If 25ml of a solution of sodium chloride requires 17.9 ml of 0.100 mol/L silver nitrate for complete reaction what is the molarity of the chloride ion?

Answer 1

First, write a balanced equation for the reaction:

NaCl (aq) + AgNO₃ (aq) → AgCl (s) + NaNO₃ (aq)

We then determine how many moles of silver nitrate (AgNO₃) must be added for a complete reaction:

mol AgNO₃ = (17.9 mL)•(0.100 mol/L) = (0.0179 L)•(0.100 mol/L) = 0.00179 mol

The stoichiometry of our balanced equation indicates that NaCl reacts with AgNO₃ at a 1-to-1 molar ratio. In other words, mol AgNO₃ = mol NaCl = CV, where C is the molar concentration of NaCl and V is the volume of solution. Solving for C, we have C = (mol NaCl)/V = (0.00179 mol)/(0.025 L) = 0.0716 mol/L. Since [NaCl] = [Cl^–], the concentration of chloride ion in the solution is 0.0716 mol/L.