The amount of chloride in an unknown sample can be determined by potentiometric titration, which use — Electrochemistry / Analytical Titration Chemistry Question
Problem Context
The amount of chloride in an unknown sample can be determined by potentiometric titration, which uses an electrochemical cell shown schematically below:
During the titration, when some of the 0.200 M AgNO3 solution has been added to the analyte, which half-cell (A or B) contains the anode of the electrochemical cell? Explain your reasoning.
Model Answer
The concentration of Ag+ ion will always be smaller in half-cell B than it is in half-cell A (it cannot exceed 0.2 M!). Oxidation of the Ag electrode will take place in the half-cell with lower Ag+ concentration, so the anode is in half-cell B.
A saline solution, consisting of NaCl dissolved in 5% dextrose solution, is analyzed using this technique. 100.0 g of the saline solution is placed in half-cell B and the voltage from the voltmeter is recorded as a function of the volume of added 0.200 M AgNO3:
Calculate the mass percentage of NaCl in the saline solution.
Model Answer
The endpoint of the titration takes place at 18.8 mL added AgNO3, so
mol Cl- = (0.0188 L)(0.200 mol L-1) = 3.76 x 10-3 mol
(3.76 x 10-3 mol NaCl)(58.44 g mol-1) = 0.220 g NaCl
(0.220 g NaCl)/(100.0 g saline) x 100% = 0.22% saline solution
Calculate the concentration of free silver ion in half-cell B when 10.00 mL of titrant has been added.
Model Answer
At 10.00 mL added titrant, the absolute value of the cell potential is 0.470 V. From the Nernst equation:
E = Eº - (0.0591/n)log([Ag+]sample/[Ag+]ref)
0.470 V = 0 V - 0.0591 * log([Ag+]sample/[1])
[Ag+] = 1.12 x 10-8 M
Calculate the Ksp of AgCl(s).
Model Answer
At the point where 10.00 mL AgNO3 has been added, essentially all the added silver ion (0.01000 L x 0.200 mol L-1 = 2.00 x 10-3 mol) has reacted to form AgCl(s) (since, according to part c, only about 10-9 mol Ag+ is still present in solution). Thus the remaining chloride ion concentration is
[Cl-] = (3.76 x 10-3 mol - 2.00 x 10-3 mol)/(0.100 L + 0.010 L) = 0.016 M
Since at this point there is some solid AgCl present, then
Ksp = [Ag+][Cl-] = [1.12 x 10-8][0.016] = 1.8 x 10-10
Of course, there is nothing special about the 10.0-mL point in the titration; any point where the [Ag+] can be determined (from the cell potential) and the [Cl-] can be determined (from the un-precipitated amount) will work. But 10.0 mL is convenient since we already figured out [Ag+] in part c.