[Jin]

Hi all,

I have to write down the equations of some reactions I made in a lab and I am very unsure of my results. Could anyone help me on this ?

Here are my datas :

a) H2SO4 + H2O + BaCl2 -----> colourless solution with a white precipitate
b) H2SO4 + KMnO4 + H2O and then + BaCl2 -----> purple solution with dark purple precipitate
b') When adding Na2S2O3 to the solution b -----> colourless solution with white precipitate again

After a few searches, I assume that equation a) is :
 a) H2SO4 + H20 + BaCl2 -----> BaSO4 + 2 H30+ + 2 Cl-

The problems come with b) :
I assume that the purple colour of the solution is due to MnO4- but what about the ppt ?

As of now, I have :
H2SO4 + KMnO4 + H2O -----> KHSO4 + MnO4- + H3O+

But what happens when I add BaCl2 ? Does it react to make Ba(MnO4)2 (purple ppt ?) and HCl ?
Or would KHSO4 be my precipitate and it would only appear purple because of the colour of the solution ?

Whatever, as a final equation I would have :
8 KHSO4 + 8 MnO4- + 3 BaCl2 + 8 H3O+ + 3 Na2S2O3 -----> 8 KHSO4 + 3 BaSO4 + 3 SO4-- + 6 NaCl + 8 MnO2 + 3 H2O

And BaSO4 would be my white precipitate once again. Would that be possible at all ?

Thanks in advance.

[Borek]

First of all - think (and write) in terms of net ionic reactions.

H2SO4 + H20 + BaCl2 -----> BaSO4 + 2 H30+ + 2 Cl-

You are probably correctly recognizing precipitate, but the real reaction is

SO₄^2- + Ba²⁺ -> BaSO₄

Everything else is unchanged.

[vijaychatare]

Hi  read this equation and come to conclusion,
6KHSO4 + 8 MnO4- + 3 BaCl2 + 8 H3O+ + 3 Na2S2O3 ----->6NaHSO4 + 3 BaSO4 + 3 SO4-- + 6 KCl + 8 MnO2 + 3 H2O.
if u hav doubts feel free to contact me.

[Borek]

6KHSO4 + 8 MnO4- + 3 BaCl2 + 8 H3O+ + 3 Na2S2O3 ----->6NaHSO4 + 3 BaSO4 + 3 SO4-- + 6 KCl + 8 MnO2 + 3 H2O

You have mixed two separate processes - thiosulfate oxidation and barium sulfate precipitation. Rewrite it as ionic and cancel out spectators, it will be more obvious.

[AWK]

Problem b is completely wrong formulated - all permanganates are well soluble in water - no purple precipitate can be formed. Instead white BaSO4 precipitate in purple-violet solution

8 KHSO4 + 8 MnO4- + 3 BaCl2 + 8 H3O+ + 3 Na2S2O3 -----> 8 KHSO4 + 3 BaSO4 + 3 SO4-- + 6 NaCl + 8 MnO2 + 3 H2O

In acidic medium MnO2 cannot exist. MnO2 is black-brown!