Finding Solublity in g/100 mL

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the solubility of the compound in \(\frac{g}{100~mL}\)?

This problem exists because we don’t have devices that measure moles directly. Therefore, knowing the molar solubility is not all that helpful. In the end, this problem is about turning molar solubility into a usable value.

We start this problem by finding the molar solubility of the compound. Given that we solved this problem on this page, we’ll jump right to the answer for that portion:

Molar solubility = “x” = 0.016 M.

The first thing we need to do is fix the unit moles, by converting it to grams. That’s a relatively simple problem using molar mass:

\(0.016 \frac{moles~PbCl_2}{L} \cdot \frac{278.10~g~PbCl_2}{1~mole~PbCl_2}=4.45 \frac{g}{L}\)

This is now a useful and workable amount. It tells us that we can dissolve, at most, 4.45 g of \(PbCl_2\) in a liter of solution.

For some reason, that is beyond the reasoning of this author, it was decided that solubilities should be recorded in terms of the grams that will dissolve in 100 mL, rather than in 1 L.

Solving this is easy...IF you think. If you try to set up a conversion process, you will get it wrong. I promise.

Here’s the simple logic. 100 mL is \(\frac{1}{10}\) of a liter. Therefore you can just divide the value we had before by 10.

So, the solubility of \(PbCl_2\) is 0.445 \(\frac{g}{100~mL}\).