Answer to Question #344018 in General Chemistry for cath

Question

What is the concentration of Br^- in a solution prepared by mixing 0.500 L of 0.0450 M KBr and 0.250 L of 0.0250 M CaBr₂?

Answer 1

Solution:

Moles of solute = Molarity × Liters of solution

Therefore,

Moles of KBr = (0.0450 M) × (0.500 L) = 0.0225 mol

Moles of CaBr₂ = (0.0250 M) × (0.250 L) = 0.00625 mol

KBr and CaBr₂ are strong electrolytes that completely dissociate in solution.

KBr(aq) → K⁺(aq) + Br⁻(aq)

According to stoichiometry: Moles of Br⁻ = Moles of KBr = 0.0225 mol

CaBr₂(aq) → Ca²⁺(aq) + 2Br⁻(aq)

According to stoichiometry: Moles of Br⁻ = 2 × Moles of CaBr₂ = 2 × 0.00625 mol = 0.0125 mol

Total number of Br⁻ moles = (0.0225 mol + 0.0125 mol) = 0.0350 mol

Total liters of solution = (0.500 L + 0.250 L) = 0.750 L

Thus,

Concentration of Br⁻ = (0.035 mol / 0.750 L) = 0.0467 M

Answer: The concentration of Br⁻ in a solution is 0.0467 M