EX21 of 24 I f a solution appears orange it is primarily absorbing its complimentary color blue COLORED SOLUTIONS EX22 of 24 Fe 3 aq SCN aq FeSCN ID: 745318
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Slide1
EXPERIMENTS 2 and 3 – AN EQUILIBRIUM CONSTANT DETERMINATION
known - x
x
Fe3+ (aq) + SCN- (aq) ⇆
Initial M’sChange in M’sEquilibrium M’s
known
known
0
- x
- x
+ x
known - x
Kc = [FeSCN2+] _______________ [Fe3+][SCN-]
Kc = x _____________________________ (known – x)(known – x)
Must have a way to find x!
FeSCN2+ (aq) is blood red
FeSCN2+ (aq)
The concentration of a colored chemical is proportional to the amount of light it absorbs
EX2-1
(of
21)Slide2
EXPERIMENTS 2 and 3 – AN EQUILIBRIUM CONSTANT DETERMINATION
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
EX2-2 (of
21)
A solution’s absorbance is directly proportional to the concentration of the colored componentSlide3
SPECTROPHOTOMETER – A device that measures the amount of light absorbed by a sample
A
light bulb emits white light
Light
passes through a slit to form a narrow beam
A prism
separates the colors
of light
Another
slit allows just one color to pass
Light
passes through the sample
A detector measures the final amount of light
EX2-3
(of
21)Slide4
If you choose a specific wavelength of light, and measure the
absorbances of FeSCN2+ solutions of known concentrations, and plot absorbance vs. concentration
C:
0.25 M 0.50 M 0.75 M 1.00 MA: 0.241 0.478 0.722 0.961
Concentration
of
FeSCN
2+
Absorbance
EX2-4
(of 21)
, the relation is linear
y = mx + b
A = mC + bSlide5
EX2-5 (of 21)If an equilibrium solution has an absorbance of 0.351 using a wavelength of light of 476.0 nm,
find its concentration of FeSCN2+ 0.351 = (3425 M-1)C – 0.021
0.372 = (3425 M-1)C 0.372 = C ___________ 3425 M-1
= 0.000109 M
A@476.0 = mx + bm(slope): 3425b(y-intercept): - 0.021
m = Δy ____ Δx
= Δ Abs. __________ Δ Conc.
= no units ___________ M = M
-1
A@476 = mC + b
known - x
x
Fe3+ (aq) + SCN- (aq) ⇆known
known0
- x
-
x
+
x
known
-
x
FeSCN
2+
(
aq
)Slide6
PART A – Preparing the STOCK SOLUTION of FeSCN2+
10.00 mL0.200 M Fe(NO3
)3
3.00 mL0.00200 M KSCN
17.00 mL
6 M HNO3
EX2-6
(of 21)STOCK SOLUTION – A solution of known concentration (in this case, of FeSCN2+)? M FeSCN2+Slide7
MCVC
= MDVD
MCVC = MD_______
VD
= (0.200 M)(10.00 mL) ________________________ (30.00 mL)
= 0.0667 M Fe(NO3)3
M
C =VC =
0.200 M10.00 mL
MD
=VD =
? M30.00 mLPART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 1: Concentration of Fe(NO3)
3 in the Stock Solution (Initial)EX2-7 (of 21)20010 001.002003 0030 00
0667Slide8
MCVC
= MDVD
MCVC = MD_______
VD
= (0.00200 M)(3.00 mL) ___________________________ (30.00 mL)
= 0.000200 M KSCN
M
C =VC =
0.00200 M3.00 mL
MD
=VD =
? M30.00 mLPART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 2: Concentration of KSCN in the Stock Solution (Initial)
EX2-8 (of 21)20010 002.002003 0030 00
0667000200Slide9
PART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 3: Concentration of Fe3+ in the Stock Solution (Initial)
EX2-9 (of 21)20010 00
3.002003 0030 000667
0002000667
0.0667 M Fe(NO3)3 = 0.0667 M Fe3+
x 1 mol
Fe3+ __________________ 1 mol Fe(NO3)3Slide10
PART A – Preparing the STOCK SOLUTION of FeSCN2+Calculation 4: Concentration of Fe3+ in the Stock Solution (Initial)
