2
$\begingroup$

I have an investigation to do for my chemistry class. I'm looking at copper complexes and using ammonia to create a precipitate in the copper (II) sulfate solution, then redissolving the precipitate to complete the ligand exchange. When I tried the experiment the first time, I was unable to make a good solution. I dissolved about $3.5$ grams of $\ce{CuSO4·5H2O}$ into $200$ mL of water, and couldn't get any more to dissolve. I titrated in $1$ mL of ammonia and the solution turned cloudy (I assumed this was the precipitate). After $4$ mL I stopped, as the solution hadn't changed at all. This was all carried out at room temperature.

I know the two best ways to make a more concentrated solution are time and heat. I don't have that much time, as I'm on winter break and I'll need to complete the experiment as soon as possible when I go back next week. It would probably take a week to get the solution concentrated enough, so I've decided to use heat to help the dissolution along. After looking around the internet, I found very little that was helpful. One page outlined how to make a supersaturated solution, which was not what I wanted. It advised heating to $100^{\circ}$ Celsius, so I suspect that will be too hot for my experiment. A video done by students showed them making a solution at $50^{\circ}$ Celsius, but they used copper (II) oxide. I'm not sure if that makes the procedure significantly different.

I'm not sure what temperature is necessary for this to work. If all else fails, perhaps I'll change my topic to an investigation of the relationship between heat and solubility rates. I'll finally know how to make a saturated solution in a reasonable amount of time.

Other notes: I'll be using a hot plate (Fisher Scientific, 11-500-4H) for the heating, not a Bunsen burner. The ammonia is stock/commercial, and the concentration isn't on the label.

Thanks in advance.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ You're much below what the solubility of the pentahydrate should be. // Copper (II) sulfate will form a hexammine complex. Initially, the light blue hydroxide precipitates, but with the addition of more ammonia the precipitate will dissolve. $\endgroup$
    – MaxW
    Commented Dec 31, 2015 at 2:06
  • $\begingroup$ copper sulfate dissolves slowly. I suggest to move to chloride if possible and work with smaller volumes (25-50 ml at most for test experiments). Also, please post concentration of ammonia solution you used. $\endgroup$
    – permeakra
    Commented Dec 31, 2015 at 8:19
  • $\begingroup$ @MaxW I know the precipitate will form. That's what I'm looking for. It's useless, though, if the precipitate appears as soon as I add a few drops of ammonia. I need a solution with a higher concentration. $\endgroup$
    – kierlani
    Commented Dec 31, 2015 at 14:53
  • $\begingroup$ @permeakra Do you mean copper chloride? And I said at the end that the concentration of the ammonia isn't on the label. I'll be able to determine the concentration through the titration, because the reaction for the formation of the precipitate and its dissolution is known. $\endgroup$
    – kierlani
    Commented Dec 31, 2015 at 14:55
  • $\begingroup$ @kierlani Yes, I mean copper chloride. And with no knowledge of ammonia concentration it is possible that you didn't use enough of it. To my knowledge, a rather concentrated solution of ammonia is required. $\endgroup$
    – permeakra
    Commented Dec 31, 2015 at 17:06

1 Answer 1

Reset to default
2
$\begingroup$

I suggest to use a combination of heating, dissolving, shock cooling and filtrating off the precipitate.

As was already stated, copper(II) sulphate dissolves slowly and is better soluble in the heat. The exact temperature doesn’t matter, but why not just go up to $80$ or $100~\mathrm{^\circ C}$? This will also make the dissolution faster. You can look up the solubility of copper(II) sulphate to calculate how much you would need and use some more to be on the safe side.

Then, cool the solution down rather quickly (but not too quickly or the glass will crack). This will initially create a supersaturated solution but the supersaturated bit will crystallise out rather quickly. You can filter off this crystalline precipitate to get your saturated copper(II) sulphate solution. Do not filtrate too quickly, or the solution will still be too warm and slightly supersaturated.

Then finally remember that if you add too much ammonia, rather than a precipitate of $\ce{Cu(OH)2}$ you might get the redissolved $\ce{[Cu(NH3)4(H2O)2]^2+}$ tetraammindiaquacopper(II) complex. You can tell this by the intensive dark blue colour.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Thank you so much. This is just what I needed. I knew the supersaturated solution would naturally precipitate for equilibrium, but it didn't occur to me to filter that out. And actually I'm titrating to both the precipitation point and the ligand exchange point. (I'm ridiculously happy about this answer, thank you.) $\endgroup$
    – kierlani
    Commented Dec 31, 2015 at 17:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.