Question

. The concentration of chloride ions can be determined indirectly using atomic absorption spectroscopy. An excess of silver ions is added to each standard and sample and a silver chloride precipitate forms. The samples are centrifuged and the supernatant is aspirated into the instrument. A silver hollow cathode lamp is used as the source. For this analysis, 5 mL of 0.1 M AgNO3 was added to each standard and sample. a. Use the following data to determine the concentration of chloride ions in a water sample. chloride, ppm Absorbance 0 0.728 250 0.554 500 0.386 750 0.214 water sample 0.327 b. Explain the behavior of the calibration curve.

Answer 1

Ksp of AgCl = 1.8 x 10^-10

[Ag+] [Cl-] = 1.8 x 10^-10

Initial mMoles of Ag+ = 5 mL * 0.1 M = 0.5

Since Ag+ is in excess,

[Ag+] after precipitation on 250 mL addition = 0.5 / (5 + 250)

= 0.00196

[Cl-] = Ksp / [Ag+] = 9.18 x 10^-8 M

Similarly for other volumes, concentrations can be obtained as shown below:

Volume, mL	Absorbance	[Ag+]	[Cl-]
0	0.728	0.1	1.80E-09
250	0.554	0.001961	9.18E-08
500	0.386	0.00099	1.82E-07
750	0.214	0.000662	2.72E-07

a)

A = 241.59 [Ag+] + 0.0937

0.327 = 241.59 [Ag+] + 0.0937

[Ag+] = 0.000966 M

[Cl-] = Ksp / [Ag+]

= 1.86 x 10^-7 M

b)

On addition of more water, Ag+ concentration will decrease because 1) more Ag+ ions are utilized for precipitation and 2) dilution effect. As the [Ag+] decreases the absorption coefficient also decreases.