Question

In the laboratory you dissolve 16.7 g of iron (III) sulfate in a volumetric flask and add water to a total volume of 250 mL.

What is the molarity of the solution?
What is the concentration of the iron(III) cation?
What is the concentration of the sulfate anion?

Answer 1

To calculate the molarity of the solution, we need to first calculate the number of moles of iron (III) sulfate present:

Number of moles = mass / molar mass

molar mass of Fe2(SO4)3 = 2(55.845) + 3(32.066) + 12(15.999) = 399.88 g/mol

Number of moles = 16.7 g / 399.88 g/mol = 0.0418 mol

Next, we need to calculate the volume of the solution in liters:

Volume = 250 mL = 0.250 L

Finally, we can calculate the molarity of the solution:

Molarity = Number of moles / Volume

Molarity = 0.0418 mol / 0.250 L = 0.167 M

To calculate the concentration of the iron(III) cation, we need to know the stoichiometry of the compound. In this case, the formula of iron (III) sulfate is Fe2(SO4)3, which means that there are 2 iron (III) cations for every 1 sulfate anion. Therefore, the concentration of the iron(III) cation is:

Concentration of Fe3+ = 2 x Molarity

Concentration of Fe3+ = 2 x 0.167 M = 0.334 M

Similarly, the concentration of the sulfate anion is:

Concentration of SO42- = Molarity

Concentration of SO42- = 0.167 M

Step-by-step explanation: