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
(5 marks) A solution containing chloride ion was analyzed by the mL sample containing chloride ion...
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