Question

The solubility of AgBr (Ksp =5.0 x 10^-13 ) in 0.230 Molar CaBR₂ is__________ (a number) Please show all work

Answer 1

CaBr2 here is Strong electrolyte

It will dissociate completely to give Br-

[Br-] = 2*[Ca(Br)2] = 2*0.230 = 0.460 M

At equilibrium:

AgBr <----> Ag+ + Br-

   s 0.46 + s

Ksp = [Ag+][Br-]

5*10^-13=(s)*(0.46+ s)

Since Ksp is small, s can be ignored as compared to 0.46

Above expression thus becomes:

5*10^-13=(s)*(0.46)

5*10^-13= (s) * 0.46

s = 1.09*10^-12 M

Answer: 1.09*10^-12 M