calculate volume of 0.500M sulfuric acid needed to react completely with 200.0ml of 1.25 M barium hydroxide? How many ml of 0.500 M phosphoric acid will be needed to neutralize the same base?

Question

anonymous · Answer

Okay, so I'm just working through this. So:

$$H _{2} SO _{4} (aq) + Ba(OH)_{2} (aq) ----> 2H_{2}O (l) + BaSO_{4} (s)$$

I check and it balances...

anonymous · Answer

Oh, wait...
Back up...

jim_thompson5910 · Answer

I'm thinking you're going to have sulfuric acid (H2SO4) along with barium hydroxide (Ba(OH)2) and you're going to have something like this 

H2SO4 + Ba(OH)2 --> ??? 

the question is 

a) what goes on the right side (I'm not sure, but I'm thinking H2O and some combo of S and BA, but idk 

b) how to balance out that equation once you have the proper right side (I've completely forgotten how to do this) 


I could be way off on this though, so not sure


It looks like ShotPutJones has it (not sure on that either lol)

anonymous · Answer

Well, I know you need to figure your molar ratio for the rxn.  Normally: 1mol H2SO4/ 1 mol Ba(OH)2, but we're working with 0.500 Molarity for H2SO4 and 1.25 Molarity for Ba(OH)2.

anonymous · Answer

You need .4 mol of 1.25M Ba(OH)2 to react with each mol of 0.500M H2SO4.

anonymous · Answer

So if you have 200.0mL of 1.25M Ba(OH)2, you will need 500.0mL of H2SO4 to completely react it.
I would not take this to the bank, just problem solving.  I'd like verification.

anonymous · Answer

sorry not really sure wish i could help