Question

When equal volumes of the following solutions are mixed, the precipitation of AgCI \(\left(K_{s p}=1.8 \times 10^{-10}\right)\) will occur with : \(\begin{array}{ll}\text { (a) } 10^{-4} M\left(\mathrm{Ag}^{+}\right) \text {and } 10^{-4} M\left(\mathrm{Cl}^{-}\right) & \text {(b) } 10^{-5} M\left(\mathrm{Ag}^{+}\right) \text {and } 10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\end{array}\) (c) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\) and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (d) \(10^{-10} M\left(\mathrm{Ag}^{+}\right)\) and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

Short answer

Precipitation of AgCl will occur with the pairs that have an ionic product greater than Ksp. This includes case (a) with a Q value of 10^-8, which is greater than Ksp, and possibly case (b), where Q equals Ksp. No precipitation will occur in cases (c) and (d) as the Q values are less than Ksp.

Step-by-step solution

- Understanding the Exercise

The exercise is about determining when the precipitation of AgCl will occur when equal volumes of two solutions are mixed. Precipitation occurs when the product of the concentrations of the ions exceeds the solubility product constant (Ksp) of the salt.

- Calculate the Ionic Product for Each Pair

Calculate the ionic product (Q) for the given concentrations by multiplying the molarity of the silver ions (Ag+) by the molarity of the chloride ions (Cl-) for each pair. The formula for ionic product is Q = [Ag+][Cl-].

- Determine Precipitation by Comparing Q and Ksp

Compare the ionic product (Q) obtained from Step 2 with the solubility product constant (Ksp) of AgCl. If Q > Ksp, precipitation will occur.

- Calculate Ionic Product for Case (a)

For case (a), the concentrations are both 10^-4 M. Therefore, Q = (10^-4) * (10^-4) = 10^-8.

- Check for Precipitation in Case (a)

Since the calculated Q value for case (a) is 10^-8, which is greater than the Ksp value of 1.8 x 10^-10, precipitation will occur in this case.

- Calculate Ionic Product for Other Cases

Repeat Step 4 for the remaining cases (b), (c), and (d) to calculate the ionic product for each case. For (b) Q = (10^-5) * (10^-5) = 10^-10, for (c) Q = (10^-5) * (10^-6) = 10^-11, and for (d) Q = (10^-10) * (10^-10) = 10^-20.

- Check for Precipitation in Other Cases

Compare the calculated Q values for cases (b), (c), and (d) with Ksp. Precipitation will occur in cases where Q > Ksp. Case (b) equals the Ksp value, so precipitation may begin to occur. In cases (c) and (d), Q is less than Ksp, so no precipitation will occur.

Key concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant (Ksp) is essential for predicting whether a salt will dissolve in a solution or form a precipitate. Ksp is a special type of equilibrium constant that applies to the dissolving of ionic compounds.

For a salt such as silver chloride (AgCl), when it dissolves in water, it dissociates into its respective ions: AgCl(s) → Ag+ (aq) + Cl- (aq).

The Ksp expression for this dissociation is given by:
\[ K_{sp} = [Ag^{+}][Cl^{-}] \]
where [Ag+] and [Cl-] represent the molar concentrations of the silver and chloride ions in the saturated solution. A higher Ksp value indicates a more soluble ionic compound. When solutions containing Ag+ and Cl- are mixed, if the product of their concentrations exceeds the substance's Ksp, a precipitate will form.

Ionic Product (Q)

The ionic product (Q) is similar to the solubility product constant, but it refers to the product of the concentrations of the dissolved ions at any moment in time, not just at equilibrium.

Q is calculated by multiplying the concentrations of the ionic species involved in the dissolution process. For instance, if you have a solution with concentrations of Ag+ and Cl- ions, you can find the ionic product using the formula:
\[ Q = [Ag^{+}][Cl^{-}] \]
Comparing the ionic product (Q) with the solubility product constant (Ksp) tells us whether a precipitate will form (Q > Ksp), the solution is exactly at equilibrium (Q = Ksp), or the solution is unsaturated (Q < Ksp).

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. This concept is pivotal in understanding precipitation reactions.

In the context of solubility, the equilibrium exists between the solid ionic compound and its dissolved ions. For example, in the case of AgCl, the equilibrium can be represented as:
\[ AgCl(s) \rightleftharpoons Ag^{+} (aq) + Cl^{-} (aq) \]
At equilibrium, the product of the ion concentrations will be constant (the Ksp), providing that the temperature remains constant. Understanding equilibrium helps us predict the outcome of mixing various solutions and whether a precipitate will form or not.

AgCl Precipitation

AgCl precipitation occurs when solutions containing silver ions (Ag+) and chloride ions (Cl-) are mixed together in such a way that the ionic product (Q) exceeds the solubility product constant (Ksp) for AgCl. This results in the formation of solid silver chloride.

The solubility of AgCl is very low, so even small concentrations of Ag+ and Cl- can result in precipitation. The Ksp of AgCl is known to be \(1.8 \times 10^{-10}\). So, in the aforementioned exercise, when the molarity of Ag+ and Cl- is \(10^{-4} M\) for each and they are mixed in equal volumes, precipitation will occur because the Q value (\(10^{-8}\)) is far greater than the Ksp. The variations in concentration will either lead to the initiation of precipitation or no precipitation at all, depending on whether Q is greater than, equal to, or less than Ksp.