# Question 2d9c7

Jan 21, 2015

When an ionic compound such as $P b S {O}_{4}$ is insoluble in water, it means that the $P {b}^{2 +}$ ions and $S {O}_{4}^{2 -}$ ions have such a strong attraction for each other that the dissolving capabilities of water cannot pull them apart! The ions will stay clumped together rather than being pulled apart.

In order to be classified as insoluble, an ionic compound must have a solubility in water lower than $\text{0.01 mol/L}$.

Since ${\text{PbSO}}_{4}$ is considered to be insoluble in water, that must mean that it has a solubility of less than the above value. Indeed, its solubility is listed at $\text{0.00443 g/100 mL}$ at $\text{20"^@"C}$. In order to express this value in $\text{mol/L}$, we'll use the compound's molar mass

"0.00443" "g"/("100 mL") * ("1000 mL")/("1 L") * ("1 mol")/("303.2 g") = 0.000142# $\text{mol"/L}$

As you can see, this value is much lower than the $\text{0.01 mol/L}$ value used as a threshold to classify insolubility.

This is why, according to the solubility rules, ${\text{PbSO}}_{4}$ is listed as insoluble. The same method can be applied for all the ionic compounds listed as insoluble, since the calculations must confirm their extremely low solubility in water.

The solubility rules:

http://www.csudh.edu/oliver/chemdata/solrules.htm