Molarity: Ion Concentration

Question

In the laboratory you dissolve 23.8 g of chromium(III) sulfate in a volumetric flask and add water to a total volume of 250 mL.

What is the molarity of the solution?  M.

What is the concentration of the chromium(III) cation?  M.

What is the concentration of the sulfate anion?  M.

Answer 1

Chromium (III) Sulfate (FW 392.16 g/mol) is insoluble in water. (So, the answers are zero)

Chromium (III) Sulfate Hydrate (FW 401.2 g/mol) is soluble in water, so in an actual lab, that is what one would have to use to get a non-trivial answer.

(28.3 g Chromium (III) Sulfate)(1 mole)/(401.2 g/mol) = 0.0705 mol

(0.0705 mol)/(0.25 liter) = 0.282 mol/liter chromium (III) sulfate

There are 2 chromium ions per formula unit, so

[Cr^+3] = (2)(0.282 mol/liter) = 0.564 mol/liter

There are 3 sulfates per formula unit, so

[SO4^-2] = (3)(0.282 mol/liter) = 0.846 mol/liter