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.

Answer 2

To determine the concentrations of silver ions (Ag+) and bromide ions (Br-) at equilibrium, you can use the formula for the solubility product constant (Ksp). For the reaction AgBr(s) ⇌ Ag+(aq) + Br-(aq), the Ksp expression would be [Ag+][Br-] = 5x10^-13. Given that each mole of AgBr dissociates into 1 mole of Ag+ and 1 mole of Br-, at equilibrium, the concentrations of Ag+ and Br- would be equal. So, you would consider them both as 'x' in the equation [x][x] = 5x10^-13 to solve for their concentrations.