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#

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!