Question

Assuming your unknown has a Cl- content of 65%. Using 0.2M AgNO3 to furnish the Ag+, calculate the number of mL of 0.2M AgNO3 needed for the precipitation.

Please explain. Thank you!

Answer 1

Answer – Given, unknown has a Cl- content = 65%

[AgNO₃] =0.2 M

We assume the unknown solution is HCl and its molarity 1.0 M and 100 mL

So the Cl^- is 65 %

Moles of HCl = 1.0 M *0.100 L = 0.100 moles

So, moles of Cl^- = 0.100 moles *65 % /100 %

                            = 0.065 moles of Cl^-

So, [Cl^-] = 0.065 moles

So, for the precipitate of AgCl the mass of Cl^- precipitate of AgCl = mass of Cl^- in the unknown

So, mass of Cl^- = 0.065 moles * 35.453 g/mol

                          = 2.30 g

Mass of Cl^- in the AgCl precipitate = 2.30 g

So, moles of Cl^- = moles of AgCl

So, moles of AgNO₃ = 0.065 moles

So, volume of AgNO₃ = 0.065 moles /0.20 M

                                     = 0.325 L

                                     = 325 mL

So, the number of 325 mL of 0.2M AgNO3 needed for the precipitation