1. A darker color means a higher concentration of the colored component
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
EX2-1 (of 21)

2. COLORED SOLUTIONS
A solution will appear a certain color if it absorbs its complementary color from the color wheel
EX2-2 (of 21)

3. COLORED SOLUTIONS
If a solution appears red, it is primarily absorbing its complimentary color, green
EX2-3 (of 21)

4. 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
EX2-4 (of 21)

5. Incident Light
100 photons
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
EX2-5 (of 21)

6. 100 photons
10 photons I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.1)
= 1
EX2-6 (of 21)

7. 100 photons 1
photon I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.01)
= 2
EX2-7 (of 21)

8. PART A – Preparing the STOCK SOLUTION
10.00 mL
0.200 M Fe(NO3)3
3.00 mL
M KSCN
17.00 mL
0.5 M HNO3
EX2-8 (of 21)

9. PART A – Preparing the STOCK SOLUTION
Concentration of Fe(NO3)3 in the Stock Solution:
MCVC = MDVD
MC =
VC =
0.200 M
10.00 mL
MD =
VD =
? M
30.00 mL
MCVC = MD
_______ VD
= (0.200 M)(10.00 mL)
________________________
(30.00 mL)
= M Fe(NO3)3
EX2-9 (of 21)

10. PART A – Preparing the STOCK SOLUTION
Concentration of KSCN in the Stock Solution:
MCVC = MDVD
MC =
VC = M
3.00 mL
MD =
VD =
? M
30.00 mL
MCVC = MD
_______ VD
= ( M)(3.00 mL)
___________________________
(30.00 mL)
= M KSCN
EX2-10 (of 21)

11. PART A – Preparing the STOCK SOLUTION
Concentration of Fe3+ in the Stock Solution:
M Fe(NO3)3 x 1
= M Fe3+
Concentration of SCN- in the Stock Solution:
M KSCN x 1
= M SCN-
EX2-11 (of 21)

12. PART A – Preparing the STOCK SOLUTION
Concentration of FeSCN2+ in the Stock Solution:
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
EX2-12 (of 21)

13. PART A – Preparing the STOCK SOLUTION
Concentration of FeSCN2+ in the Stock Solution:
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
EX2-13 (of 21)

14. PART B – Preparing the STANDARD SOLUTIONS
Must calculate the concentration of FeSCN2+ in each standard solution
Solution 0:
0 M FeSCN2+
Solution 1:
M FeSCN2+
Solutions 2-5:
MCVC = MDVD
EX2-14 (of 21)

15. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths
EX2-15 (of 21)

16. 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
EX2-16 (of 21)

17. PART C – Determining the Absorbances of the STANDARD SOLUTIONS
Measure the absorbance of FeSCN2+ solutions of known concentrations at the peak wavelength, and plot absorbance vs. concentration
C: M M M M A:
Absorbance
Concentration of FeSCN2+
EX2-17 (of 21)

18. ɛ = extinction coefficient
BEER’S LAW – The mathematical relationship between concentration and absorbance
A = ɛbc
A = absorbance
ɛ = extinction coefficient
(a constant for a given solute at a given wavelength)
b = 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+”)
b = 1.00 cm
EX2-18 (of 21)

19. This means a graph of A vs. c will produce a straight line
A = ɛbc
A = kc
y = mx + b
This means a graph of A vs. c will produce a straight line
slope = ɛb
EX2-19 (of 21)

20. This is called a CALIBRATION LINE
C: M M M M A:
A = mx + b
m(slope): 3425
b(y-intercept):
A = ɛbc
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 21)

21. C: M M M M A:
A = mx + b
m(slope): 3425
b(y-intercept):
A = ɛbc
A = (3425 M-1)c –
If an unknown solution has an absorbance of 0.351, find its concentration
= (3425 M-1)c –
0.372 = (3425 M-1)c
= c
___________
3425 M-1
= M
EX2-21 (of 21)