let x = solubility of AgCl in moles/L, then
AgCl(s) ==> Ag^+ + Cl^-
...x.........x......x
............CaCl2(aq) ==> Ca^2+ + 2Cl^-
initial......0.004..........0.......0
change......-0.004.........0.004..0.008
equil.........0............0.004..0.008
Ksp = (Ag^+)(Cl^-)
Now substitute from the ICE charts (both of them).
(Ag^+) = x
(Cl^-) = x from the AgCl and 0.008 from the CaCl2 for a total of x+0.008
Solve for x = solubility AgCl in moles/L.
Convert from 1 L soln to 100 mL soln.
Calculate the solubility of AgCl in the following solution:
100mL of .004 M CaCl2
1 answer
1 answer