Question

In the laboratory you dissolve 12.2 g of iron(II) iodide in a volumetric flask and add water to a total volume of 500 mL.

What is the molarity of the solution?
M.

What is the concentration of the iron(II) cation?
M.

What is the concentration of the iodide anion?
M.

In the laboratory you dissolve 16.8 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?
M.

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

What is the concentration of the sulfate anion?

M.
You wish to make a 0.271 M hydrobromic acid solution from a stock solution of 12.0 M hydrobromic acid. How much concentrated acid must you add to obtain a total volume of 50.0 mL of the dilute solution?

mL

Answer 1

Question 1

1) The formula of iron(II) iodide is
`\text{FeI}_(2)` and has a formula mass of 309.654 g/mol. This means that in 12.2 grams, there are

• 12.2/309.654 = 0.039398812868557 mol.

So, the molarity is (0.039398812868557)/(0.5) = 0.0788 M (to 3 sf)

2) In 0.039398812868557 mol of iron(II) iodide, there are 0.039398812868557 moles of iron(II) iodide cations, and thus the molarity is still 0.0788 M (to 3 sf)

3) In 0.039398812868557 mol of iron(II) iodide, there are

• 2(0.039398812868557) = 0.078797625737114 moles of iodide anions

Thus, the molarity is (0.078797625737114)/(0.5) = 0.158 M (to 3 sf)

Question 2

1) The formula of iron(III) sulfate is
`\text{Fe}_(2)\text{(SO}_(4))_(3)` and has a formula mass of 399.9 g/mol. This means that in 16.8 grams, there are

• 16.8/399.9 = 0.042010502625656 mol

So, the molarity is (0.042010502625656)/(0.25) = 0.168 M (to 3 sf)

2) In 0.042010502625656 moles of iron(III) sulfate, there are 2(0.042010502625656)=0.084021005251312 moles of iron(III) cations, so the molarity is:

• 0.084021005251312/0.25 = 0.336 M (to 3 sf)

3) In 0.042010502625656 moles of iron(III) sulfate, there are 3(0.042010502625656)=0.12603150787697 moles of sulfate anions, so the molarity is:

• 0.12603150787697/0.25 = 0.504 M (to 3 sf)

Question 3

Since we need 50.0 mL of solution at 0.271 M,

• 0.271 = (moles of HBr)/(0.05)
• moles of HBr = 0.01355 mol

So, we need this from 12.0 M HBr so, meaning that:

• 12.0 = (0.01355)/(liters of 12.0 M HBr stock solution)
• liters of 12.0 M HBr stock solution = 0.0011291666666667 L = 1.13 M (to 3 sf)