The tutor tackles an essential concept of grade 12 chemistry.

I introduced solubility product back in my Dec 1 post. Today the question is slightly different: How much of a salt will dissolve in a solution that already contains one of its ions?

Example 1: What is the solubility of lead(II)chloride in a 0.150 M solution of NaCl?

Solution:

The dissociations of NaCl and PbCl₂ are as follows:

NaCl(s) → Na_aq⁺ + Cl_aq^–
PbCl(s) → Pb_aq²⁺ + 2Cl_aq^–

The solubility product of PbCl₂ is given by

k_sp=[Pb²⁺][Cl^–]²=1.2×10^-5

Let the concentration of PbCl₂ in moles per litre be given by x. Then the following ion concentrations are present:

Putting the concentrations into the k_sp expression gives

k_sp=x(2x+0.150)²=1.2×10^-5

Now, we can make the approximation that 2x+0.150≈0.150. Hebden (see source below) argues as follows: First, realize that 1.2×10^-5=0.000012. If x(2x+0.150)²=0.000012, then x<0.000012/(0.150)². Therefore, x<0.000533; it follows that 2x< 0.00107. If 2x < 0.00107, then 2x + 0.150 < 0.00107 + 0.150 = 0.151 in significant digits. The 2x term is worth, at most, 0.001/0.150 or 0.7%; the approximation that 2x+0.150≈0.150 is valid. Applying the approximation, we arrive at x(2x+0.150)²≈x(0.150)²=1.2×10^-5

We continue

x(0.150)²=1.2×10^-5

x=1.2×10^-5/(0.150)²
x=0.00053

Apparently, the solubility of PbCl₂ in 0.150M NaCl is 0.00053 or 5.3×10^-4M. Perhaps next post I’ll compute its solubility in pure water by way of comparison.

Source:

Hebden, James A. Chemistry: Theory and Problems, Book Two. Toronto: McGraw-Hill   Ryerson Limited, 1980.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.