# 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.

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.

##### 1 Answer

#### Answer:

#### 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 **moles** of silver(I) cations means that the equilibrium concentration of the silver(I) cations must be raised to the power of

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 acoefficient!

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

#color(red)(2)# must be anexponent, so don't use it as a coefficient!