# Question #75478

May 31, 2017

$1.3 \cdot {10}^{- 5} \text{moles}$

#### Explanation:

The problem essentially wants you to determine the molar solubility of silver chloride, i.e. the maximum number of moles of silver chloride that can be dissolved in $\text{1 L}$ of solution at a given temperature.

So, you know that silver chloride is insoluble in water. This tells you that when you dissolve this salt in water, most of it will remain undissociated.

The equilibrium that is established between the undissolved salt and the dissolved ions looks like this

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

The solubility product constant, ${K}_{s p}$, is calculated by taking the equilibrium concentrations of the dissolved ions in a saturated solution of silver chloride.

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

Now, notice that when $1$ mole of silver chloride dissociates in aqueous solution, it produces $1$ mole of silver(I) cations and $1$ mole of chloride anions.

So you can say that, at equilibrium, the solution has

$\left[{\text{Ag"^(+)] = ["Cl}}^{-}\right]$

If you take $s$ to be the equilibrium concentrations of these ions, i.e. the molar solubility of the salt, you can say that

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

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

Rearrange to solve for $s$

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

Plug in your value to find

${K}_{s p} = \sqrt{1.8 \cdot {10}^{- 10}} = 1.3 \cdot {10}^{- 5}$

Since this number represents the number of moles of silver chloride that can be dissolved in $\text{1 L}$ of solution at the temperature at which ${K}_{p s} = 1.8 \cdot {10}^{- 10}$, you will have

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{no. of moles AgCl" = 1.3 * 10^(-5)color(white)(.)"moles}}}}$