200 mL of 0.5 M aqueous BaCl 2 solution is mixed with 300 mL of 0.2 M aqueous Na 2 SO 4 solution. A precipitate of BaSO 4 is formed. What is the final concentration of Cl − ions (in M ) in the

Question

200 mL of 0.5 M aqueous BaCl 2 ​ solution is mixed with 300 mL of 0.2 M aqueous Na 2 ​ SO 4 ​ solution. A precipitate of BaSO 4 ​ is formed. What is the final concentration of Cl − ions (in M ) in the resulting solution? (Report your answer by multiplying the molarity by 100 and rounding to the nearest integer).

TestHub · Accepted Answer

To find the final concentration of Cl⁻ ions, first calculate the initial moles of Cl⁻ from BaCl₂.

Moles of BaCl₂ = $0.200 L \times 0.5 M = 0.100 m o l$ .

Since BaCl₂ dissociates into Ba²⁺ and 2Cl⁻, moles of Cl⁻ = $2 \times 0.100 m o l = 0.200 m o l$ .

Cl⁻ ions are spectator ions and do not react.

The total volume = $200 m L + 300 m L = 500 m L = 0.500 L$ .

Final concentration of Cl⁻ = $\frac{0.200 m o l}{0.500 L} = 0.4 M$ .

Multiplying by 100 gives $0.4 \times 100 = 40$ .

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