SME
Biochemistry · Problem type
Equilibrium potential of one ion (Nernst)
Use this for “Cl⁻ is 110 mM outside a neuron and 10 mM inside. Find its equilibrium potential at 37 °C.”
Diagram
V = −89.1 mV: the inside is negative, so + charges line the outside face and − the inside.
One pair per 10 mV: 8 a side (8 at most).
Dots on one scale, one dot ≈ 3.5 mM: Ion 5 mM outside, 140 mM inside.
This is an example. Type your own values to start a new problem.
E (mV)
=
R
T (K)
F
ln
cₒᵤₜ (mM)
cᵢₙ (mM)
Formulas
E
=
8.314
×
T
z
× 96,485
×
ln(
cₒᵤₜ
cᵢₙ
)
×
1,000
−89.1
=
8.314
×
310.15
1 × 96,485
×
ln(
5
140
)
×
1,000
Assumptions
•
At the equilibrium potential the ion’s concentration push and its charge push cancel, so ΔG = 0 and there is no net flow.
•
E is the inside relative to the outside; z is the ion’s charge with its sign (K⁺ +1, Ca²⁺ +2, Cl⁻ −1).
•
E = (RT ÷ zF) ln(cₒᵤₜ ÷ cᵢₙ) with R = 8.314 J/(mol·K) and F = 96,485 C/mol gives volts; × 1000 gives mV.
•
Body temperature, 37 °C, is 310.15 K, where RT ÷ F is 26.7 mV.
Step-by-step
WE KNOW
cₒᵤₜ = 5 mM, cᵢₙ = 140 mM, z = +1, T = 310.15 K
FIND
equilibrium potential (E)
Step 1 · Find equilibrium potential (E)
Use E = 8.314 × T ÷ (z × 96,485) × ln(cₒᵤₜ ÷ cᵢₙ) × 1,000
Divide RT by zF for volts per unit of ln, times ln of outside over inside; × 1000 gives mV.
E
=
8.314
×
310.15
1 × 96,485
×
ln(
5
140
)
×
1,000
E
=
0.02673
×
ln(0.03571)
×
1,000
E
=
0.02673
×
(−3.332)
×
1,000
E
=
−89.1
mV
→
E
<
0:
with
the
inside
this
negative,
the
ion
has
no
net
flow.
Check
Put the numbers back in. Both sides should match.
−89.1
=
8.314
×
310.15
1 × 96,485
×
ln(
5
140
)
×
1,000
✓
← Previous lesson
Carbohydrates, lipids and membranes
Next lesson →
Saturated or unsaturated fatty acid
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.