# Why does silver sulfate not precipitate in this double displacement reaction?

Aqueous silver nitrate is mixed with aqueous sodium sulfate. We were asked to perform this reaction in lab in my chemistry class, and we did not observe a precipitant being formed. However, the equation indicates otherwise:

$$\ce{2AgNO3 (aq) + Na2SO4 (aq) -> Ag2SO4 (s) + 2NaNO3 (aq)}$$

This shows the formation of $\ce{Ag2SO4}$, silver sulfate. I found in the Wikipedia article for silver sulfate a reference to "ruby red illumination" and thus am led to believe that, like some silver-based substances used in photographic film, the precipitant might have been immediately been affected by light in the room and thus decomposed into something soluble in water. I have not, however, been able to find references to this anywhere else; I have found here a reference to it being classified as "sparingly soluble," but I haven't the slightest idea what this means.

• Also I remember from my ion lottery lab course that the precipitation of silver halides is not clouded by sulphate present … i.e. one can detect the absence of halides in the presence of sulphate with silver nitrate. – Jan Oct 27 '15 at 22:05
• You can check solubilities Wikipedia article you linked... – Mithoron Oct 27 '15 at 22:06
• As far as "sparing soluble" I"ll add that "soluble" and "insoluble" are really absolutes that don't really exist. We'd say that NaCl is soluble but only so much will dissolve. We'd say that SiO2 is insoluble but water flowing over it on a geological time scale will dissolve it. So as Einstein said - It's all relative. – MaxW Oct 27 '15 at 22:25
• Actually, "complete solubility" does exist; it's called miscibility. E.G. water and ethanol, or silver and gold, are miscible in any proportion. – DrMoishe Pippik Oct 28 '15 at 0:26
• The solvent must be the solute and solute must be the solvent -- all is one. A koan? – DrMoishe Pippik Oct 28 '15 at 17:39

Since the lab is past due now, I'll give what I think is the answer.

$\ce{Ag2SO4}$ is somewhat soluble in water. It is most likely that you simply didn't get enough to cause precipitation.

The other possibility, which I don't think applies here, is that you have a supersaturated solution. There are some precipitates for which crystals are just stubborn to nucleate.

• I think it really does not matter if the lab is done or not. I think the OP showed reasonable effort and subsequently asked for help to understand the concept of solubility better. – Martin - マーチン Oct 12 '16 at 7:25
• The solubility product constant for silver sulfate is: $6.0\times10^{-5}$. The concentrations of the reagents used (i.e. silver nitrate and sodium sulfate) has to be high enough: $[\ce{Ag+}]^2[\ce{SO4^2-}] > 6.0\times10^{-5}$. A better answer can be found here. However, here, the $K_\mathrm{sp}$ for silver sulfate is said to be $1.2\times10^{-5}$. – Coco Feb 22 '18 at 0:54
• @Coco Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have reformatted your answer and converted it to a comment, as it does not really answer the question on its own. – Martin - マーチン Feb 22 '18 at 7:28

It could have been the temperature and conc. At 100C 1.33g of silver sulphate will dissolve in 100mL of water. Maybe if your cooled the solution you would see the precipitate. I just dissolved metallic silver in a solution of H2SO4 and H202. I got a white precipitate that I would assume is silver sulphate.

Because in the lab, for this experiment, we used the concentrations of AgNo3 0.1M solution or Na2so4 0.1 M solution is not enough strong to make the reaction happen. Therefore, the precipitates aren't formed.