Question

A solution is prepared by mixing 49.0 mL49.0 mL of a 0.185 M0.185 M NaClNaCl solution with 88.0 mL88.0 mL of a 0.278 M0.278 M CaCl2CaCl2 solution. Determine the volume in milliliters of a 0.1313 M0.1313 M AgNO3AgNO3 solution needed to precipitate all of the Cl−Cl− as AgClAgCl.

volume AgNO3AgNO3 =?

Answer 1

We need to determine the total no.of moles of Cl^- in the mixture.

No.of moles of Cl^- from NaCl = (0.185 M) x (49 x 10^-3 L) = 0.009065 mole

No.of moles of Cl^- from CaCl₂= 2 x (0.278 M) x (88 x 10^-3 L) = 0.024464 mole

Hence, Total no.of moles of Cl^- = 0.009065 mole + 0.024464 mole = 0.033529 mole

Required reaction:

AgNO₃ + Cl^- -----------> AgCl + NO₃^-

Hence, 1 mole of AgNO₃ reacts with 1 mole of Cl^-

Therefore, No.of moles of AgNO₃ required = 0.033529 mole

Therefore, Volume of AgNO₃ solution required = (0.033529 mol) / (0.1313 mol/L) = 0.25536 L = 255.36 mL

• Volume of AgNO₃ solution required = 255.36 mL