A 4.534 g sample of a solid mixture was analyzed for barium ion. First the rock was crushed in a mortar and pestle. 100% of the sample was transferred to a beaker, where it was dissolved in 150.0 mL of d. i. water. to give 152 mL of solution. The aqueous sample was then analyzed by adding a small excess of sulfuric acid. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, rinsed, dried and weighed. If 0.5376 g of barium sulfate was obtained, what was the mass percent of barium in the original sample? What was the molar concentration barium in the 152 mL of solution prior to adding sulfuric acid? Show work clearly; include a balanced net ionic chemical equation, units and sig. figs.

6.976%

0.0152 M

Explanation:

Step 1: Write the balanced chemical equation

Ba²⁺(aq) + H₂SO₄(aq) ⇒ BaSO₄(s) + 2 H⁺(aq)

The balanced net ionic chemical equation is:

Ba²⁺(aq) + SO₄²⁻(aq) ⇒ BaSO₄(s)

Step 2: Calculate the moles corresponding to 0.5376 g of BaSO₄

The molar mass of BaSO₄ is 233.39 g/mol.

0.5376 g × (1 mol/233.39 g) = 2.303 × 10⁻³ mol

Step 3: Calculate the moles of Ba²⁺ that produced 2.303 × 10⁻³ moles of BaSO₄

The molar ratio of Ba²⁺ to BaSO₄ is 1:1. The moles of Ba²⁺ are 1/1 × 2.303 × 10⁻³ mol = 2.303 × 10⁻³ mol.

Step 4: Calculate the mass corresponding to 2.303 × 10⁻³ moles of Ba²⁺

The molar mass of Ba²⁺ is 137.33 g/mol.

2.303 × 10⁻³ mol × 137.33 g/mol = 0.3163 g

Step 5: Calculate the mass percent of barium in the original sample

0.3163 g of barium ion were in a 4.534 g-sample. The mass percent of barium ion is:

0.3163 g/4.534 g × 100% = 6.976%

Step 6: Calculate the molar concentration barium in the 152 mL of solution

2.303 × 10⁻³ moles of barium ion were in 152 mL (0.152 L) of solution. The molarity of barium ion is:

M = 2.303 × 10⁻³ mol/0.152 L = 0.0152 M