Question

What is the molar solubility of silver bromide in each of the following solutions?

a. pure water

b. 0.0550 M sodium bromide

c. 0.250 M silver nitrate

Answer 1

a) In pure water
The Ksp of AgBr =5.2*10^-13
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-

[Ag+] = x
[Br-] = x
Ksp = [Ag+][Br-] = (x) ( x )=5.2 x 10^-13
x^2=5.2 * 10^-13
molar solubility x = 7.21*10^-7 M

b. 0.0550 M sodium bromide
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-
But we have arleady 0.0550 M of Br- so
[Ag+] = x
[Br-] = x + 0.0550

Ksp = [Ag+][Br-] = (x) ( x+0.0550) = 5.2 x 10^-13

x = molar solubility =9.45*10^-12 M

c. 0.250 M silver nitrate
AgBr <-----> Ag+ + Br-
let x = moles/L of AgBr that dissolves : this will give us x mol/L of Ag+ and x mol/L of Br-
But we have arleady 0.0250 M of Ag+ so
[Ag+] = x + 0.250
[Br-] = x

Ksp = [Ag+][Br-] = (x+0.250) ( x) = 5.2*10^-13

x = molar solubility =2.08*10^-12 M