SME
Analytical Chemistry
Electroanalytical methods
Use this for “A nitrate electrode reads +0.0420 V in a 1.00 × 10⁻³ M nitrate standard and +0.0890 V in a soil extract. Find the nitrate concentration in the extract.”
Chart
E(−3.59) = −0.085
Level standard = −0.12 meets the curve at x = −3
This is an example. Type your own values to start a new problem.
Eₓ (V)
−
(
Eₛ (V)
)
=
0.05916
V
z
log
cₓ (M)
cₛ (M)
The picture’s line is E = K + S log c, with S = 0.05916 V ÷ z. Taking the standard’s reading from the sample’s cancels K and leaves this equation. Eₛ and cₛ belong to the standard, Eₓ and cₓ to the sample.
S
Slope (V per tenfold change)
worked out (not typed) · V
log(cₓ ÷ cₛ)
Log of the concentration ratio
worked out (not typed)
Formulas
S
=
0.05916
z
−0.0592
=
0.05916
−1
log(
cₓ
cₛ
)
=
Eₓ − Eₛ
S
−0.5916
=
−0.0850 − (−0.1200)
−0.0592
cₓ
=
cₛ
×
10^(log(
cₓ
cₛ
))
2.56 ×
10
-4
=
0.001 ×
10
−0.5916
Assumptions
•
At 25 °C the electrode’s potential changes by 0.05916 ÷ z volts for each tenfold change in the ion’s concentration (z is the ion’s charge, negative for an anion).
•
The standard and the sample have the same ionic strength (both in the same buffer), so concentrations stand in for activities.
•
One standard fixes the line, since the slope is the ideal Nernst slope.
Step-by-step
WE KNOW
z = −1, cₛ = 0.001 M, Eₛ = −0.1200 V, Eₓ = −0.0850 V
FIND
slope (V per tenfold change) (S), log of the concentration ratio (log(cₓ ÷ cₛ)), sample concentration (cₓ)
Step 1 · Find slope (V per tenfold change): S
Use S = 0.05916 ÷ z
The Nernst slope at 25 °C divided by the ion’s charge: negative for an anion.
S
=
0.05916
−1
S
=
−0.0592
V
Step 2 · Find log of the concentration ratio (log(cₓ ÷ cₛ))
Use log(cₓ ÷ cₛ) = (Eₓ − Eₛ) ÷ S
Each slope’s worth of change in the reading is one tenfold step in concentration.
log(
cₓ
cₛ
)
=
−0.0850 − (−0.1200)
−0.0592
log(
cₓ
cₛ
)
=
0.035
−0.0592
log(
cₓ
cₛ
)
=
−0.5916
→
log
ratio
<
0:
the
sample
is
weaker
than
the
standard.
Step 3 · Find sample concentration (cₓ)
Use cₓ = cₛ × 10log(cₓ ÷ cₛ)
Undo the log: multiply the standard’s concentration by 10 to the log ratio.
cₓ
=
0.001 ×
10
−0.5916
cₓ = 0.001 × 0.2561
cₓ
=
2.56 ×
10
-4
M
Check
Put the numbers back in. Both sides should match.
−0.0592
=
0.05916
−1
✓
−0.5916
=
−0.0850 − (−0.1200)
−0.0592
✓
2.56 ×
10
-4
=
0.001 ×
10
−0.5916
✓
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Calibrate a pH meter with two buffers
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