# Calculate the concentration of silver ions at equilibrium when "0.5 g" of "AgCl" is placed in "100 mL" of water?

## Ksp = $1.8 \cdot {10}^{-} 10$

Jan 3, 2018

Here's what I got.

#### Explanation:

The trick here is to realize that because silver chloride has a solubility of about $520$ $\mu \text{g}$ per $\text{100 g}$ of water at ${50}^{\circ} \text{C}$, there's no way you could actually dissolve $\text{0.5 g}$ in $\text{100 mL}$ of solution $\to$ see here.

This implies that a lot of the silver chloride that you're adding will remain undissolved, meaning that it will not affect the equilibrium concentrations of the silver(I) cations and of the chloride anions.

So, when silver chloride is placed in water, the following equilibrium is established.

${\text{AgCl"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

Notice that for every $1$ mole of silver chloride that dissociates, the reaction produces $1$ mole of silver(I) cations and $1$ mole of chloride anions.

By definition, the solubility product constant for this equilibrium, ${K}_{s p}$, is equal to

${K}_{s p} = \left[{\text{Ag"^(+)] * ["Cl}}^{-}\right]$

If you take $s$ to be the molar solubility of the salt in water at the specific temperature for which

${K}_{s p} = 1.8 \cdot {10}^{- 10}$

you can say that you have

${K}_{s p} = s \cdot s$

${K}_{s p} = {s}^{2}$

This will get you

$s = \sqrt{{K}_{s p}}$

$s = \sqrt{1.8 \cdot {10}^{- 10}} = 1.34 \cdot {10}^{- 5}$

This means that at the temperature used here, a saturated solution of silver chloride will have

["Ag"^(+)] = 1.34 * 10^(-5)color(white)(.)"M"

["Cl"^(-)] = 1.34 * 10^(-5)color(white)(.)"M"

In other words, you can only hope to dissolve $1.34 \cdot {10}^{- 5}$ moles of silver chloride for every $\text{1 L}$ of solution at the given temperature.

Consequently, you can say that $\text{100 mL}$ of this solution, which is what you should expect to get when you add $\text{0.5 g}$ of silver chloride to $\text{100 mL}$ of water, will be able to dissolve

100 color(red)(cancel(color(black)("mL solution"))) * (1.34 * 10^(-5)color(white)(.)"moles AgCl")/(10^3color(red)(cancel(color(black)("mL solution")))) = 1.34 * 10^(-6)color(white)(.)"moles AgCl"

Use the molar mass of silver chloride to convert this to grams

1.34 * 10^(-6) color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(1color(red)(cancel(color(black)("mole AgCl")))) = 1.92 * 10^(-4)color(white)(.)"g"

Therefore, you can say that if you add $\text{0.5 g}$ of silver chloride to $\text{100 mL}$ of water, only

$1.92 \cdot {10}^{- 4} \textcolor{w h i t e}{.} \text{g" = 192color(white)(.)mu"g}$

of silver chloride will actually dissociate to produce silver(I) cations and chloride anions. The rest will remain undissolved.