Chemistry

bonkysleuth SG

Hello. I just want to find out what happens when you add H2SO4 into potassium hydrogencarbonate. I know a colourless solution and gas are formed. Is the gas produced CO2? May I know how you can accurately make a guess regarding the products formed from reactants such as Potassium Hydrogencarbonate? I simply do not know how you get the answers with regards to the products formed. This is because Potassium hydrogencarbonate is of 3 different names. and usually, you only have to deal with compounds with 2 names. that makes giving the names of products formed an easier job.

Garrick_3658 SG

Isn't Potassium HydrogenCarbonate an acidic salt? I only learnt of their presence though, their properties are unknown to me. Sorry can't help.

UltimaOnline SG

KHCO3 is basic because the HCO3- ion can still accept a proton (H+ ion).

Adding protons (H+) from sulfuric acid (H2SO4 --> H+ + HSO4- --> 2H+ + SO4 2-) to hydrogencarbonate ion (HCO3-), you get carbonic acid :

HCO3- + H+ ---> H2CO3(aq)

H2CO3(aq) ---> CO2(g) + H2O(l)

Colourless gas is CO2.

Colourless solution is potassium hydrogen sulfate (KHSO4) and potassium sulfate (K2SO4).

ohnoez! SG

sorry for hijacking. is iron 2 sulphate a reducing agent? and why?

UltimaOnline SG

sorry for hijacking. is iron 2 sulphate a reducing agent? and why?

May act as oxidizing agent as well as reducing agent, depending on what other species are present to react with.

Fe2+ can be reduced to Fe. (hence oxidizing agent).

Fe2+ can be oxidized to Fe3+. (hence reducing agent).

SO4 2- can be reduced to S2- (hence oxidizing agent)

SO4 2- can be reduced to SO2 (hence oxidizing agent)

(and so on.)

The above are only a few brief examples of the many possibilities you have to begin considering at 'A' levels (betcha you can't wait to go JC).

At 'O' levels though, FeSO4 is usually neither a reducing nor an oxidizing agent. But if reacted with a strong reducing or oxidizing agent, it won't be too difficult to explore and work out the possibilities (even at 'O' levels).