Question

Calculate the activity coefficients and the effective concentrations of all the ionic species in a solution containing 0.01 M MgCl2 and 0.02 M Na2SO4.

Answer 1

Recall that ionic strength considers all ions in solution, and its charges. It is typically used to calculate the ionic activity of other ions. The stronger the electrolytes, the more ionic strength they will have.

The formula:

I.S. = 1/2*sum( Ci * Zi^2)

Where

I.S. = ionic strength, M (also miu / μ ) used

Ci = concentration of ion “i”

Zi = Charge of ion “i”

The exercise:

IS = 1/2* ( (0.01)(2)^2 + (0.02)(1)^2 + (0.04)*(1)^2 + (0.02)(2)^2 )

IS = 0.09

Where

γi = activity coefficient for species “i”

αi = theoretical diameter in pm (10^-12 m)

Zi = Charge of ion

I.S. = ionic Strength (usually used as μ as well)

If we wanted only γ

γ = 10^-(0.51*(Zi^2)*sqrt(I.S.) / ( 1 + (α * sqrt(I.S)/305)))

given IS = 0.09

γ = 10^-(0.51*(Zi^2)*sqrt(0.09 / ( 1 + (α * sqrt(0.09)/305)))

now... find alpha for each ion

α-Mg = 800

α-Cl = 300

α-Na = 450

α-SO4 = 400

get each coefficient

γ-Mg = 800; +2

γ = 10^-(0.51*(2^2)*sqrt(0.09 / ( 1 + (800 * sqrt(0.09)/305))) = 0.3484

γ-Cl = 300; -1

γ = 10^-(0.51*(1^2)*sqrt(0.09 / ( 1 + (300* sqrt(0.09)/305))) = 0.7337

γ-Na = 450; +1

γ = 10^-(0.51*(1^2)*sqrt(0.09 / ( 1 + (450* sqrt(0.09)/305))) = 0.7458

γ-SO4 = 400

γ = 10^-(0.51*(2^2)*sqrt(0.09 / ( 1 + (400* sqrt(0.09)/305))) = 0.3031

then... effectiv econcentrations

[Mg+2] = 0.01*0.3484 = 0.003484

[Cl-] = 0.02*0.7337= 0.014674

[Na+] = 0.04*0.7458 = 0.029832

[SO4-2] = 0.02*0.3031 = 0.006062