 \times 10^{-12}\) Q = \(2.0 \times 10^{-11}\) Since Q > Ksp (\(2.0 \times 10^{-11}\) > \(2.2 \times 10^{-12}\)), we can conclude that AgCN will precipitate in the given solution.

Key Concepts

Reaction Quotient (Q)

The reaction quotient, denoted as Q, is an important term in chemistry that helps us determine the direction in which a chemical reaction will proceed. You can think of Q as a snapshot of the concentration of products and reactants at any given moment. It is calculated using the same expression as the equilibrium constant, but for a system that is not necessarily at equilibrium.

To find Q, use the concentrations of the chemical species at the specific moment you are examining:

• First, write out the balanced chemical equation for the reaction in question.
• Next, write out the expression for Q, similar to the equilibrium constant expression, using the concentrations of the reactants and products.
• For example, in the case of dissociation of AgCN, we use: \[ Q = [Ag^+][CN^-] \]
• Substitute the known concentrations into this expression to find Q.

Comparing Q with the equilibrium constant tells us if the system has reached equilibrium:

• If Q < K, the reaction will proceed in the forward direction, forming more products.
• If Q > K, the reaction will proceed in the reverse direction, forming more reactants.
• If Q = K, the system is at equilibrium, and there is no net change.

Solubility Product Constant (Ksp)

The solubility product constant, abbreviated as Ksp, is a specialized form of equilibrium constant that applies to sparingly soluble ionic compounds. It tells us how much of the compound can dissolve in a solvent before the solution becomes saturated, which means no more solute can dissolve.

Let's understand Ksp with the example of silver cyanide, AgCN:

• The dissociation of AgCN in water is expressed as:\[ AgCN(s) \rightleftharpoons Ag^+(aq) + CN^-(aq) \]
• The solubility product expression is:\[ K_{sp} = [Ag^+][CN^-] \]
• Each ion's concentration in this expression is raised to the power of its coefficient in the balanced equation, which in this case is 1 for both.
• A higher Ksp value indicates that the substance is more soluble; a very low Ksp means the substance is not very soluble.

For AgCN, the given Ksp is \(2.2 \times 10^{-12}\), reflecting its low solubility. When the product of ion concentrations exceeds Ksp, precipitation occurs, which is a key insight for predicting the formation of a solid.

• If Q (reaction quotient) is greater than Ksp, it indicates supersaturation, and as a result, the excess ions will form a precipitate.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept signifying a state where the concentrations of reactants and products remain constant over time. It's important to note that this does not mean the reactions have stopped but that the rates of the forward and reverse reactions are equal.

There are several aspects of chemical equilibrium that are crucial to understand:

• Dynamic Equilibrium: At equilibrium, while the concentrations do not change, individual molecules continue to react, showing the dynamic nature of equilibrium.
• Reversible Reactions: Equilibrium can only occur in reversible reactions where the products can reform into reactants.
• Equilibrium Constant (K): Each reversible reaction has an equilibrium constant that is determined by the relative concentrations of the reactants and products at equilibrium. The equilibrium expression based on the balanced chemical equation is used to calculate the constant.
• Impact of Changes: Changes in concentration, pressure, and temperature can disturb an equilibrium state, as described by Le Chatelier's principle, which predicts changes the system will undergo to reach a new equilibrium.

Understanding equilibrium helps in determining conditions under which reactions form precipitates, as seen when comparing Q and Ksp to decide if a solid will form in the solution.