# 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.

Feb 20, 2018

$4.63 \cdot {10}^{- 9}$ $\text{M}$

#### Explanation:

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

${K}_{s p} = \left[{\text{Ag"^(+)]^color(red)(2) * ["SO}}_{3}^{2 -}\right]$

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

${\text{Ag"_ 2"SO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "SO}}_{3 \left(a q\right)}^{2 -}$

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

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

${K}_{s p} = \left[\textcolor{red}{2} {\text{Ag"^(+)]^color(red)(2) * ["SO}}_{3}^{2 -}\right]$

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

${\text{Ag"_ 2"SO"_ (3(s)) rightleftharpoons color(red)(2)"Ag"_ ((aq))^(+) + "SO}}_{3 \left(a q\right)}^{2 -}$

Here $\textcolor{red}{2}$ is a coefficient!

${K}_{s p} = \left[{\text{Ag"^(+)]^color(red)(2) * ["SO}}_{3}^{2 -}\right]$

Here $\textcolor{red}{2}$ must be an exponent, so don't use it as a coefficient!