EX2-10 (of 21)4.000200
20010 00002003 0030 00
06670002000667
0.000200 M KSCN= 0.000200 M SCN-
x 1
mol SCN- _______________ 1 mol KCSNSlide11
0.000200 - x
x Fe3+
(aq) + SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’sChange in M’sEquilibrium M’s
0.0667
0.000200
0
- x
- x
+ x
0.0667 - xPART A – Preparing the
STOCK SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium)Because we have 333 times more Fe3+ than SCN-, we will assume that essentially all of the SCN- is converted to FeSCN2+ at equilibrium
EX2-11 (of 21)Slide12
~0.000200 Fe3+
(aq) + SCN- (aq) ⇆ FeSCN2+ (aq)
Initial M’sChange in M’sEquilibrium M’s
0.0667
0.000200
0
- ~0.000200
- ~0.000200
+ ~0.000200
~0.0665PART A – Preparing the STOCK
SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium) the [FeSCN2+] = [SCN-] = 0.000200 M
EX2-12 (of 21)
~0Slide13
PART A – Preparing the STOCK SOLUTION of FeSCN2+Concentration of FeSCN2+ in the Stock Solution (Equilibrium)
EX2-13 (of 21)00020020010 00
002003 0030 000667000200
0667000200Slide14
PART B – Preparing the
STANDARD SOLUTIONS of FeSCN2+Must calculate the concentration of FeSCN2+ in each standard solution
0 M FeSCN2+
Solution 1:0.000200 M FeSCN2+Solutions 2-5:MCVC = MD
VDSolution 0:
EX2-14 (of 21)0
000200Slide15
PART C – Determining the
Absorbances of the STANDARD SOLUTIONS
EX2-15 (of 21)
Place a colorless solution (called the
BLANK) in the spectrometer and set it to zero absorbanceSlide16
PART C – Determining the
Absorbances of the STANDARD SOLUTIONS
EX2-16 (of 21)
ABSORBANCE SPECTRUM – A graph of the absorbance of a solution at different wavelengths
Place a colored standard solution in the spectrometer
I
f a solution appears orange, it is primarily absorbing its complimentary color, blueSlide17
PART C – Determining the
Absorbances of the STANDARD SOLUTIONSLAMBDA MAX (λmax) – The wavelength of maximum absorbance
When
measuring the absorbance of solutions, it is most accurate to measure the absorbance at λmax
EX2-17 (of 21)Peak of the absorbance graph
λmaxSlide18
This is called a CALIBRATION LINE
C: 0.25 M 0.50 M 0.75 M 1.00 M
A: 0.241 0.478 0.722 0.961
y = mx + bA@476 = (3425 M-1)C – 0.021
A@476.0 = mx + bm(slope): 3425
b(y-intercept): - 0.021R2 = 0.995
EX2-18 (of 21)
PART C – Determining the
Absorbances of the STANDARD SOLUTIONS
Determine the absorbance of each standard solution at
λmaxSlide19
This is called a CALIBRATION LINE
C: 0.25 M 0.50 M 0.75 M 1.00 M
A: 0.241 0.478 0.722 0.961A@476.0 = mx + b
m(slope): 3425b(y-intercept): - 0.021R2 = 0.995
EX2-19 (of 21)
PART C – Determining the
Absorbances of the STANDARD SOLUTIONS
Determine the absorbance of each standard solution at
λmax
y = mx + b
A@476 = (3425 M
-1)C – 0.021Slide20
EX2-20 (of 21)
1852 AUGUST BEERProposed a mathematical explanation for the linear relationship between
concentration and absorbance
A = mC + b
A = absorbance
ɛ = extinction coefficient (a constant for a given solute at a specific 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
A = ɛ
l CBEER’S LAW :
+ 0Slide21
Calculate the extinction coefficient for absorbance at a wavelength of 476 nm and using a 1.00 cm cuvet given the calibration line:
A@476 = (3425 M-1)C – 0.021
A@476 = ɛ
l C + b
A@
476 = mC + b slope = ɛ l
EX2-21 (of 21)
m = ɛ
___ l
= 3425 M-1 ____________
1.00 cm= 3430 M-1cm-1