# Question #6c52e

Apr 11, 2015

The easiest way to go about solving solubility problems is to examine the solubility rules

So, starting with mercury (I) chloride, or $H {g}_{2} C {l}_{2}$. As you can see, you're dealing with a halide or, more specifically, with a chloride. Notice that all halides are soluble with the exception of those formed with three cations, including $H {g}^{2 +}$.

As a result, mercury (I) chloride will be insoluble in water.

Use the same approach for all the compounds listed.

Sodium sulfide, or $N {a}_{2} S$, is soluble in water because all compounds that contain the sodium cation, $N {a}^{+}$, are soluble in water. The same is true for compounds that contain the ammonium ion, $N {H}_{4}^{+}$, so ammonium phosphate, ${\left(N {H}_{4}\right)}_{3} P {O}_{4}$ is soluble in water.

Cadmium carbonate, $C \mathrm{dC} {O}_{3}$, and lead (II) sulfate, $P b S {O}_{4}$, are both insoluble in water because they do no represent soluble exceptions for compounds formed with the carbonate, $C {O}_{3}^{2 -}$, and phosphate, $P {O}_{4}^{3 -}$ anions.