ANALYTICAL CHEMISTRY TUTORIAL 3 Kusniar Deny Permana
[email protected] 01115639556
1 At the maximum absorption band, the transmittance is only 5.00%. What is the corresponding absorbance?
1 A 2 log %T A 2 log 5 A 0.699
2 a.
What value of absorbance corresponds to 45 % T?
b. If a 0.01 M solution exhibits 45 % T at some wavelength, what will be the % T for a 0.02 M solution of the same substance?
2a What value of absorbance corresponds to 45 % T?
A 2 log %T A 2 log 45
A 0.437
2b A bc b
A 2 log %T
A
log %T 2 A
c
Same substance & cuvet
b A1 c1
b
b
1 1 2 2
A2
A2
A2 c2 c2 A1 c1 0.02
log %T 2 0.694 log %T 1.306
%T 20.23%
0.347
0.01 A2 0.694
3
Chloroaniline in a sample is determined as the amine picrate. A 0.0265 g sample is reacted with picric acid and diluted to 1.00 L. The solution exhibits an absorbance of 0.368 in a 1.00 cm cell. What is the % chloroaniline in the sample? (Given: MW of Chloroaniline = 127.6 g/mol, ϵ= 1.25 x 104cm1mol-1L)
3 A bc A c b 0.368 c 1.25 x 104 x 1 c 2.944 x10 5 M
The reaction is 1:1:1
mol chloroanil ine mol AMINEPICRATE mol chloroanil ine 2.944 x10 5 mol / L x 1 L
mol chloroanil ine 2.944 x10 5 mol
mass chloroanil ine 2.944 x10 5 mol x MW mass chloroanil ine 2.944 x10 5 mol x 127 .6 g / mol mass chloroanil ine 3.8 x10 3 g
3.8 x10 3 g % chloroanil ine x100% 0.0265 % chloroanil ine 14.34%
4 a. A 3.96 x 10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1.00 cm cuvet; a blank solution containing only solvent has an absorbance of 0.0290 at the same wavelength. Find the molar absorbtivity of compound A. b. The absorbance of an unknown solution of compound A in the same solvent and cuvet was 0.375 at 238 nm. Find the concentration of A in the unknown. c. A concentrated solution of compound A in the same solvent was diluted form an initial volume of 2.00 mL to a final volume of 25.0 mL and then had an absorbance of 0.733. What is the concentration of A in the concentrated solution?
4a A
b
b
(x2,y2)
b
(x1,y1)
b
y2 y1 x2 x1
0.624 0.0290 3.96 x10
0.595
3.96 x10
4
1502 .53 M
4
0 1
1502 .53M
1
1cm
1502 .53 M 1cm
1
c
x1=0; x2=3.6x10-4; y1=0.0290; y2=0.624 y=mx+C A=(εb)c+C
The slope is εb
3
1
1
1.5 x10 mol cm L
4b The equation for calibration curve y mx C y (1.5 x103 ) X 0.029 unknown, y 0.375 0.375 (1.5 x103 ) X 0.029 X 2.31 x10 4 M
4c The equation for calibration curve y (1.5 x103 ) X 0.029
sample, y 0.733 0.733 (1.5 x103 ) X 0.029
X 4.693 x10 4 M 25mL 4 mol A (25mL) 4.693 x10 M x x1 L 1000 mL mol A (25mL) 1.173 x10 5 mol 1.173 x10 5 mol concentrat ion A (5mL) 2mL x1 L 1000 mL
concentrat ion A (5mL) 5.865 x10 3 M
