Question #6c52e

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Answer 1 · 2015-04-11T10:02:30

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

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So, starting with mercury (I) chloride, or $H g_{2} C l_{2}$ $Hg_2Cl_2$ . 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 $H g^{2 +}$ $Hg^(2+)$ .

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

Use the same approach for all the compounds listed.

Sodium sulfide, or $N a_{2} S$ $Na_2S$ , is soluble in water because all compounds that contain the sodium cation, $N a^{+}$ $Na^(+)$ , are soluble in water. The same is true for compounds that contain the ammonium ion, $N H_{4}^{+}$ $NH_4^(+)$ , so ammonium phosphate, ${(N H_{4})}_{3} P O_{4}$ $(NH_4)_3PO_4$ is soluble in water.

Cadmium carbonate, $C d C O_{3}$ $CdCO_3$ , and lead (II) sulfate, $P b S O_{4}$ $PbSO_4$ , are both insoluble in water because they do no represent soluble exceptions for compounds formed with the carbonate, $C O_{3}^{2 -}$ $CO_3^(2-)$ , and phosphate, $P O_{4}^{3 -}$ $PO_4^(3-)$ anions.

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