What is the solubility of AgCl when it is in a solution of 0.15 M NaCl? (AgCl has Ksp = 1.8 x 10-10.) Make a reaction table. Include rows for initial concentration, change in concentration, and equilibrium concentration. Write down the equation for the Ksp of the reaction. What is the concentration of Cl-? Plug that number into the Ksp equation to determine the solubility of AgCl when it is in a solution of 0.15 M NaCl
...........AgCl ==> Ag^+ + Cl^-
I............x........0........0
C...........-x.........x........x
E...........-x..........x......x
........NaCl ==> Na^+ + Cl^-
I.......0.15M.....0......0
C.......-0.15M....0.15....0.15
E........0.........0.15...0.15
Ksp + (Ag^+)(Cl^-)
1.8E-10 = (x)(x+0.15) note:(Cl^-) = x+ 0.15 or x from the AgCl + 0.15 from the NaCl. This a problem illustrating the common ion effect. The solubility of AgCl is DECREASED significantly by the addition of a common ion.
Solve for x.
To determine the solubility of AgCl when it is in a solution of 0.15 M NaCl, we need to set up a reaction table and use the Ksp equation.
1. Reaction Table:
We will assume that x moles of AgCl dissolve in the 0.15 M NaCl solution.
AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)
Initial concentration: AgCl = x M, Ag⁺ = 0 M, Cl⁻ = 0.15 M
Change in concentration: AgCl = -x M, Ag⁺ = +x M, Cl⁻ = +0.15 M
Equilibrium concentration: AgCl = (x-x) M = 0 M, Ag⁺ = x M, Cl⁻ = 0.15 + 0.15 M = 0.3 M
2. Ksp equation:
The Ksp expression for AgCl is: Ksp = [Ag⁺][Cl⁻]
Given that Ksp = 1.8 x 10⁻¹⁰, we can substitute the equilibrium concentrations into the Ksp equation:
1.8 x 10⁻¹⁰ = (x)(0.3)
3. Solving for x:
Rearranging the equation, we get:
x = (1.8 x 10⁻¹⁰) / (0.3)
x ≈ 6 x 10⁻¹² M
Therefore, the solubility of AgCl when it is in a solution of 0.15 M NaCl is approximately 6 x 10⁻¹² M.