Question

How do you determine the maximum amount of AgCl (Ksp= 1.8x10^-10) that can be dissolved in 250 mL of a 0.0100 M solution of NaCl?

Please provide the steps for solving the problem, thank you! :)

Answer 1

`"Solubility"_"AgCl"-=2.6*mug*L^-1`

Explanation:

We interrogate the solubility equilibrium...

`AgCl(s) rightleftharpoonsAg^+ +Cl^(-)`

For which `K_"sp"=[Ag^+][Cl^(-)]=1.8xx10^-10`

Now clearly, the chloride ion that derives from the added `NaCl`...and so...if we put the concentration of `AgCl(aq)=x`...

then....`x xx(0.0100+x)=1.8xx10^-10`...and now if `x` is small...then we can make the approx. that `0.0100+x~=0.01`

And so `x~=(1.8xx10^-10)/(0.0100)`

`x_1=1.8xx10^-8`...

`x_2=1.7xx10^-8`...

`x_3=1.7xx10^-8*mol*L^-1`...

And so ....................

`"Solubility"_"AgCl"-=1.7xx10^-8*mol*L^-1xx143.32*g*mol^-1=2.58xx10^-6*g*L^-1`...and thus in a `250*mL` volume..`6.4xx10^-7*g` silver chloride would dissolve.

This is example of salting out....in pure water....silver chloride would be 100 times as soluble....