Question

What concentration of `"SO"_3^(2–)` is in equilibrium with `"Ag"_2"SO"_3(s)` and `1.80 * 10^-3` `"M"` `"Ag"^(+)` ? The `K_(sp)` of `"Ag"_2"SO"_3` is `1.50 * 10^–14`

One last question from me. I thought it would be Ksp=[2Ag^+]^2[SO3^2-], but plugging in 1.80×10^-3 for Ag and doing algebra doesn't seem to get me the correct answer when I put it on the online homework tool we use. Thanks for your help! I still can't seem to understand concentrations.

Answer 1

`4.63 * 10^(-9)` `"M"`

Explanation:

The expression of the solubility product constant for silver sulfite looks like this

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

This is the case because silver sulfite will partially ionize to produce silver(I) cations and sulfite anions in aqueous solution.

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

Notice that the fact that every mole of silver sulfite that dissociates in aqueous solution produce `color(red)(2)` moles of silver(I) cations means that the equilibrium concentration of the silver(I) cations must be raised to the power of `color(red)(2)`.

Mind you, the coefficients of the ions are used as expoents in the expression of the solubility product constant, not as coefficients.

So something like

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

is not correct because the coefficient must only be used as an exponent in the expression of the solubility proeduct constant.

In your case, you have--I'll do the calculations with the added units for the solubility product constant!

`1.50 * 10^(-14) quad "M"^3 = (1.80 * 10^(-3))^color(red)(2) quad "M"^color(red)(2) * ["SO"_3^(2-)]`

So you can say that the equilibrium concentration of the sulfite anions will be equal to

`[S"O"_3^(2-)] = (1.50 * 10^(-14) quad "M"^color(red)(cancel(color(black)(3))))/((1.80 * 10^(-3))^color(red)(2) quad color(red)(cancel(color(black)("M"^2)))) = color(darkgreen)(ul(color(black)(4.63 * 10^(-9) quad "M")))`

The answer is rounded to three sig figs.

So, a quick recap

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

Here `color(red)(2)` is a coefficient!

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

Here `color(red)(2)` must be an exponent, so don't use it as a coefficient!