Question

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, and an excess of aqueous BaCl2 is slowly added to the mixture resulting in the formation of a white precipitate.
1. Assuming that 0.433 g of precipitate is recovered calculate the percent by mass of SO4 (2-) in the unknown salt.

2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

Answer 1

Ans. #1. SO₄^2-(aq) + Ba²⁺(aq) ---------> BaSO₄(s)

Stoichiometry: 1 mol sulfate ion is precipitated by 1 mol barium ion.

It’s assumed that all SO₄^2- is precipitated in form of BaSO₄ (white precipitate).

Given,

            Mass of BaSO4 precipitated = 0.433 g

            Moles of BaSO4 precipitated = Mass / Molar mass

                                                            = 0.433 g / (233.3906 g/ mol)

                                                            = 0.0018533 mol

1 mol BaSO4 consists of 1 mol sulfate ion. So, number of moles of sulfate ion in the unknown sample must be equal to the moles of BaSO4 precipitated.

So,

            Moles of SO₄^2- in unknown sample = 0.0018533 mol

            Mass of SO₄^2- in unknown sample = 0.0018533 mol x (96.0636 g/mol)

                                                                        = 0.178223 g

Now,

            % SO₄^2- in salt sample = (Mass of sulfate / Mass of salt) x 100

                                                            = (0.178223 g / 0.323 g) x 100

                                                            = 55.17735 %

#2. Group 1 elements are alkali metals. Group 1 elements have a valency of 1. So, 2 alkali atoms combine with 1 sulfate ion to form alkali sulfate.

That is,

Alkali sulfate = X₂SO₄

Amount of alkali metal in salt sample = Mass of salt sample – Mass of sulfate in it

                                                            = 0.323 g – 0.178223 g

                                                            = 0.1447771 g

Moles of alkali metal in salt sample = 2 x moles of sulfate ion

                                                                        = 2 x 0.0018533 mol

                                                                        = 0.0037105 mol

Now,

Atomic mass of alkali metal = Mass of alkali metal in salt / Moles of alkali metal in salt

                                                = 0.1447771 g / 0.0037105 mol

                                                = 39.018 g/ mol

The calculated atomic mass of alkali metal atom is closest to potassium (K, atomic mass = 39.0983 g/mol).

Thus, the element is Potassium (K).

Or, the alkali cation is K⁺