What is the final concentration of chloride ions?

Question

A chemist dilutes 35.0 mL of a 0.811 M solution of calcium chloride into 515.0 mL of a 0.578 M solution of sodium chloride. What is the final concentration of chloride ions?

Answer 1

Let's find the moles of Cl- in each solution.

CaCl₂: 35.0 ml x 1 L/1000 ml x 0.811 mol CaCl₂/L x 2 mol Cl^-/mol CaCl₂ = 0.05677 mol Cl^-

NaCl: 515.0 ml x 1 L/1000 ml x 0.578 mol NaCl/L x 1 mol Cl^-/mol NaCl = 0.2977 mol Cl^-

Now we can add these to obtain total moles of Cl^-: 0.05677 + 0.2977 = 0.3545 mol Cl^- total

Next, we find the total volume and then divide moles by volume to obtain final concentration of Cl^-

Total volume = 35.0 ml + 515.0 ml = 550.0 ml = 0.5500 L

Final Cl- concentration = 0.3545 mol Cl- / 0.5500 L = 0.645 M Cl^- (to 3 sig. figs.)