SOLUTION STOICHIOMETRY [URGENT QUESTION !]

Question

An unknown solution was prepared by dissolving 0.420 g of sodium chloride or calcium chloride in enough water to make 0.100 mL of solution. 15.0 mL of 0.500 mol/L silver nitrate was required to precipitate out all the chloride ions from this solution. Was the unknown solute calcium chloride or sodium chloride? Prove your answer using stoichiometry.

Answer 1

An unknown solution was prepared by dissolving 0.420 g of sodium chloride or calcium chloride in enough water to make 0.100 L of solution. 15.0 mL of 0.500 mol/L silver nitrate was required to precipitate out all the chloride ions from this solution. Was the unknown solute calcium chloride or sodium chloride? Prove your answer using stoichiometry.

What a Great Question! We first need to analyze what is happening at the microscopic level (the reaction).

Step 1: Calculate experimental moles of Cl^- ions from the given amount of silver nitrate

When ionic substances are added to water it dissociates into its ions:

NaCl(s) -> Na⁺(aq) + Cl^-(aq)

CaCl₂(s) -> Ca²⁺(aq) + 2 Cl^-(aq)

AgNO₃(aq) -> Ag⁺(aq) + NO₃^-(aq)

A precipitation reaction between the silver nitrate and the chloride ion (from NaCl and CaCl₂) occurs:

Ag⁺(aq) + Cl^-(aq) -> AgCl(s)

Ag⁺ and Cl^- react in a 1:1 molar ratio...this becomes important

Molarity = moles solute / L of solution OR Moles solute = Molarity times volume of solution

15.0 mL x 1 L x 0.500 Mol AgNO₃ x 1 mol Ag⁺

1000 mL 1 L 1 mol AgNO₃

= 0.00750 mol Ag⁺ = 0.00750 mol Cl^-

Step 2: Determine the moles of Cl^- from the unknown compound

Sodium Chloride (NaCl):

0.420 g NaCl x 1 mol NaCl x 1 mol Cl^-

58.44 g NaCl 1 mol NaCl) = 7.19 x 10^-3 mol Cl-

Calcium Chloride (CaCl₂):

0.420 g CaCl₂ x 1 mol CaCl₂ x 2 mol Cl^-

110.98 g CaCl₂ 1 mol CaCl₂ = 0.00757 mol Cl-

Experimentally determined Cl^- (0.00750 mol Cl-) matches closely with the theoretical value for calcium chloride (0.00757 mol Cl^-).