If 50.0 mL of 0.100 M sodium chloride is added to 25.0 mL of 0.100 M calcium chloride,what is the resulting chloride ion concentration?

Question

Quizplus · Accepted Answer

The total volume of the solution after mixing is 50.0 mL + 25.0 mL = 75.0 mL. Sodium chloride (NaCl) dissociates into Na+ and Cl- ions, and calcium chloride (CaCl2) dissociates into Ca2+ and 2Cl- ions. Therefore, 50.0 mL of 0.100 M NaCl provides 0.0050 moles of Cl- ions, and 25.0 mL of 0.100 M CaCl2 provides 0.0025 moles of Ca2+ and 0.0050 moles of Cl- ions (since each CaCl2 unit provides two Cl- ions). The total moles of Cl- ions is 0.0050 (from NaCl) + 0.0050 (from CaCl2) = 0.0100 moles. The concentration of Cl- ions in the final solution is then 0.0100 moles / 0.075 L = 0.133 M.

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