Question

Question #32388

Answer 1

For part (b) `s = 1.2 * 10^(-4)"M"`

Explanation:

I'll show you how to find the molar solubility of silver carboinate, `"Ag"_2"CO"_3`.

Silver carbonate is considered insoluble in aqueous solution, which means that it does not dissociate completely into silver cations, `"Ag"^(2+)`, and carbonate anions, `"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

`"Ag"_2"CO"_text(3(s]) rightleftharpoons 2"Ag"_text((aq])^(+) + "CO"_text(3(aq])^(2-)`

The solubility product constant, `K_(sp)`, essentially tells you the extent of dissociation.

The smaller the value of `K_(sp)`, 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

`"Ag"_2"CO"_text(3(s]) " "rightleftharpoons" " color(red)(2)"Ag"_text((aq])^(+)" " + " " "CO"_text(3(aq])^(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_(sp)` will be

`K_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]`

`K_(sp) = (2s)^2 * s = 4s^2 * s = 4s^3`

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

`s = root(3)(K_(sp)/4) = root(3)( (6.2 * 10^(-12))/4) = color(green)(1.2 * 10^(-4)"M")`