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50 mL of 20.8% w/v BaCl2(aq) solution and 100 mL of 9.8% w/v H2SO4(aq) solution are mixed. What are the concentrations of the ions in the final solution? (Atomic mass of Ba = 137)

1. [Cl-] = 0.66 M
2. [H+] = 1.33 M
3. [Ba²+] is approximately 0 M
4. [SO4²-] = 0.33 M

Correct answer: [Cl-] = 0.66 M

Solution

BaCl2 molar mass = 137+71=208 g/mol. In 50 mL of 20.8% w/v: mass = 20.8*50/100 = 10.4 g => moles = 10.4/208 = 0.05 mol. H2SO4 molar mass = 98 g/mol. In 100 mL of 9.8% w/v: mass = 9.8*100/100 = 9.8 g => moles = 9.8/98 = 0.1 mol. Reaction: BaCl2 + H2SO4 -> BaSO4(s) + 2HCl. BaCl2 is limiting (0.05 mol vs 0.1 mol H2SO4). BaCl2 fully consumed: 0.05 mol BaCl2 reacts with 0.05 mol H2SO4. Remaining H2SO4 = 0.1 - 0.05 = 0.05 mol. Total volume = 150 mL = 0.15 L. Cl- produced: 0.05 mol BaCl2 gives 0.10 mol Cl-. [Cl-] = 0.10/0.15 = 0.667 M ≈ 0.66 M (correct). [H+] from HCl: 2*0.05=0.10 mol from reaction + from excess H2SO4: 2*0.05=0.10 mol. Total H+ = 0.20 mol. [H+]=0.20/0.15=1.33 M (also correct). [Ba²+] ≈ 0 (BaCl2 fully consumed, BaSO4 precipitates). [SO4²-] from excess H2SO4 = 0.05 mol; [SO4²-] = 0.05/0.15 = 0.33 M. So options A, B, C, D are ALL correct! This is a multi-correct question.

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