......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.
The solubility of Agcl with solubility product 1.6 × 10^-10 in 0.1 M Nacl solution will be ?
1 answer
1 answer