Problem 101 - Chemical Principles: The Quest for Insight

Question

Will \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) precipitate from a solution formed from a mixture of \(100 \mathrm{~mL}\) of \(1.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{AgNO}_{3}(\) aq \()\) and \(100 \mathrm{~mL}\) of \(1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) ?

Step-by-Step Solution

Answer

No, \(Ag_2CO_3\) will not precipitate from the solution because the reaction quotient \(Q\) is less than the solubility product constant \(Ksp\).

Step 1: Calculate the final concentrations of Ag+ and CO3 2- ions

Since the volumes of the solutions are equal, when they are mixed, the concentration of each ion is halved because the total volume doubles. Calculate the concentration of \(Ag^+\) and \(CO_3^{2-}\) ions in the new solution: \[\frac{1.0 \times 10^{-4} \text{M}}{2} = 5.0 \times 10^{-5} \text{M}\]

Step 2: Write the solubility product (Ksp) expression for Ag2CO3

Write the Ksp expression: \[\text{Ksp} = [Ag^+]^2[CO_3^{2-}]\] Use the Ksp value for \(Ag_2CO_3\) from a table of solubility products. For \(Ag_2CO_3\), the Ksp is typically given as \(8.1 \times 10^{-12}\).

Step 3: Calculate the reaction quotient (Q)

Calculate the reaction quotient using the concentrations found in Step 1: \[Q = (5.0 \times 10^{-5})^2(5.0 \times 10^{-5}) = 1.25 \times 10^{-13}\]

Step 4: Compare Q with Ksp

Compare the calculated Q value with the given Ksp value. If \(Q < Ksp\), the solution is unsaturated, and no precipitate will form. If \(Q = Ksp\), the solution is at equilibrium, and if \(Q > Ksp\), a precipitate will form. In this case, \(Q = 1.25 \times 10^{-13} < Ksp = 8.1 \times 10^{-12}\), so no precipitate will form.

Key Concepts

Concentration Calculation

Understanding concentration calculations is crucial when you deal with chemical solutions. Concentration generally tells us how much of a substance is present in a certain volume of solution, usually denoted in molarity (M), which is moles per liter.

For the exercise, we know that we've got two 100 mL solutions with an initial concentration of 1.0 x 10^-4 M. When these two solutions are mixed, the volume doubles, leading to a halving of each ion's concentration. It's like pouring two equal glasses of juice into a larger jug; you have more to drink, but the flavor (concentration) is less intense because you’ve essentially ‘diluted’ it.

The concentration of the ions becomes 5.0 x 10^-5 M after mixing. It's essential to understand this dilution effect because it directly affects the next steps in predicting whether a precipitate will form in the solution.

Reaction Quotient (Q)

The reaction quotient, Q, plays a significant role in predicting the direction a reaction will proceed. It is calculated in the same way as the equilibrium constant (K), using the concentrations of the products and reactants at any point in time, not just at equilibrium. Think of Q as a snapshot of a reaction's progress, telling you where you are on the reaction path at that specific moment.

For our calculation, we used the formula
Q = [Ag+]^2[CO_3^{2-}],

where [Ag+] and [CO_3^{2-}] are the concentrations we calculated earlier. By comparing Q to the Ksp value, we get to know if the system will form a precipitate (if Q > Ksp), remain the same (if Q = Ksp), or dissolve more solute (if Q < Ksp). In this case, the value of Q was less than Ksp, telling us that the solution is unsaturated, and hence, no precipitate will form.

Precipitation Reaction

Precipitation reactions involve the formation of a solid from a solution when two soluble salts react. As ions meet and exceed the product of their solubility (a value that determines just how much can dissolve without forming a solid), they form a precipitate, which is the insoluble solid.

In the context of our exercise, we're looking at whether
Ag_{2}CO_{3} will precipitate when AgNO_{3} and Na_{2}CO_{3} are mixed. Precipitation happens when the reaction quotient Q exceeds the solubility product, Ksp. However, as we've already calculated, the current state of the mixed solution does not favor precipitation, since Q is less than Ksp, meaning that no solid
Ag_{2}CO_{3} will form under these conditions.

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of the reactants and products. This doesn't mean the reactions have stopped, but that they're occurring at the same rate in both directions.

To express this equilibrium, we use the equilibrium constant, K, with its specific case for solubility being the Ksp, the solubility product constant. This constant helps predict the extent of a substance's solubility in a solution. In our silver carbonate scenario, we have Ksp, and if the reaction reached equilibrium, Q would equal Ksp. In this case, since Q is not greater than Ksp, the system hasn't reached the point where a precipitate starts forming, indicating that it remains in a state where dissolution is favored, rather than reaching equilibrium with solid formation.