Question

1. Calculate the activity coefficient and activity of each ion in a solution containing 500 mg/L NaNO₃ and 150 mg/L CaSO₄.

2. One gram of solid silver chloride, AgCl_(s), is added to 1.0 L of estuarine water that contains 0.10 mol/L Cl^- and has an ionic strength of 0.12 mol/L. the solid begins to dissolve according to the following reaction, which has an equilibrium constant of
10^-9.75.

AgCl_(s)  Ag⁺ + Cl^-

Estimate the mass of AgCl_(s) that dissolves per 1.0 L of solution.

Answer 1

1: To calculate the activity coefficient we need to calcualte the ionic strength.

Given concentration of NaNO3 = 500 mg/L = 0.5 g/L

Mlecular mass of NaNO3 = 85.0 g/mol

Hence molarity of NaNO3, C = 0.5 g / 85.0 g/mol / L = 0.0059 M

Given concentration of CaSO4 = 150 mg/L = 0.150 g/L

Mlecular mass of CaSO4 = 136.1 g/mol

Hence molarity of CaSO4 = 0.150 g / 136.1 g/mol / L = 0.0011 M

Now ionic strength of the solution can be calculated as, I = 1/2 x sum(C_iZ_i²) = 1/2 x{[Na+] x Z_Na+² + [NO3-] x Z_NO3-² + [Ca2+] x Z_Ca2+² + [SO42-] x Z_SO42-²}

= 1/2 x {0.0059M x 1² + 0.0059M x 1²+ 0.0011M x (+2)² + 0.0011M x (-2)²} = 0.0103 M

Now we can calculate activity coefficient using Davies equation as follows