In the laboratory you dissolve 19.2 g of chromium(II) sulfate in a volumetric flask and add water to a total volume of 375 mL.

Question

What is the molarity of the solution?  

What is the concentration of the chromium(II) cation?  

What is the concentration of the sulfate anion?  

Answer 1

The first thing we need to do is determine the amount of chromium (II) sulfate present in the solution in moles. To do that, we will need the molar mass of the chromium (II) sulfate, and to determine the molar mass, we need to know the chemical formula. Sulfate ions have a -2 charge, and the "(II)" in "chromium (II)" means that the chromium has a +2 charge. Therefore, a neutral compound will have equal amounts of chromium and sulfate, and so we should expect a chemical formula of CrSO₄. From there, we can calculate the molar mass using the molar masses of the individual elements and the number of atoms in the compound:

Cr => 51.996 g/mol

S => 32.065 g/mol

O => 15.999 g/mol

Molar Mass = 1 * (51.996 g/mol) + 1 * (32.065 g/mol) + 4 * (15.999 g/mol) = 148.06 g/mol

Now, to calculate the moles in solution, we take the mass added to the solution and divide by the molar mass:

Moles = 19.2 g / 148.06 g/mol = 0.12968 mols.

To find the molarity, we divide the number of moles by the volume of the solution in liters. To get from milliliters to liters, we divide the number of milliliters by 1000, so 375 mL is 0.375 L.

Molarity = 0.12968 mols / 0.375 L = 0.34581 M. We should report this to the number of significant figures given in the question, which is 3, and so our answer is 0.346 M.

Now, if we want to know the concentration of the individual ions, we need to know how many of each ion is in the undissolved compound. We already figured this out earlier when we determined the chemical formula, and so we can see that one chromium ion and one sulfate ion form the chromium (II) sulfate compound. Therefore, when the compound dissolves, we would expect one chromium ion and one sulfate ion for every chromium (II) sulfate molecule. If a dozen chromium (II) sulfate molecules dissolved, we would expect a dozen chromium ions and a dozen sulfate ions to be in solution. If, per se, 0.12968 moles of chromium sulfate molecules dissolved, we would expect 0.12968 moles of chromium ions and 0.12968 moles of sulfate ions to be in solution. And so, by dividing by the solution volume again, we can see that the concentrations of both the chromium cation and the sulfate anion are the same as that of the overall solution, or 0.346 M.

Somewhat Interesting Side-Note:

The problem states that the compound is dissolved and then water is added until the volume of the solution is 375 mL, but I was curious about the solubility of the compound in water. After googling the compound, I found that the pentahydrate was soluble in water, and that chromium (III) sulfate was insoluble, but couldn't seem to find much on the anhydrous form of the compound. Regardless, if you are ever doing one of these problems and you are dealing with a hydrate, you would take that into account during the molar mass calculation. In this case, if the problem stated that we were dealing with the pentahydrate, we would have to add five times the molar mass of water to our result for the anhydrous molar mass to get the total molar mass of the hydrate (148.06 g/mol + 18.02 g/mol * 5 = 238.16 g/mol for the hydrate). We would use this molar mass just like we did above to calculate the moles of the compound in solution.