1. COLORED SOLUTIONS
A solution will appear a certain color if it absorbs its complementary color from the color wheel
EX2-1 (of 24)

2. COLORED SOLUTIONS
If a solution appears orange, it is primarily absorbing its complimentary color, blue
EX2-2 (of 24)

3. Fe3+ (aq) + SCN- (aq) ⇆ FeSCN2+ (aq)
EQUILIBRIUM CONSTANT DETERMINATION
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
(blood red)
A darker color means a higher concentration of the colored component
The “darkness” can be determined by measuring the amount of light absorbed by the solution, called its ABSORBANCE
EX2-3 (of 24)

4. Concentration of FeSCN2+
If you measure the absorbance of FeSCN2+ solutions of known concentrations, and plot absorbance vs. concentration
, the relation is linear
C: M M M M A:
Absorbance
Concentration of FeSCN2+
EX2-4 (of 24)

5. A light bulb emits white light
SPECTROPHOTOMETER – A device that measures the amount of light absorbed by a sample
A light bulb emits white light
A diffraction grating separates the colors
of light
Light passes through the sample
Light passes through a slit to form a narrow beam
Another slit allows just one color to pass
A detector measures the final amount of light
EX2-5 (of 24)

6. 100 photons
Incident Light
Transmitted Light
10 photons I0 It
TRANSMITTANCE (T) – the fraction of the incident light that passes through the sample
T = It / I0
T = photons = 0.1
_________________
100 photons
EX2-6 (of 24)

7. 100 photons
10 photons I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.1)
= 1
EX2-7 (of 24)

8. 100 photons 1
photon I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.01)
= 2
EX2-8 (of 24)

9. PART A – Preparing the STOCK SOLUTION of FeSCN2+
10.00 mL
0.200 M Fe(NO3)3
3.00 mL
M KSCN
17.00 mL
6 M HNO3
EX2-9 (of 24)

10. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Calculation 1: Concentration of Fe(NO3)3 in the Stock Solution (Initial) 1.
200
MCVC = MDVD
10 00
MC =
VC =
0.200 M
10.00 mL
MD =
VD =
? M
30.00 mL
00200
3 00
MCVC = MD
_______ VD
= (0.200 M)(10.00 mL)
________________________
(30.00 mL)
30 00
0667
= M Fe(NO3)3
EX2-10 (of 24)

11. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Calculation 2: Concentration of KSCN in the Stock Solution (Initial) 2.
200
MCVC = MDVD
10 00
MC =
VC = M
3.00 mL
MD =
VD =
? M
30.00 mL
00200
3 00
MCVC = MD
_______ VD
= ( M)(3.00 mL)
___________________________
(30.00 mL)
30 00
0667
000200
= M KSCN
EX2-11 (of 24)

12. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Calculation 3: Concentration of Fe3+ in the Stock Solution (Initial) 3.
200
M Fe(NO3)3 x 1 = M Fe3+
10 00
00200
3 00
30 00
0667
000200
0667
EX2-12 (of 24)

13. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Calculation 4: Concentration of Fe3+ in the Stock Solution (Initial) 4.
200
M KSCN x 1 = M SCN-
10 00
00200
3 00
30 00
0667
000200
0667
000200
EX2-13 (of 24)

14. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Concentration of FeSCN2+ in the Stock Solution (Equilibrium)
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.0667
- x
- x
+ x x x x
We will assume all of the SCN- is converted to FeSCN2+ at equilibrium
EX2-14 (of 24)

15. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Concentration of FeSCN2+ in the Stock Solution (Equilibrium)
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.0667 – –
We will assume all of the SCN- is converted to FeSCN2+ at equilibrium
 the [FeSCN2+] = M
EX2-15 (of 24)

16. PART A – Preparing the STOCK SOLUTION of FeSCN2+
Concentration of FeSCN3+ in the Stock Solution (Equilibrium)
200
10 00
00200
3 00
30 00
0667
000200
0667
000200
000200
EX2-16 (of 24)

17. PART B – Preparing the STANDARD SOLUTIONS of FeSCN2+
000200
Must calculate the concentration of FeSCN2+ in each standard solution
Solution 1:
M FeSCN2+
Solution 2-5:
MCVC = MDVD
Solutions 0:
0 M FeSCN2+
EX2-17 (of 24)

18. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
Place a colored standard solution in the spectrometer
ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths
EX2-18 (of 24)

19. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
LAMBDA MAX (λmax) – The wavelength of maximum absorbance
When measuring the absorbance of solutions, it is most accurate to measure the absorbance at λmax
EX2-19 (of 24)

20. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
Determine the absorbance of each standard solution
A = mx + b
m(slope): 3425
b(y-intercept):
C: M M M M A:
y = mx + b
A = mc + b
A = (3425 M-1)c –
This is called a CALIBRATION LINE
m = Δy
____ Δx
= Δ Absorbance
_____________________
Δ Concentration
= no units
___________ M
= M-1
EX2-20 (of 24)

21. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
Determine the absorbance of each standard solution
A = mx + b
m(slope): 3425
b(y-intercept):
C: M M M M A:
y = mx + b
A = mc + b
A = (3425 M-1)c –
This is called a CALIBRATION LINE
m = Δy
____ Δx
= Δ Absorbance
_____________________
Δ Concentration
= no units
___________ M
= M-1
EX2-21 (of 24)

22. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
If an unknown solution has an absorbance of 0.351, find its concentration of FeSCN2+
A = mx + b
m(slope): 3425
b(y-intercept):
= (3425 M-1)c –
0.372 = (3425 M-1)c
= c
___________
3425 M-1
= M
EX2-22 (of 24)

23. ɛ = extinction coefficient
AUGUST BEER
Proposed a mathematical explanation for the linear relationship between concentration and absorbance
BEER’S LAW
: A = ɛ l c
A = absorbance
ɛ = extinction coefficient
(a constant for a given solute at a given λ)
l = width of the cuvet holding the sample
(for our cuvets it is 1.00 cm)
c = concentration
(in our lab it’s in “M FeSCN2+”)
l = 1.00 cm
EX2-23 (of 24)

24. A = ɛ l c
+ b
A = mc + b
slope = ɛ l
Calculate the extinction coefficient for absorbance at a wavelength of 546 nm and using a 1.00 cm cuvet given the calibration line:
= (3425 M-1)c –
m = ɛ l
m = ɛ
___ l
= M-1
____________
1.00 cm
= M-1cm-1
EX2-24 (of 24)