If 20.5 g of BaSO4(s) are precipitated from the reaction, what was the molar concentration of Ba2+ in the original sample?

Question

Excess Na₂SO₄(aq) is added to a 3.07×10² mL sample of industrial waste containing Ba²⁺ ions. If 20.5 g of BaSO₄(s) are precipitated from the reaction, what was the molar concentration of Ba²⁺ in the original sample?

Ba²⁺(aq) + Na₂SO₄ (aq) --> BaSO₄ (s) + 2Na⁺ (aq)

Answer 1

The limiting reagent in this problem will be the Ba²⁺, since the sodium sulfate was added in excess. Therefore, the amount of BaSO₄ you can make is dependent on the amount of Ba²⁺ in solution. Assuming all of the reagents are used up, you can calculate the amount of Ba²⁺ in the original sample by calculating the amount of Ba²⁺ in the product. Therefore, if there is 20.5g of BaSO₄, which has a MW of 233.38, you can calculate the moles for this and determine the moles for Ba, as it is a 1:1 ratio, to get an answer of approximately 0.08784mol. Once you have the moles for Ba, you can divide by the volume in liters (0.307L) to get molarity (0.286M), since M = mol/V.