# Question #f9692

May 10, 2017

$\text{0.028 M}$

#### Explanation:

Lead(II) iodide is considered insoluble in water, which means that the equilibrium that is established when this salt is dissolved in water will lie significantly to the left, i.e. most of the salt will remain undissolved.

${\text{PbI"_ (color(red)(2)(s)) rightleftharpoons "Pb"_ ((aq))^(2+) + color(red)(2)"I}}_{\left(a q\right)}^{-}$

The solubility product constant, ${K}_{s p}$, for this solubility equilibrium is defined as

${K}_{s p} = {\left[{\text{Pb"^(2+)] * ["I}}^{-}\right]}^{\textcolor{red}{2}}$

Now, the molar solubility of the salt represents the maximum concentration of lead(II) iodide that dissociates in water to produce a saturated solution.

If you take $s$ to be the molar solubility, you can say that the saturated solution will contain $s$ lead(II) cations and $\textcolor{red}{2} \cdot s$ iodide anions.

The solubility product constant will thus be equal to

${K}_{s p} = s \cdot {\left(\textcolor{red}{2} \cdot s\right)}^{\textcolor{red}{2}}$

${K}_{s p} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt{{K}_{s p} / 4}$

Plug in your value to find

$s = \sqrt{\frac{8.8 \cdot {10}^{- 5}}{4}} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{0.028 M}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the ${K}_{s p}$ of the salt.

So, what does this mean?

This tells you that at room temperature, you can only dissolve $0.028$ moles of lead(II) iodide per $\text{1 L}$ of solution to produce a saturated solution.