# What is the molar solubility of silver sulfite in a 0.112 M silver nitrate solution?

## Question:

What is the molar solubility of silver sulfite in a 0.112 M silver nitrate solution?

## Molar Solubility:

The molar solubility of a solution is the number of moles of the solute that are soluble in a given solution. If we pass the molar solubility of a solution, then the solution will become saturated. Molar solubility is usually measured in moles per liter, which are the same units as that of molarity.

## Answer and Explanation:

Given:

- {eq}\displaystyle \rm M = 0.112\ M {/eq} is the concentration of the silver nitrate

Since silver nitrate is a completely soluble compound, we can use its concentration as its amount that is soluble. That is, we can write:

{eq}\displaystyle \rm \Big[Ag^+_{aq}\Big] = \Big[AgNO_3\Big] = 0.112\ M {/eq}

We can write out the reaction of silver sulfite and silver nitrate as:

{eq}\displaystyle \rm Ag_2 SO_{3(s)}\ \to\ 2Ag^+_{(aq)} + SO_{3(aq)}^{-2} {/eq}

The solubility product constant for silver sulfite is:

- {eq}\displaystyle \rm K_{sp} = 1.5\ \times\ 10^{-14} {/eq}

So by using the equation:

{eq}\displaystyle \rm K_{sp} = \Big[Ag^+_{(aq)}\Big]\Big[SO_{3(aq)}^{-2}\Big] {/eq}

We can re-write this as:

{eq}\displaystyle \rm K_{sp} = \Big[0.112\ M\Big]^2 S {/eq}

We isolate the molar solubility *S* here and substitute our constant:

{eq}\displaystyle \rm S = \frac{1.5\ \times\ 10^{-14}}{(0.112\ M)^2} {/eq}

We get:

{eq}\displaystyle \rm \boxed{\rm S = 1.196\ \times\ 10^{-12}\ M} {/eq}

#### Learn more about this topic:

from

Chapter 11 / Lesson 5Learn about solubility product constant. Understand the definition of Ksp, the Ksp formula, how to calculate Ksp, and how to find molar solubility from Ksp.