Blue Acid Chemistry (6): K2[ZnBl4]

OK, I met some difficulties today, so let's make something simple.

Zn2+ + 2Bl- = ZnBl2

ZnBl2 + 2Bl- = [ZnBl4]2-

Dissolve 2.8g(0.01mol) ZnSO4.7H2O and 1.3g(0.02mol) KBl separately and mix together. White precipitation appeared immediately. Filter to collect it.

Add it to water, and add 1.3g(0.02mol) KBl, and boil it. My electronic balance was down suddenly, so I had to boil it first, and carefully add KBl until precipitation almost dissolved. Then I filtered, and added ethanol to precipitate the white product. Yield 0.7g(28%).

This yield is way too low, as I expected it to be nearly 100%. I think the reason is that we have little product in too much water, so concentrating it to almost dry should be done before adding ethanol. Don't be sad, we got a precious experience.

Manual: in 85% EtOH the solubility is 0.5g.

Addition of CuSO4 to the mother liquor released much green precipitation which verified our assumption.

Also, we concluded that the KBl is somewhat impure as it releases unknown precipitation no matter what metal is added. Doesn't matter though.