10ml of approximately ~0.1M HCH3CO2 with 40ml of DI water is titrated with 0.0759 M of NaOH.

Find the original molarity of acetic acid with the given information

a) the equivalence point is 13.75ml of NaOH
b) half equivalence point is 6.9ml of NaOH
c) half equivalence point pH is 4.61
d) pka = 4.61

Can someone direct me into figuring out the problem?

Question

10ml of approximately ~0.1M HCH3CO2 with 40ml of DI water is titrated with 0.0759 M of NaOH. 

Find the original molarity of acetic acid with the given information

a)	the equivalence point is 13.75ml of NaOH
b)	half equivalence point is 6.9ml of NaOH
c)	half equivalence point pH is 4.61
d)	pka = 4.61

Can someone direct me into figuring out the problem?

aaronq · Answer

why don't you just add the volume of NaOH used, then using it's molarity find the moles. These are equivalent to the moles of acetic acid.

anonymous · Answer

so would it be as simple as converting the equivalence point ml of NaOH to liters and then multiplying it to its molarity to get the moles?

0.01375L * 0.0759mol/ L = 0.00104 mol ?

aaronq · Answer

yeah, i don't see why that wouldn't work.

aaronq · Answer

once you found the molarity of the 50 mL portion, shrink down to the original volume.

anonymous · Answer

got it. thanks for the help @aaronq!

aaronq · Answer

no problem !