Take approximately 15 grams of cerium (IV) ammonium sulfate dehydrate, add 12.5

Question

Take approximately 15 grams of cerium (IV) ammonium sulfate dehydrate, add 12.5 mL of concentrated sulfuric acid and then carefully add 12.5 mL of distilled water. Stir the mixture for 30 minutes. Add 25 mL of distilled water and stir again for 10 minutes. Continue the addition of 25 mL portions of distilled water with intervening stirring until the solution reaches a volume of 125 mL. Allow the solution to cool to room temperature, filter if necessary and, then dilute to 250 mL
and store in a glass stoppered bottle. The resulting solution will be approximately 0.1 M in Ce (IV).
Weigh out the appropriate amount of primary standard ferrous ammonium sulfate. You should calculate the amount by assuming that you want your equivalence point to take about 40 mL of your
Ce (IV) solution.

How much primary standard should be weighed out?

Explanation / Answer

Ans. Molar mass of Cerium ammonium sulphate (NH4)4Ce(SO4)4 = 596.5234 g/ mol

Note that the given salt is dehydrate but not dihydrate.

# Moles of Cerium ammonium sulphate taken = Mass / Molar mass

                                                = 15.0 g / (596.5234 g/ mol)

                                                = 0.0251457 mol

Therefore, the final solution of 250.0 mL has 0.0251457 mol Ce(IV).

# Moles of Ce(IV) in 40.0 mL = Required volume x [Ce(IV)] in final solution

                                                = 40.0 mL x (0.0251457 mol / 250.0 mL)

                                                = 0.004023312 mol

# Reduction of Ce(IV) with Fe(II):

            Ce4+(aq) + Fe2+(aq) -------------> Ce3+(aq) + Fe3+(aq)

According to the stoichiometry of balanced reaction, 1 mol Ce(IV) is reduced by 1 mol Fe(II).

So,

            Moles of Fe(II) required = 0.004023312 mol = moles of Ce(IV) in 40.0 mL

# Molar mass of Ferrous ammonium sulfate, (NH4)2Fe(SO4)2 = 284.0512 g/ mol

1 mol of (NH4)2Fe(SO4)2 consists of 1 mol Fe(II).

So, to get 0.004023312 mol Fe(II) as required in above step, the required moles of the salt = 0.004023312 mol .

Therefore, required moles of (NH4)2Fe(SO4)2 = 0.004023312 mol

            Required mass of the salt = Required moles x Molar mass

                                                            = 0.004023312 mol x (284.0512 g/ mol)

                                                            = 1.143 g

Therefore, required mass of primary standard, (NH4)2Fe(SO4)2 = 1.143 g

# Note: The hydration state of primary standard is NOT specified. Hence, it’s assumed to be in dehydrated form.

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