What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?

I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.

My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {->} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$

But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid?

Please help, thanks :)

  • 1
    $\begingroup$ You're right, it is soluble, given enough water. However, potassium nitrate is WAYYY more soluble, so it will dissociate first. If there's not enough water for both of them to dissociate, the barium sulfide will precipitate out first. $\endgroup$
    – wes3449
    Feb 5 '15 at 14:54
  • $\begingroup$ Oh, okay. So in general, when doing these types of questions, what guidelines should I abide by to figure out if a compound in the product is aqueous or not (other than solubility rules)? $\endgroup$ Feb 5 '15 at 15:10
  • $\begingroup$ Did the question specify which product was the precipitate? $\endgroup$
    – wes3449
    Feb 5 '15 at 18:20
  • $\begingroup$ @wes3449 it didn't, unfortunately. the question was what i typed verbatim $\endgroup$ Feb 5 '15 at 19:32
  • $\begingroup$ See my answer for a moderately detailed explanation. $\endgroup$
    – wes3449
    Feb 5 '15 at 19:52

So basically this problem is bad and incomplete (they tend to be at this level) in that it doesn't give you the concentrations/volumes of the reagents/solutions, so we end up making several assumptions. First of all, precipitates cannot be aqueous, by definition they are solids. So the way you end up solving this question is by assuming one of the products precipitates out completely, and it is the least soluble one that precipitates out (barium sulfide in this case). Unfortunately, this isn't exactly accurate.

In a real world scenario, given enough water you could have no precipitate at all, but most likely you would have a mix (not all of the barium sulfide would precipitate out of solution). At higher levels of chemistry (AP/IB -> post-secondary) you'll learn that none of these reactions (including the solubility ones) ever go to completion (some just go so far that you can assume they go to completion), and what you end up doing is using solubility constants to find out how much of each product is dissolved and/or precipitated. If you're interested in more info google "solubility constants" and "common ion effect".

  • $\begingroup$ How do you know which compound is "more soluble" than another? $\endgroup$ Feb 5 '15 at 23:57
  • $\begingroup$ Use a solubility chart such as this one $\endgroup$
    – wes3449
    Feb 6 '15 at 13:56

Generally speaking, if the problem doesn't give concentration or enough information to find concentration, you can't know whether all or only a portion of the product will dissolve. You need to know the solubility product constant (Ksp), the final volume, and the amount of each ion to calculate the amount that will dissolve.

However, this particular question has a complicating issue because:

$\ce{2BaS2 + 2H2O -> Ba(OH)2 + Ba(SH)2}$


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