Q.2 Electrolytes (ions) in a solution interfere with each other reducing the “effective
concentration” of dissolved ions. The extent of this interference depends on the “ionic
strength”. Calculate the ionic strengths of each aqueous environment given that:
I = ∑0.5MZ2 (or ∑0.5 x molarity x charge2
2
where “∑” tells you to sum together all the individual “0.5 x molarity x charge2
”s).
Ion or complex Ionic strength
groundwater
Ionic strength
seawater
Ionic strength
rainwater
sodium (Na+
)
magnesium (Mg2+)
calcium (Ca2+)
potassium (K+
)
chloride (Cl-
)
sulphate (SO4
2-
)
bicarbonate (HCO3
-
)
TOTAL IONIC
STRENGTH
Now, it would really be useful if we could estimate to what extent minerals might change
their solubilities when moving between aqueous environments of different ionic strengths.
Converting concentrations to activities (or effective concentrations) allows us to do just
that. Let’s take the fairly insoluble mineral barite as an example.
The equation for barite dissolution is: BaSO4 = Ba2+ (aq) + SO4
2-
(aq)
From the law of mass action, the equilibrium constant = a Ba2+
.a SO4
2-
. This is also called
the solubility product constant or Ksp. Because barite is so insoluble, the actual value of
Ksp is very low, about 10-10. This means that when the product of the activities of dissolved
barium and sulphate ions exceeds 10-10 mol L-1
, barite will no longer be soluble and will
begin to precipitate out from solution.
Now we know that: (a Ba2+
. a SO4
2-
) = 10-10
= (γ Ba2+
. γ SO4
2-
).( [Ba2+].[SO4
2-
] )
and we can work out the activity coefficients (γ) from the ionic strengths using:
-log γ = Az2 {[I
1/2 / (1 + I
1/2)] – 0.2I} (Davies equation)
Given the ionic strengths of each aqueous environment:
Groundwater:
I = (0.5)(1)(1)^2 + (0.5)(0)(2)^2 + (0.5)(0)(2)^2 + (0.5)(1)(1)^2 + (0.5)(1)(-1)^2 + (0.5)(1)(-2)^2 + (0.5)(1)(-1)^2
I = 0.5 + 0 + 0 + 0.5 + 0.5 + 1 + 0.5
I = 3
Seawater:
I = (0.5)(0.5)(1)^2 + (0.5)(0.12)(2)^2 + (0.5)(0.01)(2)^2 + (0.5)(0.016)(1)^2 + (0.5)(0.55)(-1)^2 + (0.5)(0.28)(-2)^2 + (0.5)(2.8)(-1)^2
I = 0.25 + 0.12 + 0.01 + 0.016 + 0.27 + 1.12 + 1.4
I = 3.23
Rainwater:
I = (0.5)(0.001)(1)^2 + (0.5)(0.005)(2)^2 + (0.5)(0.012)(2)^2 + (0.5)(0.001)(1)^2 + (0.5)(0.005)(-1)^2 + (0.5)(0.016)(-2)^2 + (0.5)(0.017)(-1)^2
I = 0.0005 + 0.01 + 0.06 + 0.0005 + 0.0025 + 0.128 + 0.017
I = 0.2185
Now we can calculate the activity coefficients for barium and sulphate ions in each of these environments using the Davies equation. By plugging in the respective ionic strengths into the equation, we can determine the activity coefficients and then use them to calculate the product of the activities of barium and sulphate ions to see if barite would precipitate out.