1. (6 points) Chloride ion is determined by a potentiometric titration using 0.2
ID: 1030804 • Letter: 1
Question
1. (6 points) Chloride ion is determined by a potentiometric titration using 0.200 M AgNO, as t titrant. AgClI (s) is formed during the titration by the following reaction: Ag, (aq) + cr (aq) AgCl (s) A 1:10 dilution of an unknown water sample is made and a 25 mL aliquot of the diluted samples is placed into a 100-mL beaker. Two electrodes (Ag wire and Ag/AgCl reference electrode) are immersed, the voltage reading between the electrodes is allowed to stabilize, and the initial voltage recorded. Nineteen titration aliquots of 0.100 mL of 0.200 M AgNO, are made with an automatic pipettor. After each aliquot, the solution is stirred, the voltage reading stabilizes, and the voltage in m V recorded. The following graph of Potential (m V) versus volume of AgNO, titrant is made. Potentiometric titration of chloride 380.0 360.0 340.0 320.0 300.0 S 280.0 260.0 240.0 E 220.0 200.0 5 180.0 160.0 140.0 120.0 100.0 80.0 8 6 8 8 8 8 8 Volume of AgNO, (mL) a) The endpoint of this titration is the midpoint of the steepest vertical rise in potential. Estimate the volume of titrant added to reach the endpoint. b) How many moles of AgNO, have been added at the endpoint? How many moles of CI have reacted at the endpoint?Explanation / Answer
Potentiometric titration of chloride ion with AgNO3 solution
From the graph shown above,
a)
The end point volume of titrant (AgNO3) added = 1.30 ml
b)
mols of AgNO3 added = mols of Cl- present in diluted sample
= 0.200 M x 0.0013 L
= 0.00026 mol
c)
molar concentration of Cl- in diluted sample = mols/L of solution
= 0.00026 mol/0.025 L
= 0.0104 M
d)
molar concentration of Cl- in original solution = 0.0104 M x 10
= 0.104 M
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