# Question 32388

Sep 10, 2015

For part (b) $s = 1.2 \cdot {10}^{- 4} \text{M}$

#### Explanation:

I'll show you how to find the molar solubility of silver carboinate, ${\text{Ag"_2"CO}}_{3}$.

Silver carbonate is considered insoluble in aqueous solution, which means that it does not dissociate completely into silver cations, ${\text{Ag}}^{2 +}$, and carbonate anions, ${\text{CO}}_{3}^{2 -}$, when dissolved in water.

Actually, the fact that it doesn't dissociate completely is an understatement. When silver carbonate is placed in aqueous solution, an equilibrium reaction is established

${\text{Ag"_2"CO"_text(3(s]) rightleftharpoons 2"Ag"_text((aq])^(+) + "CO}}_{\textrm{3 \left(a q\right]}}^{2 -}$

The solubility product constant, ${K}_{s p}$, essentially tells you the extent of dissociation.

The smaller the value of ${K}_{s p}$, the fewer ions will dissociate in solution. Of course, this implies a lower solubility, since most of the solid will remain undissolved.

To find the molar solubility of silver carbonate, use an ICE table

${\text{Ag"_2"CO"_text(3(s]) " "rightleftharpoons" " color(red)(2)"Ag"_text((aq])^(+)" " + " " "CO}}_{\textrm{3 \left(a q\right]}}^{2 -}$

color(purple)("I")" " " " - " " " " " " " " " " " " 0 " " " " " " " " " " 0
color(purple)("C")" " " " - " " " " " " " " " " (+color(red)(2)s) " " " " " " "(+s)
color(purple)("E")" " " " - " " " " " " " " " " " " color(red)(2)s " " " " " " " " " " s#

By definition, ${K}_{s p}$ will be

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

${K}_{s p} = {\left(2 s\right)}^{2} \cdot s = 4 {s}^{2} \cdot s = 4 {s}^{3}$

This means that $s$, which is the molar solubility of silver chloride, is equal to

$s = \sqrt{{K}_{s p} / 4} = \sqrt{\frac{6.2 \cdot {10}^{- 12}}{4}} = \textcolor{g r e e n}{1.2 \cdot {10}^{- 4} \text{M}}$