Question

An aqueous solution is prepared by dissolving 100 mg/L ofCaCl2 and 230 mg/L of Na2SO4 in water. Assuming that these salts dissociate completely,and that the dissociated ions do not undergo any additional reactions, calculate(a) the molar concentration of each ionic species;(b) the ionic strength of the solution;(c) the activity coefficient for each species (with the Davies equation);(d) the activity of each species

Answer 1

(a) The molar masses of CaCl2 and Na2SO4 are 111 g/mol and 142 g/mol respectively.

The number of moles of CaCl2 are $\frac {100mg \times 1g}{111g/mol \times 1000mg}= 0.000901 mol$

Thus

$[Cl^-]=2 \times 0.000901 mol/L =0.00180 mol/L$

The number of moles of Na2SO4 are $\frac {230mg \times 1g} {142 g/mol \times 1000mg}=0.00162mol$

Hence, $[SO_4^{2-}]=0.00162mol/L$

$[Na^+]= 2 \times 0.00162mol/L = 0.00324 mol/L$

(b) The ionic strength of the solution is

$I = 0.000901 \times (2)^2 + 0.00180 \times (1)^2 + 0.00324 \times (1)^2 + 0.00162 \times (2)^2 = 0.0151$

(c) The activity coefficent fro Na2SO4 will be same as the activity coefficient fro CaCl2. It will be

$-log f = 0.5 z1z2(\frac {\sqrt {I}}{1+\sqrt{I}}-0.15I)$

$-log f = 0.5 \times 2 (-1) (\frac {\sqrt {0.0151}}{1+\sqrt {0.0151}}-0.15 \times 0.0151)=0.1072$

$f=1.28$

(d) The activity of Ca²⁺ ion is $1.28 \times 0.000901= 0.00115$

The activity of Cl^- ion is $1.28 \times 0.00180 = 0.00230$

The activity of Na⁺ ion is $1.28 \times 0.00324= 0.00415$

The activity of SO₄^2- ion is $1.28 \times 0.00162 = 0.00207$