Chapter 10, Problem 15E

The hard water scum that forms a ring around the bathtub is an insoluble soap

${\text{Ca(C}}_{\text{18}}{\text{H}}_{\text{35}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}$. It is formed when a soluble soap, ${\text{NaC}}_{\text{18}}{\text{H}}_{\text{35}}{\text{O}}_{\text{2}}$, reacts with the calcium ion that is responsible for the hardness in water: ${\text{2 NaC}}_{\text{18}}{\text{H}}_{\text{35}}{\text{O}}_{\text{2}}+{\text{Ca}}^{\text{2+}}\to {\text{Ca(C}}_{\text{18}}{\text{H}}_{\text{35}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}+{\text{2Na}}^{\text{+}}$. How many milligrams of scum can form from $\text{616 mg}$ of ${\text{NaC}}_{\text{18}}{\text{H}}_{\text{35}}{\text{O}}_{\text{2}}$?

Hard water contains calcium ion, which reacts with the stearate ion in soap to form bathtub scum.