The Ksp of AgCl is 1.8 * 10-10. How do you know that mixing equal amounts of 0.04M solution of AgNO3 and 0.002M solution of NaCl will lead to the formation of a precipitate?

a. [Ag+]*[Cl-] = Ksp AgCl

b. [Ag+]*[Cl-] < Ksp AgCl

c. [Ag+]*[Cl-] > Ksp AgCl

option c. [Ag⁺] * [Cl⁻] > Ksp AgCl,

Explanation:

Ksp is the product solubility constant.

The solubility equation for AgCl is:

AgCl ⇄ Ag⁺ + Cl⁻

The Ksp expression is:

Ksp = [Ag⁺] [Cl⁻]

Since Ksp = 1.8 * 10⁻¹⁰, at equilibrium [Ag⁺] [Cl⁻] = 1.8 * 10⁻¹⁰.

When AgNO₃ dissolves in water it completely ionizes into Ag⁺ and NO₃⁻.

When NaCl dissolves in water it completely inoizes into Na⁺ and Cl⁻.

Uisng the molarity formula, M = n / V, you get n = M * V.

So, a volume V of a 0.04 M solution of AgNO₃ will yield 0.04V moles of Ag⁺ and the same volume V of a 0.002 M solution of NaCl will yield 0.02V moles of Cl⁻⁻

So the concentrations will be:

[Ag⁺] = 0.04V / (2V) = 0.02 M

[Cl⁻] = 0.002V / (2V) = 0.001 M

And the product of the two concentrations will be:

0.02 * 0.0001 = 0.00002 = 2.0 * 10⁻⁵

That product is much greater than 1.8 * 10⁻¹⁰ meaning that the concentrations are higher than the equilibrium concentrations and the mixture will lead to the formation of a precipitate.

The option c., [Ag⁺] * [Cl⁻] > Ksp AgCl, describes that situation.