1. S as an energy relationship
The relationship between a supersaturated (or subsaturated) system and a system at equilibrium

2. Energy released when NaCl (S=1.5) precipitates @25C/1atm
This energy release drives the precipitation reaction forward
SNaCl=1.5
−91,588.4 𝑐𝑎𝑙
Energy in solution
243.4 cal released when NaCl precipitates
SNaCl=1.0
−91,831.8 𝑐𝑎𝑙
Concentration
Energy released when NaCl (S=1.5)

3. Saturation Ratio and Energy
There is a difference however with respect to amount of solids forming
Solids, umol/kg
S=1
S=2
S=10
S=100
Calcite
48.2
274.8
1,620
Barite
4.4
23.5
101.6
Saturation Ratio and Energy

4. The reason energy is the same for a given S
𝐾 𝑠𝑝 𝑁𝑎𝐶𝑙, ℎ𝑎𝑙𝑖𝑡𝑒 = 𝑦 𝑁𝑎 + ∗ 𝑚 𝑁𝑎 + ∗ 𝛾 𝐶𝑙 − ∗ 𝑚 𝐶𝑙 − 𝑎 𝑁𝑎𝐶𝑙 (𝑠)
𝑺= 𝑰𝑨𝑷 𝑲 𝒔𝒑 = 𝒆 − ∆𝑮 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑹𝑻 𝒆 − ∆𝑮𝑹 𝒂𝒕 𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 𝑹𝑻
∆𝑮 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑎𝑡 𝑠𝑢𝑝𝑒𝑟𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛
∆𝑮𝑹 𝒂𝒕 𝒆𝒒𝒖𝒊𝒍𝒊𝒃𝒓𝒊𝒖𝒎 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 − ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑎𝑡 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
The reason energy is the same for a given S

5. Saturation Ratio and Scale Mass
There is a relationship between the energy released and the saturation ratio
GS=x -GS=1
S=1
S=2
S=10
S=100
Calcite
407
1,361
2,707
Barite
408
1,362
2,724
Saturation Ratio and Scale Mass

6. Relationship of S to scale formation
S>1, supersaturated
S<1, subsaturated
Relationship of S to scale formation

7. Three Elements to Developing Scale Risk
Thermodynamic
“Saturation Ratio” or “Scale Tendency”
“Excess Solute” or “Scale Mass”
Kinetic
Nucleation induction time
Crystal Growth rates
Mass Transfer
Transport to surface
Surface Adhesion
Three Elements to Developing Scale Risk

8. NaCl-H2O phase balance 𝑁𝑎𝐶𝑙,𝐻𝑎𝑙𝑖𝑡𝑒 ↔ 𝑁𝑎 + + 𝐶𝑙 −
the rate expression for dissolving or precipitating NaCl is:
𝑟𝑎𝑡𝑒= 𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡
NaCl-H2O phase balance

9. 𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 <0 :𝑁𝑎𝐶𝑙 𝑖𝑠 𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑖𝑛𝑔
water is Supersaturated with NaCl 𝑑𝑡
𝑑 𝑁𝑎 +
concentration at equilibrium
𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 =0
water is Subsaturated with NaCl 𝑑𝑡
𝑑 𝑁𝑎 + 𝑑𝑡 = 𝑑 𝐶𝑙 − 𝑑𝑡 =− 𝑑 𝐻𝑎𝑙𝑖𝑡𝑒 𝑑𝑡 >0 :𝑁𝑎𝐶𝑙 𝑖𝑠 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑖𝑛𝑔
t=
Plotted on a time curve

10. Three parts of the curve
@t0, rate0
@t=0, rate=0
@t=, rate=0
Scale Tendency value
Excess Solute value
Supersaturated
Equilibrium
Subsaturated
t=
Two boundaries and one curve
At t=0, rate=0 (by definition) – we obtain the Scale tendency
At t=, rate=0 (equilibrium) – we obtain the Excess Solute
Between t=0 and , rate0 (reaction occurs) – nothing obtained
Three parts of the curve

11. How is Ksp calculated? 𝐾 𝑠𝑝 = 𝑒 − ∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇
The fundamental equation for Ksp
𝐾 𝑠𝑝 = 𝑒 − ∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇
Definition: The Solubility Product constant of a solid that is in equilibrium with its corresponding dissolved species in water
It is derived from the thermodynamics properties of the solid and of the dissolved ions
∆𝐺 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 - ∆𝐺 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
For NaCl (halite):
𝑁𝑎𝐶𝑙 = 𝑁𝑎 + + 𝐶𝑙 −
∆𝐺 𝑅,𝑁𝑎𝐶𝑙 = ∆𝐺 𝑓,𝑁𝑎 + + ∆𝐺 𝑓,𝐶𝑙 − - ∆𝐺 𝑓,𝑁𝑎𝐶𝑙,(𝐻𝑎𝑙𝑖𝑡𝑒)
How is Ksp calculated?

12. Thermodynamic K 𝐾 𝑟𝑥𝑛 = 𝑒 − ∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇
The fundamental equation
𝐾 𝑟𝑥𝑛 = 𝑒 − ∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝑇
when used with a rigorous Equation of State, is accurate to 300C and 1500atm
∆𝐺 (𝑇,𝑃) 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =𝑓(𝐻, 𝑆, 𝐶 𝑝 , 𝑉 𝑚 )
Log K for HCO3-1
Thermodynamic K

13. A simplified equation for Ksp
An simplified thermodynamic equation that is useful to ~350K (75C) and 1atm. This is called the Van’t Hoff equation.
ln 𝐾 𝑟𝑥𝑛, 𝑇 𝐾 𝑟𝑥𝑛,298 = −∆ 𝐻 𝜃 𝑅 1 𝑇 2 − 1 298
A simplified equation for Ksp

14. Curve fitting equation used when solubility data is available
𝑙𝑜𝑔𝐾=𝑎+ 𝑏 𝑇 +𝑐∗𝑇+𝑑∗ 𝑇 2 +𝑒∗𝑃+𝑓 𝑃 2
Useful tool and effective within the experimental region
Empirical Ksp

15. Temperature effects on  and Ksp