1. EQUILIBRIUM CONSTANT DETERMINATION
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
(blood red)

2. COLORED SOLUTIONS
A solution will appear a certain color if it absorbs its complementary color from the color wheel

3. COLORED SOLUTIONS
If a solution appears red, it is primarily absorbing its complimentary color, green

4. a combination of all colors
White light is
a combination of all colors
Sample absorbs green, but transmits all other colors
Your eye sees the
remaining combination
of colors as red

5. ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths

6. LAMBDA MAX (λmax) – The wavelength of maximum absorbance
For best accuracy, when measuring the absorbance of solutions, it is best to measure the absorbance at λmax

7. 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

8. The fraction of light that gets through is the TRANSMITTANCE
Incident Light
100 photons
Transmitted Light
10 photons
0 photons I0 It
TRANSMITTANCE (T) – the fraction of the incident light that passes through the sample
T = It / I0
T = 0 photons = 0
________________
100 photons
T = photons =
________________
100 photons

9. The fraction of light that gets through is the TRANSMITTANCE
100 photons
10 photons
The fraction of light that doesn’t get through (is blocked or absorbed) is the ABSORBANCE I0 It
ABSORBANCE (A) – negative logarithm of the transmittance
A = -log (T)
A = -log (0.1)
= 1
A = -log (0.01)
= 2
A = -log (1)
= 0

10. The darker the color, the higher concentration of the colored component, the higher the absorbance of the solution

11. 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
In our lab ɛ and b are constants, so A and c are the variables

12. This means a graph of A vs. c will produce a straight line
A = ɛbc A c
+ 0
y = mx + b
This means a graph of A vs. c will produce a straight line

13. You will produce several solutions of known concentration:
A = ɛbc A c
+ 0
y = mx + b
You will produce several solutions of known concentration:
C: M M M M A:

14. _____________________ Δ Concentration = no units ___________ M = M-1
C: M M M M A:
A = ɛbc
A = ( M-1)c
This is called a CALIBRATION LINE because it shows the relationship between the measured absorbance and the concentration of FeSCN2+
m = Δy
____ Δx
= Δ Absorbance
_____________________
Δ Concentration
= no units
___________ M
= M-1

15. C: M M M M A:
A = ɛbc
A = ( M-1)c
If an unknown solution has an absorbance of 0.351, find its concentration
= ( M-1)c
= c
_____________
M-1
0.365 M = c

16. Calculate the extinction coefficient of this substance, with units
C: M M M M A:
A = ɛbc
A = ( M-1)c
Calculate the extinction coefficient of this substance, with units
m = ɛb
m = ɛ
___ b
= M-1
_____________
1.00 cm
= M-1cm-1

17. PART A – Preparing the STOCK SOLUTION
10.00 mL
0.200 M Fe(NO3)3
3.00 mL
M KSCN
17.00 mL
6 M HNO3

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

19. PART A – Preparing the STOCK SOLUTION
Concentration of Fe3+ in the Stock Solution:
mol Fe(NO3)3
__________________________ L
x mol Fe3+
___________________
1 mol Fe(NO3)3
= M Fe3+

20. 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

21. 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

22. 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+
Solution 2:
MCVC = MDVD

24. EXPERIMENT 3 – DETERMINATION OF THE Keq
MCVC = MD
_______ VD
= ( M)(5.00 mL)
__________________________
(10.00 mL)
= M Fe(NO3)3
= M Fe3+
MCVC = MD
_______ VD
= ( M)(2.00 mL)
__________________________
(10.00 mL)
= M KSCN
= M SCN-

25. EXPERIMENT 3 – DETERMINATION OF THE Keq
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
- x
- x
+ x x x x
Kc = [FeSCN2+]
_______________
[Fe3+][SCN-]
Kc = x
_________________________________
( – x)( – x)

26. EXPERIMENT 3 – DETERMINATION OF THE Keq
Measured absorbance of solution:
A = (3425 M-1)c
= (3425 M-1)c
= (3425 M-1)c
= c
__________
3425 M-1
M = c

27. EXPERIMENT 3 – DETERMINATION OF THE Keq
Fe3+ (aq) SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
- x
- x
+ x x x x
Kc = [FeSCN2+]
_______________
[Fe3+][SCN-]
[Fe3+]eq =
[SCN-]eq =
[FeSCN-]eq =
– x
– x x
= M
= M
= M
x = M
Kc = ( M)
_________________________________
( M)( M)
= M-1