you measure 5.3g of KOH and dissolve it in enough water to make 400 ml of solution. what is the molarity of this base? You want to neutralize this base with a 3.7M H2SO4. How much sulfuric acid is necessary to do this?

Question

you measure 5.3g of KOH and dissolve it in enough water to make 400 ml of solution. what is the molarity of this base? You want to neutralize this base with a 3.7M H2SO4. How much sulfuric acid is necessary to do this?

x=5.3g/39+16+1 = 5.3/56=.095m

M=moles/L of solution = .095/.4 = .2375
is this right so far? im not sure what else to do, please help.

Answer 1

No,not quite. I would have been a bit more careful with organization.

To Wit:

MolesKOH=massKOH/molmassKOH=5.3/56
MolarityBase=above/.4=.095M

Now, the titration equation:
NormalityBase*volumebase=NormalityH2SO4*VolumeAcid

.095N*.4Liters=VolumeAcid*7.4N

Then, solve for volume of acid.