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Multiple Choice
What is the concentration (in M) of the chloride ion when 15.3 mL of a 0.629 M solution of barium chloride is combined with 18.9 mL of a 0.624 M solution of aluminum chloride? Assume the volumes are additive.
A
0.626 M
B
1.254 M
C
0.942 M
D
0.312 M
Verified step by step guidance
1
First, understand that barium chloride (BaCl₂) and aluminum chloride (AlCl₃) both dissociate in water to release chloride ions. BaCl₂ releases two chloride ions per formula unit, while AlCl₃ releases three chloride ions per formula unit.
Calculate the moles of chloride ions from the barium chloride solution. Use the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert 15.3 mL to liters by dividing by 1000, then multiply by the molarity (0.629 M) and the number of chloride ions per formula unit (2).
Calculate the moles of chloride ions from the aluminum chloride solution. Convert 18.9 mL to liters, multiply by the molarity (0.624 M), and then multiply by the number of chloride ions per formula unit (3).
Add the moles of chloride ions from both solutions to find the total moles of chloride ions present.
Calculate the total volume of the combined solution in liters by adding the volumes of the two solutions (15.3 mL + 18.9 mL) and converting to liters. Finally, find the concentration of chloride ions by dividing the total moles of chloride ions by the total volume in liters.