Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. For AgCl, K_sp = 1.8 × 10^{- 10}. A) 1.8 × 10^-10 mol Ag⁺ B) 9.0 × 10^-10mol Ag⁺ C) 9.0 × 10^-9mol Ag⁺ D) 6.7 × 10^-8mol Ag⁺ E) 1.3 × 10^-5mol Ag⁺

Question

Quizplus · Accepted Answer

First, write the chemical equation for the dissolution of AgCl in water:

AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

The dissociation equilibrium constant for this reaction is given:

Ksp = [Ag+][Cl-]

The addition of KCl (which fully dissociates into K+ and Cl-) will cause an increase in the concentration of Cl- ions in the solution. This can lead to more AgCl dissolving according to the above equation. However, since the solution was initially saturated with AgCl, any additional AgCl that dissolves will immediately precipitate out again. This means that the concentration of Ag+ ions will remain the same, but the concentration of Cl- ions will increase until the solubility product for AgCl is reached again:

Ksp = [Ag+][Cl-] = 1.8 × 10^-10

Let x be the increase in the concentration of Cl- ions due to the addition of KCl. Then, the equilibrium concentrations of Ag+ and Cl- ions are:

[Ag+] = [Cl-] = x

Since the total volume of the solution does not change, the total number of moles of Ag+ and Cl- ions before and after the addition of KCl must be the same. Initially, there is 1.8 × 10^-10 mol of Ag+ ion in the solution (from the dissolution of AgCl), and 1.8 × 10^-10 mol of Cl- ion (from the dissociation of AgCl). After the addition of KCl, there will be an additional x mol of Cl- ion, giving a total of:

1.8 × 10^-10 mol Ag+ ion
(1.8 × 10^-10 + x) mol Cl- ion

Since KCl fully dissociates in water, the additional x mol of Cl- ion comes from the dissociation of 0.200 g of KCl. First, calculate the molar mass of KCl:

K = 39.10 g/mol
Cl = 35.45 g/mol
Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol

Convert the mass of KCl to moles:

0.200 g KCl × 1 mol KCl/74.55 g KCl = 0.00268 mol KCl

Since KCl dissociates completely into K+ and Cl- ions, there will be 0.00268 mol of Cl- ion added to the solution. This gives:

(1.8 × 10^-10 + x) mol Cl- ion = 1.8 × 10^-10 mol Cl- ion + 0.00268 mol Cl- ion

Solving for x:

x = 0.00268 mol Cl- ion

Since [Ag+] = [Cl-] = x, the concentration of Ag+ ion after equilibrium has been reestablished is also 0.00268 mol/L. The total number of moles of Ag+ and Cl- ions in the solution is:

2 × 0.00268 mol = 0.00536 mol

However, the question asks for the number of moles of Ag+ and Cl- ions COMBINED in the form of AgCl. Since AgCl dissociates into one Ag+ ion and one Cl- ion, the number of moles of AgCl present at equilibrium is:

0.00536 mol Ag+ ion × (1 mol AgCl/1 mol Ag+ ion) = 0.00536 mol AgCl

Therefore, the number of moles of Ag+ and Cl- ions in the solution after equilibrium has been reestablished is 0.00536 mol AgCl, or approximately 6.7 × 10^-8 mol AgCl. The answer is D.

Assuming That the Total Volume Does Not Change After 0