SME
Analytical Chemistry · Problem type
Two absorbers at two wavelengths
Use this for “Co²⁺ and Ni²⁺ both absorb at 510 nm and 395 nm. From their ε at each wavelength and the mixture’s two absorbances in a 1 cm cell, find both concentrations.”
Diagram
D = 5,000 × 4,000 − 1,000 × 500
D
=
1.95 ×
10
7
cX
=
D₁
D
=
780
1.95
×
10
7
cX
=
4.00 ×
10
-5
M
cY
=
D₂
D
=
1,560
1.95
×
10
7
cY
=
8.00 ×
10
-5
M
D₁ and D₂ put the right sides in that unknown’s column (shaded).
This is an example. Type your own values to start a new problem.
A₁
=
εX1 (L/(mol·cm))
×
cX (M)
+
εY1 (L/(mol·cm))
×
cY (M)
A₂
=
εX2 (L/(mol·cm))
×
cX (M)
+
εY2 (L/(mol·cm))
×
cY (M)
D
Determinant of the ε matrix
worked out (not typed)
Formulas
D
=
εX1
×
εY2
−
εY1
×
εX2
1.95 ×
10
7
=
5,000
×
4,000
−
1,000
×
500
cX
=
A₁
× εY2 −
A₂
× εY1
D
4.00 ×
10
-5
=
(0.2800
×
4,000
−
0.3400
×
1,000)
÷
(1.95
×
10
7
)
cY
=
εX1 ×
A₂
− εX2 ×
A₁
D
8.00 ×
10
-5
=
(5,000
×
0.3400
−
500
×
0.2800)
÷
(1.95
×
10
7
)
A₁
=
εX1
×
cX
+
εY1
×
cY
0.2800
=
5,000
×
(4.00
×
10
-5
)
+
1,000
×
(8.00
×
10
-5
)
A₂
=
εX2
×
cX
+
εY2
×
cY
0.3400
=
500
×
(4.00
×
10
-5
)
+
4,000
×
(8.00
×
10
-5
)
Assumptions
•
Absorbances add: at each wavelength A = εX·b·cX + εY·b·cY, with b = 1.00 cm, so there are two equations in the two concentrations.
•
Each ε comes from a pure standard of that absorber at that wavelength.
•
Cramer’s rule solves the pair: the determinant D = εX1εY2 − εY1εX2 must not be near 0, so pick wavelengths where X and Y absorb differently.
Step-by-step
WE KNOW
εX1 = 5,000 L/(mol·cm), εY1 = 1,000 L/(mol·cm), εX2 = 500 L/(mol·cm), εY2 = 4,000 L/(mol·cm), A₁ = 0.2800, A₂ = 0.3400
FIND
determinant of the ε matrix (D), concentration of X (cX), concentration of Y (cY)
Step 1 · Find determinant of the ε matrix (D)
Use D = εX1 × εY2 − εY1 × εX2
Cramer’s rule starts from the determinant of the ε matrix: down one diagonal, less the other.
D
=
5,000
×
4,000
−
1,000
×
500
D
=
(2
×
10
7
)
−
500,000
D
=
1.95 ×
10
7
Step 2 · Find concentration of X (cX)
Use cX = (A₁ × εY2 − A₂ × εY1) ÷ D
Put
the
absorbances
in
X’s
column,
take
that
determinant,
and
divide
by
D
.
cX
=
(0.2800
×
4,000
−
0.3400
×
1,000)
÷
(1.95
×
10
7
)
cX
=
1,120 − 340
1.95 ×
10
7
cX
=
780
1.95 ×
10
7
cX
=
4.00 ×
10
-5
M
Step 3 · Find concentration of Y (cY)
Use cY = (εX1 × A₂ − εX2 × A₁) ÷ D
Put
the
absorbances
in
Y’s
column
instead,
take
that
determinant,
and
divide
by
D
.
cY
=
(5,000
×
0.3400
−
500
×
0.2800)
÷
(1.95
×
10
7
)
cY
=
1,700 − 140
1.95 ×
10
7
cY
=
1,560
1.95 ×
10
7
cY
=
8.00 ×
10
-5
M
Check
Put the numbers back in. Both sides should match.
1.95 ×
10
7
=
5,000
×
4,000
−
1,000
×
500
✓
4.00 ×
10
-5
=
(0.2800
×
4,000
−
0.3400
×
1,000)
÷
(1.95
×
10
7
)
✓
8.00 ×
10
-5
=
(5,000
×
0.3400
−
500
×
0.2800)
÷
(1.95
×
10
7
)
✓
0.2800
=
5,000
×
(4.00
×
10
-5
)
+
1,000
×
(8.00
×
10
-5
)
✓
0.3400
=
500
×
(4.00
×
10
-5
)
+
4,000
×
(8.00
×
10
-5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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