Question

1. At 25°C, 298.25 mL of an aqueous solution of NiCl2 with a molar analytical concentration of 0.00004873 M is added to 487.37 mL of an aqueous solution of AlBr3with a molar analytical concentration of 0.00005000 M.

What is the equilibrium concentration of Ni2+ in the resulting solution?

2. What is the equilibrium concentration of Al3+ in the resulting solution?

3. What is the ionic strength of the resulting solution?

4. What is the activity coefficient of Ni2+ in the resulting solution?

5. What is the activity coefficient of Al3+ in the resulting solution?

Answer 1

1. moles of NiCl2 = molarity x volume = 4.87 x 10^-5 x 0.29825 = 1.45 x 10^-5 mols

moles of AlBr3 = 5.0 x 10^-5 x 0.48737 = 2.44 x 10^-5 mols

Total volume of solution = 0.29825 + 0.48737 = 0.786 L

equilibrium concentration of Ni2+ in solution = 1.45 x 10^-5/0.786 = 1.84 x 10^-5 M

2. Equlibrium concentration of Al3+ in solution = 2.44 x 10^-5/0.786 = 3.10 x 10^-5 M

3. Ionic strength of solution (I) = 1/2 sum of (CiZi^2)

= 1/2 [(1.84 x 10^-5 x 2^2 + 1.84 x 10^-5 x 2) + (3.10 x 10^-5 x 3^2 + 3.10 x 10^-5 x 3)]

= 1/2(7.36 x 10^-5 + 3.68 x 10^-5 + 2.79 x 10^-4 + 9.30 x 10^-5)

= 2.41 x 10^-4

4. Activity coefficient for Ni2+

effective diameter of Ni2+ = 0.6 nm

activity coefficient y can be written as,

log y = -0.51 x 2^2 x sq.rt(2.41 x 10^-4)/1 + 3.3 x 0.6 x sq.rt(2.41 x 10^-4)

         = -0.0323/1.03

       y = 0.930

5. AlCl3 is a strong electrolyte, so we would used the Debye-Huckel extended rule,

log y = -0.51 x Z1^2 x sq.rt(I)/1 + 3.3 x effective diameter of ion x sq.rt(I)

effective diameter of Al3+ = 0.9 nm

activity coefficient y can be written as,

log y = -0.51 x 3^2 x sq.rt(2.41 x 10^-4)/1 + 3.3 x 0.9 x sq.rt(2.41 x 10^-4)

         = -0.713/1.046

      y = 0.208