Question

Determine the maximum amount of AgCl (Ksp= 1.8 x 10^-10) that can be dissolved in 250 ml of a 0.0100 M solution of NaCl?

Please include the steps to solve the equation, I am reviewing for an exam... need all the help I can get. Thank you!

Answer 1

Common Ion Effect problem - see below

Explanation:

So, AgCl comes apart according to:
`AgCl = Ag^+ + Cl^-1`
Your `Cl^-1` ion is already 0.01M. When you lose a small amount of AgCl as "-s", you gain a small amount of `Ag^+1` as "+s", and you gain a small amount more of `Cl^-1` ion as "0.01+s".

The Ksp expression becomes:

`K_sp = [s][0.01+s]`. We make the simplifying assumption that s<<0.01M (we should check after), so Ksp becomes:
`K_sp = [s][0.01]`
`1.8xx10^-10 = [s][0.01]`

`s = 1.8xx10^-8 M Ag^+` ions. This all took place in 0.25L, so the number of moles of `Ag^+` ions present is:
`(1.8xx10^-8 M)xx0.25L` = `4.5xx10^-9 "moles" Ag^+` ions

`Ag^+` and AgCl are 1:1 molar ratio, so this means
`4.5xx10^-9 "moles" AgCl` was lost (or dissolved).
If we multiply by the molar mass, we find that `6.45xx10^-7`g of AgCl dissolved.