# Question 9e5cc

Aug 3, 2017

${K}_{s p} = 4.2 \cdot {10}^{- 13}$

#### Explanation:

You know that when silver carbonate, a salt that is considered insoluble in water, is dissolved in water, the following dissociation equilibrium is established in aqueous solution.

${\text{Ag"_ 2"CO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CO}}_{3 \left(a q\right)}^{2 -}$

You also know that the molar solubility of silver carbonate, $s$, in water at ${25}^{\circ} \text{C}$ is equal to "4.7 * 10^(-5) $\text{M}$.

This tells you that a saturated solution of silver carbonate will contain $4.7 \cdot {10}^{- 5}$ moles of dissociated silver carbonate for every $\text{1 L}$ of solution.

In other words, you have

["Ag"_ 2"CO"_ 3]_ "that dissociates" = s

In your case, you know that

["Ag"_ 2"CO"_ 3]_ "that dissociates" = s = 4.7 * 10^(-5) $\text{M}$

Now, notice that every mole of silver carbonate that dissociates in aqueous solution produces $\textcolor{red}{2}$ moles of silver(I) cations and $1$ mole of carbonate anions.

This means that the saturated solution of silver carbonate will contain

$\left[{\text{Ag}}^{+}\right] = \textcolor{red}{2} \times s$

$\left[{\text{CO}}_{3}^{2 -}\right] = s$

You can thus say that at ${25}^{\circ} \text{C}$, the saturated solution will contain

["Ag"^(+)] = color(red)(2) xx 4.7 * 10^(-5)color(white)(.)"M" = 9.4 * 10^(-5)color(white)(.)"M"

["CO"_3^(2-)] = 4.7 * 10^(-5)color(white)(.)"M"#

By definition, the solubility product constant for this dissociation equilibrium is equal to

${K}_{s p} = \left[{\text{Ag"^(+)]^color(red)(2) * ["CO}}_{3}^{2 -}\right]$

In your case, you will have--I'll leave the value without added units

${K}_{s p} = {\left(9.4 \cdot {10}^{- 5}\right)}^{\textcolor{red}{2}} \cdot \left(4.7 \cdot {10}^{- 5}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{s p} = 4.2 \cdot {10}^{- 13}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the molar solubility of the salt.