SME
Physical Chemistry I: Thermodynamics & Kinetics · Problem type
Ionic strength and activity (Debye–Hückel)
Use this for “Find the ionic strength of 0.010 M CaCl₂ in water at 25 °C, and the mean activity coefficient from the Debye–Hückel limiting law.”
This is an example. Type your own values to start a new problem.
I (M)
=
½
×
c (M)
×
(
ν₊
×
z₊
2
+
ν₋
×
z₋
2
)
log(
γ±
)
=
−0.509
×
z₊
×
z₋
×
I
log γ±
Log of the activity coefficient
calculated
Formulas
I
=
0.5
×
(ν₊
×
c
×
z₊²
+
ν₋
×
c
×
z₋²)
0.03
=
0.5
×
(1
×
0.01
×
2
2
+
2
×
0.01
×
1
2
)
log
γ±
=
−0.509
×
z₊
×
z₋
×
√
I
−0.176
=
−0.509
×
2
×
1
×
√
0.03
γ±
=
10
log γ±
0.6663
=
10
−0.176
Assumptions
•
The salt dissolves fully into ν₊ cations of charge z₊ and ν₋ anions of charge z₋ per formula; charges are entered as sizes, without signs.
•
The constant 0.509 is for water at 25 °C. Ionic strength and concentration are in mol/L (M).
•
The limiting law holds well up to about I = 0.01 M; at higher I it overstates the drop in γ±.
Step-by-step
WE KNOW
c = 0.01 M, z₊ = 2, z₋ = 1, ν₊ = 1, ν₋ = 2
FIND
ionic strength (I), log of the activity coefficient (log γ±), mean activity coefficient (γ±)
Step 1 · Find ionic strength (I)
Use I = 0.5 × (ν₊ × c × z₊² + ν₋ × c × z₋²)
Each ion’s concentration times its charge squared, added, then halved.
I
=
0.5
×
(1
×
0.01
×
2
2
+
2
×
0.01
×
1
2
)
I
=
0.5
×
(0.01
×
4
+
2
×
0.01)
I
=
0.5
×
(0.04
+
0.02)
I
=
0.5
×
0.06
I
=
0.03
M
Step 2 · Find log of the activity coefficient (log γ±)
Use log γ± = −0.509 × z₊ × z₋ × √(I)
The ions’ atmosphere of opposite charge lowers their activity, more for higher charges and stronger solutions.
log
γ±
=
−0.509
×
2
×
1
×
√
0.03
log γ± = −0.509 × 2 × 0.1732
log γ± = −1.018 × 0.1732
log γ± = −0.1763
Step 3 · Find mean activity coefficient (γ±)
Use γ± = 10log γ±
Undo the log: raise 10 to log γ±.
γ±
=
10
−0.1763
γ±
=
0.6663
→
I
is
above
0.01
M:
the
real
γ±
is
larger
than
the
limiting
law
says.
Check
Put the numbers back in. Both sides should match.
0.03
=
0.5
×
(1
×
0.01
×
2
2
+
2
×
0.01
×
1
2
)
✓
−0.1763
=
−0.509
×
2
×
1
×
√
0.03
✓
0.6663
=
10
−0.1763
✓
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