Analytical Chemistry · Problem type

Standard addition

Use this for “A 5.00 mL water sample gives a lead signal of 0.150. With 0.50 mL of 50 mg/L lead standard added, both made up to 25.00 mL, it reads 0.270. Find the lead concentration.”

Chart

S(Δc) = 0.01Δc + 0.2−20−15−10−5051000.050.10.150.20.250.30.35Standard added Δc (mg/L)Signal S(−20, 0)(0, 0.2)(10, 0.3)
Zero: Δc = −20
y-intercept (0, 0.2)
S(10) = 0.3
This is an example. Type your own values to start a new problem.
cₓ (mg/L)
=
×
cₛ (mg/L)
×
Vₛ (mL)
(
−
)
×
Vₓ (mL)
cₛ is the standard’s concentration and Vₛ the volume of it added. The standard raises the sample’s concentration by Δc = cₛ × Vₛ ÷ Vₓ, the picture’s step along the axis, and the signal by S₂ − S₁. So cₓ ÷ Δc = S₁ ÷ (S₂ − S₁).
Δc
Standard added, per sample volume
worked out (not typed) · mg/L
k
Slope (signal per mg/L)
worked out (not typed)

Formulas

Δc
=
cₛ
×
Vₛ
Vₓ
10
=
100
×
1
10
cₓ
=
S₁
×
Δc
S₂
−
S₁
20
=
0.2
×
10
0.3 − 0.2
k
=
S₂
−
S₁
Δc
0.01
=
0.3 − 0.2
10

Assumptions

•
The same volume of sample goes into two flasks, the standard into one, and both are filled to the same mark.
•
The signal (an absorbance, an emission or a current) is proportional to the analyte’s concentration, with nothing from the blank.
•
Adding the standard inside the sample’s own matrix cancels anything in the sample that changes the slope.

Step-by-step

WE KNOW
Vₓ = 10 mL, Vₛ = 1 mL, cₛ = 100 mg/L, S₁ = 0.2, S₂ = 0.3
FIND
standard added, per sample volume (Δc), analyte concentration in the sample (cₓ), slope (signal per mg/L) (k)
Step 1 · Find standard added, per sample volume (Δc)
Use Δc = cₛ × Vₛ ÷ Vₓ
The standard’s mass, cₛVₛ, spread over the sample’s volume: what the addition is worth in the sample.
Δc
=
100
×
1
10
Δc = 10 mg/L
Step 2 · Find analyte concentration in the sample (cₓ)
Use cₓ = S₁ × Δc ÷ (S₂ − S₁)
The
rise
S₂
−
S₁
came
from
Δc,
so
S₁
comes
from
S₁
S₂
−
S₁
times
as
much
analyte.
cₓ
=
0.2
×
10
0.3 − 0.2
cₓ
=
0.2
×
10
0.1
cₓ
=
2
0.1
cₓ = 20 mg/L → On the graph, the line through both readings crosses 0 at −cₓ.
Step 3 · Find slope (signal per mg/L): k
Use k = (S₂ − S₁) ÷ Δc
The slope of the line: the rise in signal per mg/L added.
k
=
0.3 − 0.2
10
k
=
0.1
10
k
=
0.01
Check
Put the numbers back in. Both sides should match.
10
=
100
×
1
10
✓
20
=
0.2
×
10
0.3 − 0.2
✓
0.01
=
0.3 − 0.2
10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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