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)
COLORED SOLUTIONS
A solution will appear a certain color if it absorbs its complementary color
from the color wheel
EX2-2 (of 21)
COLORED SOLUTIONS
If a solution appears red, it is primarily absorbing its complimentary color,
green
EX2-3 (of 21)
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 a slit to
form a narrow beam
EX2-4 (of 21)
Light passes
through the
sample
Another slit
allows just one
color to pass
A detector
measures the final
amount of light
Incident
100
photons
Light
Transmitted
10
photons
Light
I0
It
TRANSMITTANCE (T) – the fraction of the incident light that passes
through the sample
T = It /I0
T =
10 photons
_________________
100 photons
EX2-5 (of 21)
= 0.1
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)
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)
PART A – Preparing the STOCK SOLUTION
10.00 mL
0.200 M Fe(NO3)3
EX2-8 (of 21)
3.00 mL
0.00200 M KSCN
17.00 mL
0.5 M HNO3
PART A – Preparing the STOCK SOLUTION
Concentration of Fe(NO3)3 in the Stock Solution:
MCVC = MDVD
MC = 0.200 M
MD = ? M
VC = 10.00 mL
VD = 30.00 mL
MCVC = MD = (0.200 M)(10.00 mL) = 0.0667 M Fe(NO3)3
_______
________________________
VD
(30.00 mL)
EX2-9 (of 21)
PART A – Preparing the STOCK SOLUTION
Concentration of KSCN in the Stock Solution:
MCVC = MDVD
MC = 0.00200 M
MD = ? M
VC = 3.00 mL
VD = 30.00 mL
MCVC = MD = (0.00200 M)(3.00 mL)
_______
VD
EX2-10 (of 21)
___________________________
(30.00 mL)
= 0.000200 M KSCN
PART A – Preparing the STOCK SOLUTION
Concentration of Fe3+ in the Stock Solution:
0.0667 M Fe(NO3)3 x 1
= 0.0667 M Fe3+
Concentration of SCN- in the Stock Solution:
0.000200 M KSCN x 1
EX2-11 (of 21)
= 0.000200 M SCN-
PART A – Preparing the STOCK SOLUTION
Concentration of FeSCN2+ in the Stock Solution:
Fe3+ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.0667
-x
0.0667 - x
+
SCN- (aq)
0.000200
-x
0.000200 - x
⇆
FeSCN2+ (aq)
0
+x
x
We will assume all of the SCN- is converted to FeSCN2+ at equilibrium
EX2-12 (of 21)
PART A – Preparing the STOCK SOLUTION
Concentration of FeSCN2+ in the Stock Solution:
Fe3+ (aq)
Initial M’s
+
0.0667
- 0.000200
Change in M’s
Equilibrium M’s 0.0667 – 0.000200
SCN- (aq)
⇆
0.000200
-0.000200
0.000200 – 0.000200
FeSCN2+ (aq)
0
+ 0.000200
0.000200
We will assume all of the SCN- is converted to FeSCN2+ at equilibrium
 the [FeSCN2+] = 0.000200 M
EX2-13 (of 21)
PART B – Preparing the STANDARD SOLUTIONS
Must calculate the concentration of FeSCN2+ in each standard solution
Solution 0:
0 M FeSCN2+
Solution 1:
0.000200 M FeSCN2+
Solutions 2-5: MCVC = MDVD
EX2-14 (of 21)
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)
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)
PART C – Determining the Absorbances of the STANDARD SOLUTIONS
Absorbance
Measure the absorbance of FeSCN2+ solutions of known concentrations at
the peak wavelength, and plot absorbance vs. concentration
C: 0.25 M
A: 0.241
Concentration of FeSCN2+
EX2-17 (of 21)
0.50 M
0.478
0.75 M
0.722
1.00 M
0.961
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+”)
EX2-18 (of 21)
b = 1.00 cm
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)
Beer's Law Graph
1.2
Absorbance
1
A-546.0 = mx + b
m(slope): 3425
b(y-intercept): - 0.021
0.8
C: 0.25 M
A: 0.241
0.50 M
0.478
0.75 M
0.722
1.00 M
0.961
0.6
0.4
A = ɛbc
A = (3425 M-1)c – 0.021
0.2
0
0
0.2
0.4
0.6
M
0.8
1
1.2
FeSCN2+
This is called a CALIBRATION LINE
m = Δy
= Δ Absorbance
= no units
____
_____________________
____________
Δx
Δ Concentration
M
EX2-20 (of 21)
= M-1
Beer's Law Graph
1.2
Absorbance
1
A-546.0 = mx + b
m(slope): 3425
b(y-intercept): - 0.021
0.8
C: 0.25 M
A: 0.241
0.50 M
0.478
0.75 M
0.722
1.00 M
0.961
0.6
0.4
A = ɛbc
A = (3425 M-1)c – 0.021
0.2
0
0
0.2
0.4
0.6
M
0.8
1
1.2
FeSCN2+
If an unknown solution has an absorbance of 0.351, find its concentration
0.351
= (3425 M-1)c – 0.021
0.372
= (3425 M-1)c
0.372
= c
___________
3425 M-1
EX2-21 (of 21)
= 0.000109 M