# Question #a40f4

Jul 29, 2017

$7 \cdot {10}^{- 2}$ $\text{moles AgBr}$

#### Explanation:

The first thing that you need to do here is to figure out the molar solubility of silver bromide in water, i.e. the number of moles of silver bromide that can dissolve in $\text{1 L}$ of solution to produce a saturated solution.

So, you know that silver bromide is considered insoluble in water, so you can say that when this salt is dissolved in water, an equilibrium is established between the undissolved solid and the solvated ions.

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

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

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

Now, if you take $s$ to represent the concentration of silver bromide that dissociates to produce silver cations and bromide anions in the saturated solution, you can say that you have

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

which, in your case, is equal to

$5.0 \cdot {10}^{- 7} = {s}^{2}$

Solve for $s$ to find

$s = \sqrt{5.0 \cdot {10}^{- 7}} = 7.07 \cdot {10}^{- 4}$

This tells you that you can only hope to dissolve $7.07 \cdot {10}^{- 4}$ moles of silver bromide for every $\text{1 L}$ of solution at the given temperature, i.e. that sodium bromide has a molar solubility of $7.07 \cdot {10}^{- 4}$ ${\text{mol L}}^{- 1}$.

Consequently, you can say that $\text{100 L}$ of water, which you can easily approximate to be equal to the volume of the solution, of a saturated silver bromide solution, will hold

$100 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{L solution"))) * (7.07 * 10^(-4)color(white)(.)"moles AgBr")/(1color(red)(cancel(color(black)("L solution")))) = color(darkgreen)(ul(color(black)(7 * 10^(-2)color(white)(.)"moles AgBr}}}}$

The answer is rounded to one significant figure, the number of sig figs you have for the volume of water.