The solubility of Agcl with solubility product 1.6 × 10^-10 in 0.1 M Nacl solution will be ?
......AgCl ==> Ag^+ + Cl^-
I.....solid.....0......0
C.....solid.....x......x
E.....xolid.....x......x
NaCl is complely ionized (100%) so
....NaCl ==> Na^+ + Cl^-
I...0.1M.....0.......0
C...-0.1....0.1......0.1
E....0......0.1......0.1
Ksp = (Ag^+))Cl^-)
Ksp you know.
(Ag^+) = x from the E line.
(Cl^-) = 0.1+x (0.1 from the NaCl and x from the AgCl)
Solve for x = (Ag^+) = solubility.
Post your work if you get stuck.
To determine the solubility of AgCl in a 0.1 M NaCl solution, we need to understand the concept of solubility product constant (Ksp). The Ksp is an equilibrium constant that indicates the extent to which a solid compound dissolves in a solution.
The solubility product constant (Ksp) for AgCl is given as 1.6 × 10^-10. It represents the product of the concentrations of the ions when AgCl is at its saturation point in a solution.
AgCl(s) ⇌ Ag+(aq) + Cl^-(aq)
The equilibrium expression for AgCl dissolution can be written as:
Ksp = [Ag+][Cl-]
Since AgCl fully dissociates into Ag+ and Cl- ions, the concentration of Ag+ or Cl- ions at equilibrium (assuming equal concentrations) can be represented by 'x'. Therefore, the equilibrium expression can be written as:
Ksp = x * x
To determine the value of 'x', we can assume that the concentration of AgCl that dissolves is 's' M. This is because Ag+ and Cl- ions have a 1:1 stoichiometric ratio.
Thus, [Ag+] = [Cl-] = s M.
Plugging these values into the equilibrium expression, we get:
Ksp = (s)(s) = s^2
Since the concentration of NaCl does not affect the Ag+ or Cl- concentration, the concentration of NaCl is not involved in calculating the solubility of AgCl.
Now, we can solve for 's' using the given Ksp value:
1.6 × 10^-10 = (s)^2
Taking the square root of both sides, we get:
s = √(1.6 × 10^-10)
Evaluating this, we find:
s ≈ 1.3 × 10^-5 M
Therefore, the solubility of AgCl in a 0.1 M NaCl solution would be approximately 1.3 × 10^-5 M.