Question

Question #9e5cc

Answer 1

`K_(sp) = 4.2 * 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.

`"Ag"_ 2"CO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "CO"_ (3(aq))^(2-)`

You also know that the molar solubility of silver carbonate, `s`, in water at `25^@"C"` is equal to `"4.7 * 10^(-5)` `"M"`.

This tells you that a saturated solution of silver carbonate will contain `4.7 * 10^(-5)` moles of dissociated silver carbonate for every `"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)` `"M"`

Now, notice that every mole of silver carbonate that dissociates in aqueous solution produces `color(red)(2)` moles of silver(I) cations and `1` mole of carbonate anions.

This means that the saturated solution of silver carbonate will contain

`["Ag"^(+)] = color(red)(2) xx s`

`["CO"_3^(2-)] = s`

You can thus say that at `25^@"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_(sp) = ["Ag"^(+)]^color(red)(2) * ["CO"_3^(2-)]`

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

`K_(sp) = (9.4 * 10^(-5))^color(red)(2) * (4.7 * 10^(-5))`

`color(darkgreen)(ul(color(black)(K_(sp) = 4.2 * 10^(-13))))`

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