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^-10 mol Ag⁺ C) 9.0 × 10^-9 mol Ag⁺ D) 6.7 × 10^-8 mol Ag⁺ E) 1.3 × 10^-5 mol Ag⁺

Question

Quizplus · Accepted Answer

The balanced equation for the dissolution of AgCl is:

AgCl(s) ↔ Ag+(aq) + Cl-(aq)

The dissolution reaction is in equilibrium when the rate of the forward reaction, dissolution of AgCl(s), is equal to the rate of the reverse reaction, precipitation of AgCl(s). When the solution is saturated, the rate of dissolution is equal to the rate of precipitation, and the concentration of Ag+ and Cl- in the solution is constant.

When Ksp is known, the equilibrium concentrations of the ions in the solution can be calculated. The equilibrium constant expression for AgCl is:

Ksp = [Ag+][Cl-] = 1.8 × 10^-10

Assuming that no other sources of Ag+ or Cl- are present, the ion concentrations in the saturated AgCl solution are:

[Ag+] = [Cl-] = sqrt(Ksp) = sqrt(1.8 × 10^-10) = 1.34 × 10^-5 M

Now that we have established the equilibrium concentrations of Ag+ and Cl- in the solution, we can consider the effect of adding KCl to the solution. When KCl dissolves in water, it dissociates into K+ and Cl- ions:

KCl(s) → K+(aq) + Cl-(aq)

The addition of KCl does not affect the concentration of Cl- ions in the solution, because the concentration of Cl- is already at its maximum (saturated). The addition of more Cl- ions would cause more AgCl to precipitate until the concentration of Cl- ions was brought back to its original value.

The addition of K+ ions, however, does affect the equilibrium. K+ ions can combine with the Cl- ions already present to form KCl(s), reducing the concentration of Cl- ions in the solution. This causes more AgCl to dissolve until the concentration of Cl- ions is restored to its original value.

The amount of K+ ions added to the solution is:

nK+ = mKCl / MK+

where mKCl is the mass of KCl added and MK+ is the molar mass of K+.

mKCl = 0.200 g
MK+ = 39.1 g/mol

nK+ = (0.200 g / 39.1 g/mol) = 0.00511 mol

Since KCl is a 1:1 electrolyte, the number of moles of Cl- ions that are combined with K+ ions and removed from the solution is also 0.00511 mol. Thus, the new concentration of Cl- ions in the solution is:

[Cl-] = [Cl-]initial - nCl-

where [Cl-]initial is the initial concentration of Cl- in the solution, and nCl- is the number of moles of Cl- ions removed from the solution.

[Cl-]initial = sqrt(Ksp) = 1.34 × 10^-5 M
nCl- = 0.00511 mol

[Cl-] = (1.34 × 10^-5 M) - (0.00511 mol / 1 L) = 1.2889 × 10^-5 M

Since [Ag+] and [Cl-] are now different from their initial equilibrium values, the dissolution of AgCl must occur until a new equilibrium is established. The concentration of Ag+ ions in the solution will increase until the product of [Ag+] and [Cl-] is equal to Ksp.

[Ag+] = Ksp / [Cl-] = (1.8 × 10^-10) / (1.2889 × 10^-5 M) = 1.3967 × 10^-6 M

Therefore, the number of moles of Ag+ ions in the solution is:

nAg+ = [Ag+] × V = (1.3967 × 10^-6 mol / 1 L) × (1 L) = 1.3967 × 10^-6 mol

The number of moles of Cl- ions is:

nCl- = [Cl-] × V = (1.2889 × 10^-5 mol / 1 L) × (1 L) = 1.2889 × 10^-5 mol

The total number of moles of Ag+ and Cl- ions in the solution is:

nAg+ + nCl- = (1.3967 × 10^-6 mol) + (1.2889 × 10^-5 mol) = 1.4286 × 10^-5 mol

However, we were asked for the number of moles of AgCl ion (Ag+ + Cl-), not just the separate ions. So we need to add nAg+ and nCl-:

nAgCl = nAg+ + nCl- = 1.4286 × 10^-5 mol

Thus, the answer is D) 6.7 × 10^-8 mol AgCl.

Assuming That the Total Volume Does Not Change After 0