Activity Calculation Examples

Solubility of calcium fluoride in 0.01 F sodium sulfate

Shown here is a case of the dissolution of an 'insoluble' salt in a solution of known ionic strength. The ionic strength is known because of the presence of a soluble salt. The solubility of CaF₂ will be calculated in 0.01 F Na₂SO₄. The solubility product constant for CaF₂ is K_sp = 4.0´10^-11.

Consider first the molar solubility without taking activity effects into account. The chemical equilibrium is

CaF₂ (s) Û Ca²⁺ (aq) + 2 F^- (aq)

The solubility product equation is

K_sp = A_CaA_F²

^{The activities are assumed to be equal to concentrations (the activity coefficients are unity), thus}

K_sp = [Ca²⁺][F^-]²

Using the table method for molar solubility

Substitution into the solubility product equation

K_sp = (x)(2x)² = 4´10^-11

Solving for x, the molar solubility is found to be x = 2.15´10^-4.

Now we account for the lower activity due to the presence of ions in solutions. To account for activity changes, we need to know the ionic strength of solution and the ionic radii of species participating in the chemistry, i.e., Ca²⁺ and F^-. Since the molar solubility of CaF₂ is low, the presence of Ca²⁺ and F^- in solution are assumed to have negligible effects on the ionic strength relative to the 0.01 F Na₂SO₄ that the CaF₂ is dissolved in. Sodium sulfate is a soluble salt. In solution there is 0.02 M Na⁺ and 0.01 M SO₄^2-. The ionic strength is

m = ½[(.02)(+1)² + (.01)(-2)²]

m = 0.03

The ionic radius of Ca²⁺ is a_Ca = 600 pm and that for F^- is a_F = 350 pm. Using the Debye-Hückle equation

log₁₀g = -0.51z²Öm/[1+[aÖm/305)]

one obtains g_Ca = 0.54 and g_F = 0.84

Now one solves for the molar solubility of CaF₂. Again, the solubility product equation and the solubility table are used. But this time, substitute the activity coefficient-concentration product for the activities

K_sp = A_CaA_F²

^{Making the substitutions}

K_sp = [Ca²⁺]g_Ca[F^-]²g_F²

Rearranging and substitution of the molar solubility, or 'x' values, from the solubility table yields

K_sp / (g_Cag_F²) = (x)(2x)²

Putting in the numbers,

4´10^-11 / [0.54´(0.84)²] = 4 x³

The solution is x = 3´10^-4.

Taking into account the activity effect of having a high ionic strength solution of Na₂SO₄ increases the molar solubility of CaF₂ to x = 3´10^-4 M, a ~50% increase due to an ion that does not directly participate in the chemistry!

Solubility of lithium fluoride

The following example is more complicated than that shown the above. In particular, the ionic strength changes with the dissolution of the salt. This is important when the salt is relatively soluble. This is a difficult problem since the ion from dissolution of the salt are responsible for increasing the ionic strength of solution. The problem is solved using iteration methods.

Question: What is the molar solubility of LiF in pure water given only K_sp?

Solution

First, write out the appropriate chemical and algebraic equations

LiF (s) Û Li⁺ (aq) + F^- (aq) K_sp = 1.7´10^-3

K_sp = [Li⁺][F^-]g_Lig_F

_{Second, solve for the molar solubility using iteration methods;}

Step 1: Assuming that g_Li = 1 and g_F = 1, calculate [Li⁺] and [F^-]

Then use

K_sp = x², x = ÖK_sp, x = 0.041 M

Step 2: Calculate g_Li and g_F based on concentrations of Li⁺ and F^- from previous step, and with a_Li = 600 pm and a_F = 350 pm

m = ½[(.041)(+1)² + (.041)(-1)²]

m = 0.041

using the Debye-Hückle equation

log₁₀g = -0.51z²Öm/[1+[aÖm/305)]

one obtains g_Li = 0.85 and g_F = 0.83

Step 3: Calculate new x = [F^-] = [Li⁺] using

K_sp/(g_Lig_F) = 2.4´10^-3

x = ÖK_sp/(g_Lig_F) = 4.9´10^-2

Step 4: Calculate new g_Li and g_F based on concentrations of Li⁺ and F^- from step 3

Step 5: Continue to calculate x = ÖK_sp/(g_Lig_F) using steps 2, 3, and 4 until the value of x does not change. This generally takes 4 or more iterations.

The calculations are most easily set up in a spreadsheet program. In this case, one need only copy and paste the column(s) updating x as a function of solution concentrations.

By the way, the iteration calculations converge with g_Li = 0.84 and g_F = 0.81, and with x = 0.05 M. Thus, the molar solubility is 0.05 M. Without accounting for chemical activity, the molar solubility was found to be 0.041 M (Step 1). Thus, there is a ~25% increase in molar solubility over that predicted from the solubility product equation neglecting activity effects.

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This page was created by Professor Stephen Bialkowski, Utah State University.

Last Updated Monday, August 28, 2006