Question

Excess NA₂SO₄ (aq) is added to a 4.47x 102 mL sample of industrial waste containing Ba²⁺ ions. If 15.5 g of BaSO₄ (s) are precipitated from the reaction, what was the molar concentration of Ba²⁺ in the original sample?

Ba²⁺ (aq)+Na₂SO₄ (aq) → BaSO(s)+2Na⁺ (ag)

Answer 1

To find the molar concentration of Ba²⁺ in the original sample, calculate the moles of BaSO₄ precipitated and equate them to moles of Ba²⁺. Then, divide by the volume of the sample in liters.

Step-by-step explanation:

The question asks for the determination of the molar concentration of Ba²⁺ ions in an industrial waste sample after the precipitation of BaSO4.

To solve this, use the mass of BaSO4 precipitated and the molar mass of BaSO4 (233.43 g/mol) to calculate the moles of BaSO4. Since the reaction between Ba²⁺ and SO₄²⁺ to form BaSO4 is a 1:1 stoichiometry, the moles of Ba²⁺ originally present in the sample are equal to the moles of BaSO4 precipitated. Then, convert the volume of the sample from mL to L and use it along with the moles of Ba²⁺ to compute the molar concentration.

Here's the calculation:

• Calculate moles of BaSO4: 15.5 g BaSO4 × (1 mol BaSO4 / 233.43 g BaSO4) = 0.0664 mol BaSO4
• Since each mole of BaSO4 comes from one mole of Ba²⁺, there were 0.0664 mol of Ba²⁺ present in the original sample.
• Molar concentration of Ba²⁺: 0.0664 mol Ba²⁺ / 0.447 L (conversion of 447 mL to L) = 0.1485 M Ba²⁺