Question

What mass of NaCl will result in the precipitation of PbCl₂ from a 1.47 x 10^-3 M solution of PbSO₄? K_sp = 1.7 x 10^-5.

A) 6.52 g

B) 6.38 g

C) 0.676 g

D) 0.729 g

Answer 1

The equation for the precipitation of PbCl₂ by the reaction of NaCl with PbSO₄ is

Given, the solubility product of PbCl₂ = 1.7*10^-5 and Concentration of PbSO₄ solution = 1.47*10^-3 M, i.e. [Pb²⁺] = [SO₄^2-] = 1.47*10^-3 M

The ionic product product of PbCl₂ can be written as [Pb²⁺][Cl^-]²

Condition for the precipitation of PbCl₂ from the solution of PbSO₄ using NaCl is

[Pb²⁺][Cl^-]² > Ksp

1.47*10^-3 [Cl^-]² > 1.7*10^-5

i.e. [Cl^-]² > (1.7*10^-5) / (1.47*10^-3)

  [Cl^-]² > 1.156*10^-2 M

i.e. [Cl^-] > 1.075*10^-1 M

Molarity of a substance is the no. of moles of that substance present in 1 L of the solution.

Hence, no. of mol of Cl^- > 1.075*10^-1

no. of moles of a substance = Mass of the substance (g) / molecular mass (g.mol^-1) of the substance

So, mass of Cl^- > 1.075*10^-1*58.5

i.e. mass of Cl^- > 6.291 g

In the given options, C (0.676 g) and D (0.729 g) are less than 6.291, whereas A (6.52 g) and B (6.38 g) are greater than 6.291.

Therefore, both A and B options are the possible answers.