How do you set up the equation to determine the concentrations of silver ions and bromide ions given the balanced equation AgBr Ag Br- the Ksp value of silver bromide at 25 degrees is 5x10-13?

Answer 1

[Note: The answer has been corrected as follows:]

The equilibrium reaction is: AgBr(solid) <<---> Ag+aq + Br-aq

A solution equilibrium constant, Ksp, is the product of the ion concentrations in water (aq):

Ksp = [Ag+aq][Br-aq] = 5x10-13

and since [Ag+aq] = [Br-aq] when only AgBr is dissolved (pure solution),

you can calculate (by taking square root of Ksp) that:

[Ag+aq] = [Br-aq] = SQRT(5x10-13) = 7.1*10-7 mol/L

By having such a small solution constant, you can hopefully see that there will be MUCH more AgBr(solid) than ions(aq), because silver bromide is rather insoluble.