A solution is prepared by dissolving 3.0 × 10-4 mol of AgNO3 and 3.0 × 10-5 mol of K2CrO4 in 1.00 liter of solution. With respect to Ag2CrO4 (Ksp = 1.9 × 10-12), is the resulting solution saturated, supersaturated, or unsaturated?

Question

A solution is prepared by dissolving 3.0 × 10-4 mol of AgNO3 and 3.0 × 10-5 mol of K2CrO4 in 1.00 liter of solution. With respect to Ag2CrO4 (Ksp = 1.9 × 10-12), is the resulting solution saturated, supersaturated, or unsaturated?

I know you do [Ag][CrO4} but do I square the Ag concentration? If I don't I get supersaturated, and if I do, I get unsaturated?

Answer 1

To determine whether the resulting solution is saturated, supersaturated, or unsaturated with respect to Ag2CrO4, you need to calculate the ionic product (Qsp) and compare it to the solubility product constant (Ksp).

The solubility product constant (Ksp) for Ag2CrO4 is given as 1.9 × 10^-12.

First, let's write the balanced equation for the dissociation of Ag2CrO4:

Ag2CrO4 (s) ⇌ 2Ag+ (aq) + CrO4^-2 (aq)

From the equation, you can see that the stoichiometry of Ag2CrO4 is such that one mole of Ag2CrO4 produces two moles of Ag+ ions.

Now, let's calculate the concentration of Ag+ ions in the solution. We are given that the solution contains 3.0 × 10^-4 mol of AgNO3 dissolved in 1.00 liter of solution. Therefore, the concentration of Ag+ ions is:

[Ag+] = (3.0 × 10^-4 mol) / (1.00 L) = 3.0 × 10^-4 M

However, since Ag2CrO4 dissociates to form two Ag+ ions, the concentration of Ag+ ions should be multiplied by 2:

[Ag+] = 2 × (3.0 × 10^-4 M) = 6.0 × 10^-4 M

Now, let's calculate the concentration of CrO4^-2 ions in the solution. We are given that the solution contains 3.0 × 10^-5 mol of K2CrO4 dissolved in 1.00 liter of solution. Therefore, the concentration of CrO4^-2 ions is:

[CrO4^-2] = (3.0 × 10^-5 mol) / (1.00 L) = 3.0 × 10^-5 M

Now that we have the concentrations of Ag+ and CrO4^-2 ions in the solution, we can calculate the ionic product (Qsp) using the following expression:

Qsp = [Ag+]^2 × [CrO4^-2]

Substituting the values we calculated:

Qsp = (6.0 × 10^-4 M)^2 × (3.0 × 10^-5 M) = 1.08 × 10^-11

Comparing the calculated ionic product (Qsp) to the solubility product constant (Ksp), we find that Qsp (1.08 × 10^-11) is greater than Ksp (1.9 × 10^-12).

When Qsp is greater than Ksp, the solution is considered to be supersaturated, meaning it contains more dissolved ions than it would normally be able to hold at that temperature.

So, in this case, the resulting solution is supersaturated with respect to Ag2CrO4.