SME
Analytical Chemistry · Problem type
An unknown from a calibration line
Use this for “Iron standards give the line A = 0.0820c + 0.0030 (c in mg/L). A sample diluted 5 times reads A = 0.415. How much iron was in the sample?”
Chart
A(5.47) = 0.524
This is an example. Type your own values to start a new problem.
A
=
m
×
c (mg/L)
+
b
c₀ (mg/L)
=
D
×
c (mg/L)
Formulas
A
=
m
×
c
+
b
0.5240 = 0.095 × 5.474 + 0.0040
c₀
=
D
×
c
54.74 = 10 × 5.474
Assumptions
•
The standards were fitted to a straight line, A = mc + b, and the unknown reads on the same instrument at the same wavelength.
•
The unknown’s absorbance lies inside the standards’ range, so reading across the line is safe.
•
A dilution factor D means 1 part sample in D parts in all (10 mL made up to 100 mL is D = 10), so the sample is D times stronger.
Step-by-step
WE KNOW
m = 0.095, b = 0.0040, A = 0.5240, D = 10
FIND
concentration measured (c), concentration in the sample (c₀)
Step 1 · Find concentration measured (c)
Use A = m × c + b
Read across the line: take the intercept from the absorbance, then divide by the slope.
c
=
A
−
b
m
c
=
0.5240 − 0.0040
0.095
c
=
0.52
0.095
c
=
5.474
mg/L
Step 2 · Find concentration in the sample (c₀)
Use c₀ = D × c
The
measured
solution
was
diluted
D
times,
so
the
sample
is
D
times
as
strong.
c₀
=
10
×
5.474
c₀
=
54.74
mg/L
Check
Put the numbers back in. Both sides should match.
0.5240 = 0.095 × 5.474 + 0.0040 ✓
54.74 = 10 × 5.474 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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