What is the final concentration of chloride ions?

Question

A chemist dilutes 20.0 mL of 0.902 M solution of calcium chloride into 680.0 mL of a 0.328 M solution of sodium chloride. What is the final concentration of chloride ions?

Answer 1

Calcium chloride = CaCl₂

CaCl₂ in solution => Ca²⁺ + 2Cl^- (note moles of Cl^- is 2 x that of CaCl₂)

moles CaCl₂ = 20.0 ml x 1 L/1000 ml x 0.902 mol/L = 0.01804 moles CaCl₂

moles Cl^- = 2 x 0.01804 = 0.03608 moles Cl^-

Sodium chloride = NaCl

NaCl in soluito = Na⁺ + Cl^-

moles NaCl = 0.680 L x 0.328 mol/L = 0.2230 moles NaCl

moles Cl- = 0.2230 moles Cl^-

Total final volume = 20.0 ml + 680.0 ml = 700.0 ml = 0.700 L

Final concentration of Cl^- ions = 0.01804 mol Cl^- + 0.2230 mol Cl^- / 0.700 L = 0.345 M Cl^-