What is the molar solubility of AgBr in 0.10 M AlBr3? Ksp(AgBr) = 5.0 × 10^-13
AlBr3 is completely soluble and ionizes 100% like this.
...................AlBr3 ==> Al^3+ +3Br^-
I..................0.1M.............0........0
C................-0.1M............0.1M...0.3M
E....................0................0.1M....0.3 M
..........AgBr ==> Ag^+ + Br^-
I..........solid..........0............0
C.........solid-x........x...........x
E..........solid...........x...........x
Ksp = (Ag^+)(Br^-)
Fill in Ksp. (Ag^+) = x from the AgBr
(Br^-) = x from AgBr and 0.3 M from AlBr3 or (x + 0.3)
Ksp = 5.0E-13 = (x)(x+0.3)
Solve that to find x = solubility AgBr in the 0.1M AlCl3.
Note: You will find this is a quadratic equation and although those are not difficult most students avoid quadratics. It turns out that x is a very small number so that X + 0.3 = very nearly 0.3. By making that assumption the equation is
5.0E-13 = (x)(0.3). Post your work if you get stuck.