# 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!

Apr 17, 2018

Common Ion Effect problem - see below

#### Explanation:

So, AgCl comes apart according to:
$A g C l = A {g}^{+} + C {l}^{-} 1$
Your $C {l}^{-} 1$ ion is already 0.01M. When you lose a small amount of AgCl as "-s", you gain a small amount of $A {g}^{+} 1$ as "+s", and you gain a small amount more of $C {l}^{-} 1$ ion as "0.01+s".

The Ksp expression becomes:

${K}_{s} p = \left[s\right] \left[0.01 + s\right]$. We make the simplifying assumption that s<<0.01M (we should check after), so Ksp becomes:
${K}_{s} p = \left[s\right] \left[0.01\right]$
$1.8 \times {10}^{-} 10 = \left[s\right] \left[0.01\right]$

$s = 1.8 \times {10}^{-} 8 M A {g}^{+}$ ions. This all took place in 0.25L, so the number of moles of $A {g}^{+}$ ions present is:
$\left(1.8 \times {10}^{-} 8 M\right) \times 0.25 L$ = $4.5 \times {10}^{-} 9 \text{moles} A {g}^{+}$ ions

$A {g}^{+}$ and AgCl are 1:1 molar ratio, so this means
$4.5 \times {10}^{-} 9 \text{moles} A g C l$ was lost (or dissolved).
If we multiply by the molar mass, we find that $6.45 \times {10}^{-} 7$g of AgCl dissolved.