Indicate the concentration of each ion present in the solution formed by mixing the following: 3.50 g of NaCl in 55.4 mL of 0.577 M CaCl2 solution

Answer 1

NaCl(aq) + CaCl₂(aq) ==> NO REACTION

Na⁺(aq) + Cl^-(aq) + Ca²⁺(aq) + 2Cl^-(aq)

We will find moles of each ion and then divide by the liters of solution to get molar concentrations:

molar mass NaCl = 58.4 g / mol

mols NaCl = 3.50 g NaCl x 1 mol NaCl / 58.4 g = 0.05993 mols

mols Na⁺ = 0.05993 mols

mols Cl^- = 0.05993 mols

mols CaCl₂ = 55.4 mls x 1 L / 1000 mls x 0.577 mols / L = 0.03197 mols

mols Ca²⁺ = 0.03197 mols

mols Cl^- = 2 x 0.03197 mols = 0.06394 mols

Final volume of solution is 55.4 mls (assuming adding the 3.50 g NaCl didn't add to the volume).

Final volume of solution in liters = 0.0554 L

Final concentration of ions is as follows:

[Na⁺] = 0.05993 mols / 0.0554 L = 1.08 M

[Ca²⁺] = 0.03197 mols / 0.0554 L = 0.577 M

[Cl^-] = 0.05993 mols + 0.06394 mols / 0.0554 L = 2.24 M