3
$\begingroup$

While attempting to make Schweizer's reagent in household conditions, using $\ce{CuSO4 + NaOH + 25\% NH4OH}$ route (all at least 95% pure, either technical or lab grade purity), I've stumbled upon a problem. While trying to precipitate $\ce{Cu(OH)2}$, the temperature of the solutions rose to about 50 Celsius, and instead of expected blue precipitate, I got an almost black (with a hint of violet) one instead, which I assumed is $\ce{CuO}$. However, I was able to dissolve some of the precipitate in $\ce{NH4OH}$ slightly (a suspension was made from about 10 g of precipitate in 200 ml ammonia), resulting in dark blue solution with only traces of suspended black particles ($\ce{CuO}$?), appearing very like the expected Schweizer's reagent itself.

The resulting substance was able to dissolve regular cotton & cellulose samples overnight, however the dissolution was quite slow and unimpressive, with some of the sample material remaining almost intact; a chalk-coated paper was quite resistant to it (although it also dissolved somewhat eventually).

I don't have prior experience with Schweizer's reagent, so I don't know if the rate of dissolution I got here was the proper, expected behaviour here or a sign of dud.

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

As side questions:

  1. was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
  2. how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\% NH4OH}$ route) in dissolving various celluloses in practice?
$\endgroup$
2
$\begingroup$

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.

  1. was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?

Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water. To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.

  1. how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\% NH4OH}$ route) in dissolving various celluloses in practice?

This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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