The Ksp of AgCl is 1.7 × 10^-10. How many moles of MnCl₂ can be dissolved in one liter of a solution in which [AgNO₃] = 3.4 × 10^-4 M before a precipitate appears?

Question

Quizplus · Accepted Answer

The equation for the dissolution of AgCl is:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The Ksp expression for this reaction is:

Ksp = [Ag+][Cl-]

We are given the value of Ksp, so we can use it to find the concentrations of Ag+ and Cl- in a saturated solution of AgCl:

Ksp = [Ag+][Cl-] = 1.7 × 10^-10

[Ag+] = [Cl-] = √Ksp = 1.3 × 10^-5 M

This means that the solubility of AgCl in water is 1.3 × 10^-5 mol/L. However, the problem asks us to calculate the maximum amount of MnCl2 that can be dissolved in a solution that already contains Ag+ ions, before AgCl begins to precipitate.

The addition of Ag+ ions to a solution containing Cl- ions will cause AgCl to precipitate, according to the following reaction:

Ag+(aq) + Cl-(aq) ⇌ AgCl(s)

The solubility of AgCl will decrease as more Ag+ is added to the solution. In other words, as the concentration of Ag+ ions in the solution increases, the amount of MnCl2 that can be dissolved without precipitating AgCl will decrease.

To find the maximum amount of MnCl2 that can be dissolved in the solution, we need to calculate the concentration of Cl- ions that will be present when AgCl begins to precipitate. This will occur when the product of [Ag+] and [Cl-] equals Ksp:

Ksp = [Ag+][Cl-] = 1.3 × 10^-5 × [Cl-] = 1.7 × 10^-10

[Cl-] = 1.3 × 10^-5 M / 1.7 × 10^-10 = 7.6 × 10^4 M

Now we can use the concentration of Cl- ions in the solution to calculate the maximum amount of MnCl2 that can be dissolved, using the Ksp expression for MnCl2:

Ksp = [Mn2+][Cl-]^2 = 4.0 × 10^-6

[Mn2+] = Ksp / [Cl-]^2 = 4.0 × 10^-6 / (7.6 × 10^4)^2 = 5.0 × 10^-7 M

Finally, we can convert from concentration to moles by multiplying by the volume of the solution:

moles of MnCl2 = 5.0 × 10^-7 mol/L × 1 L = 5.0 × 10^-7 mol

Therefore, the answer is E, 2.5 × 10^-7 mol.

The Ksp of AgCl Is 1