Problem 55 - General Chemistry: Principles and Modern Applications

Question

Can the following ion concentrations be maintained in the same solution without a precipitate forming: \(\left[\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\right]=0.048 \mathrm{M},\left[\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right]=0.76 \mathrm{M},\) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\)

Step-by-Step Solution

Answer

No, the mentioned ion concentrations cannot be maintained in the same solution without a precipitate forming. The precipitate that forms is Silver Iodide (AgI).

Step 1 - Understanding the Problem

Recognize that we are dealing with a question of solubility and the formation of precipitates. Understand that the ions given are \([\mathrm{Ag(S_{2}O_{3})_{2}}]^{3-}, [\mathrm{S_{2}O_{3}]^{2-}\), and \([\mathrm{I^{-}}]\). They can react to form AgI (Silver Iodide) or stay as \([\mathrm{Ag(S_{2}O_{3})_{2}}]^{3-}\). We will check whether the formation of AgI leads to precipitation or not.

Step 2 - Calculation of Reaction Quotient (Q)

The reaction in concern here is: Ag+ + I- ⇌ AgI. For this reaction, the equilibrium constant expression, Q can be written as: Q = [AgI], as the concentration of solid AgI is not taken into account. The concentration of Ag+ ions can be calculated using the equation: [Ag+] = 3 * [\([Ag(S_2O_3)_2]^{3-}\)]= 3 * 0.048M = 0.144M. Using these values, Q = [Ag+][I-] = 0.144 M * 2.0 M = 0.288.

Step 3 - Comparison of Q and Ksp

Next, the value of Q is compared with the Ksp of AgI. If Q > Ksp, the system is supersaturated and a precipitate will form. If Q = Ksp, the system is at equilibrium and no change will occur. If Q < Ksp, the system is unsaturated and no precipitate will form. The Ksp of AgI is given as \(8.3 \times 10^{-17}\). Q \(= 0.288\) for this system is much greater than Ksp. Therefore, a precipitate of AgI will form.

Step 4 - Conclusion

Based on the comparison of Q versus Ksp, it is concluded that the concentrations of ions given cannot exist together in the solution without forming a precipitate. Specifically, the precipitate that forms is Silver Iodide, AgI. This reaction happens because the system tends towards achieving equilibrium and a lower energy state.

Key Concepts

Solubility Product Constant (Ksp)

To understand when a substance will precipitate out of a solution, it’s essential to know about its solubility product constant, commonly referred to as Ksp. Ksp is a special type of equilibrium constant that applies to the dissolution of solids. It quantifies the degree to which a compound dissolves in water to form ions.

For a general reaction where a solid salt AB dissolves into its constituent ions A+ and B-, the Ksp expression is written as:
\[ Ksp = [A^+][B^-] \]
Importantly, Ksp values only include the concentrations of dissolved ions, and solid compounds are omitted since their concentration remains constant at a given temperature. Low values of Ksp suggest a compound is poorly soluble, while high Ksp values indicate a more soluble substance. In the context of our reaction with Silver Iodide (AgI), the Ksp expression would be very low, indicating AgI has low solubility in water.

Reaction Quotient (Q)

The reaction quotient, Q, plays a pivotal role in predicting whether a precipitate will form in a solution under a given set of conditions. It is calculated in the same way as the equilibrium constant, Ksp, using the current concentrations of the ions in the solution.

For the dissolution of AgI into Ag+ and I-, the reaction quotient is given as:
\[ Q = [Ag^+][I^-] \]
In practice, you calculate Q using the initial concentrations before any reaction has reached equilibrium. If Q is greater than Ksp, it indicates that the product of the ionic concentrations exceeds the solubility, favoring precipitation. Alternatively, if Q is less than Ksp, the system has not reached the saturation point, and precipitation is not favored. Should Q equal Ksp, the system is at dynamic equilibrium, and no net change occurs.

Silver Iodide Precipitation

Precipitation reactions are fascinating chemical processes, and the formation of Silver Iodide (AgI) from silver (Ag+) and iodide (I-) ions in solution is a prime example. AgI is known for its remarkably low solubility in water, which is reflected in its very low Ksp value.

In the presence of Ag+ and I- ions, if the product of their concentrations at any moment (the reaction quotient, Q) exceeds the Ksp value of AgI, a precipitate will form. This behavior is predicted by the law of mass action, which states that the rate of a chemical reaction is directly proportional to the product of the masses of the reacting substances. For AgI, this means precipitation occurs rapidly when the conditions are right, demonstrating a shift in the equilibrium towards the formation of a solid.

Equilibrium Principles

Equilibrium is a state in chemical reactions where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. This is a dynamic state, meaning that even though there is no change when viewed from a macroscopic level, the molecules are still reacting with one another.

Le Châtelier’s Principle is fundamental to understanding changes to equilibria. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In the context of solubility, if a solution is initially unsaturated (Q < Ksp) and more solute is added, the system will shift towards forming more precipitate until a new equilibrium is established. Conversely, if the solution is supersaturated (Q > Ksp), the excess solute will precipitate out until equilibrium is achieved.