molar solubility?

Question

Calculate the molar solubility of AgBr in pure water and in 0.12 F KBr. Explain the reason for the difference between these values.

i know the ksp for AgBr is 5.0 x 10^-11.
the answer is x = 4.2 x 10^-12

can someone teach me how come that's the answer.

Answer 1

Let's assume that the molar solubility of AgBr in pure water is x.
Then

.. .. .. .. .. .. .. AgBr <=> Ag+ + Br-
Dissolve .. .. .. x
Produce .. .. .. .. .. .. .. .. x .. .. ..x
At Equilibrium .. .. .. .. .. x .. .. .. x

Ksp=[Ag+][Br-]=x^2 => x= squareroot(Ksp) =>
x=SQRT(5*10^-11) =7.07 *10^-6 M

KBr is highly soluble so 0.12 M KBr will give 0.12 M Br-.
According to Le Chatelier's principle, since we are adding products, the equilibrium will shift to the left, thus decreasing the solubility of AgBr. If the solubility under these conditions is y, then

.. .. .. .. .. .. .. AgBr <=> Ag+ + Br-
Initial .. .. .. .. .. .. .. .. .. .. .. .. .. 0.12
Dissolve .. .. .. y
Produce .. .. .. .. .. .. .. .. y .. .. ..y
At Equilibrium .. .. .. .. .. y .. .. .. 0.12+y

Ksp= [Ag+][Br-] =y(0.12+y)
To get an exact solution you need to solve the quadratic.
We can make an approximation and simplify things.
Let's assume that 0.12>>y so that practically 0.12-y=0.12
Then the equation becomes
Ksp=y*0.12 => y=Ksp/0.12= (5*10^-11)/0.12 = 4.2*10^-10 M

you see that y is 8 orders of magnitude smaller than 0.12 so our assumption is valid. (If we had found a value like 0.01 then we would have had to solve the quadratic)

I think you have mis-typed the answer or one of the data (maybe the Ksp) since I find 4.2*10^-10 instead of 4.2*10^-12