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Thursday, July 22, 2021

Finding Solublity in g/100 mL

Let’s solve this problem:

The Ksp of PbCl2 is 1.7105, what is the solubility of the compound in g100 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:

0.016moles PbCl2L278.10 g PbCl21 mole PbCl2=4.45gL

This is now a useful and workable amount. It tells us that we can dissolve, at most, 4.45 g of PbCl2 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 110 of a liter. Therefore you can just divide the value we had before by 10.

So, the solubility of PbCl2 is 0.445 g100 mL.

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