Question

First, he standardized the AgNO₃ titrant with 0.1242 g dry NaCl dissolved in 25.0 ml DI water, and 26.25 ml of titrant was consumed. Then he used the standardized titrant to determine the concentration of chloride in the solution. 11.71 ml of AgNO₃solution was used to titrate 25.00 ml of solution

a) Calculate the concentration (M) of AgNO₃ titrant.

b) Determine the content of chloride in the unknown solution (M)

Answer 1

First of all, this kind of calculations must be done using normality instead molarity. This particular case doesn't have any problem, since the equivalent weight is equal to molecular mass for all chemical. So, we just need to use the formula C₁V₁=C₂V₂ (C: concentration, V: Volume) for both cases.

AgNO₃+NaClAgCl+NaNO3

V₁=26.25 ml, m₂=0.1242 g, V₂=25 ml

C₂=(0.1242 g NaCl*(1 mol/39.98 g NaCl))/(0.025 L)=0.12 M.

C₁=(0.12 M*0.025 L)/(0.02625) =0.12 M.

Answer a. [AgNO₃]=0.12 M.

Unknown solution:

C₁=0.12 M. V₁=0.01171 L. V₂=0.025

C₂=(0.12*0.01171)/0.025=0.056 M.