Question

(5 marks) A solution containing chloride ion was analyzed by the mL sample containing chloride ion was treated with 25.00 mL of 0.2500 M silver nitrate. The precipitated AgCl was removed from the solution by filtration, and 1.00 mL of 0.1000 M Fe3 was added to the filtrate in order to titrate excess Ag with SCN. A volume of 10.67 mL of 0.2380 M KSCN was required for solution to turn red colour. What is the concentration of chloride ion in the original solution? 2. Volhard method. A 50.00

Answer 1

This solution was titrated with 10.67 mL of 0.2380 M KSCN

Moles of KSCN = molarity * volume

                        = 0.2380 M * 0.01067 L

                        = 0.00254 moles

The solution contains Ag+ and Fe3+ ions. As soon as all the Ag+ gets consumed, Fe3+ reacts with SCN- to give a red color end point.

Ag+ + SCN- -----> AgSCN

So moles of Ag+ in solution (filtrate) = 0.00254 moles

When the chloride solution is treated with silver nitrate, moles of chloride reacted is equal to moles of silver nitrate added.

Moles of AgNO3 added = 0.2500 M * 0.025 L

                                     = 0.00625 moles

So moles of chloride which were present in the original solution

= moles of Ag+ added - moles of Ag+ (excess) in filtrate

= 0.00625 - 0.00254

= 0.00371 moles

Volume = 50.00 mL

So concentration of chloride in original solution = 0.00371 moles / 0.050 L

                                                                              = 0.074 M