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​?

Please show your work/equations.

Answer 1

In this case, I think that for the effect of the common ion, the concentration of the solution of PbSO4 would be concentration of Pb in PbCl2 so:

PbCl_2(s) ----------> Pb²⁺_(aq) + 2Cl^-_(aq)  Ksp = 1.7x10^-5

With this, we can calculate the concentration of Cl, and with that the mass of NaCl:

Ksp = [Pb][Cl]²
[Cl] = (Ksp/[Pb])^1/2
[Cl^-] = (1.7x10^-5 / 1.47x10^-3)^1/2 = 0.1075 M

With this, we know that it can be the same concentration of the Na⁺ in solution, which means that the mass of NaCl (and assuming 1 L of solution) will be:

mNaCl = 0.1075 mol/L * 1L * (23+35.5)g/mol = 6.2888 g

Hope this helps