[kentucky_girl]

This is a three part problem, and I got A correct. But b and c are wrong and I don't know why.

How many milliliters of .141M HCl are needed to titrate each of the following solutions to the equivalence point?

a). 39.9 mL of .0950M sodium hydroxide (This is the one I got correct)

moles NaOH = .0399L x .0950 = .0037905
V = .0037905/.141 = .0268829787 L = 26.8829787 mL

b). 40.7mL of .149M barium hydroxide
moles BaOH = .0407L x .149 = .0060643
V = .006043 / .141 = .0430092199 L = 43.0092199 mL

c). 50.0 mL of a solution that contains 1.46g of sodium hydroxide per liter
mol HCl = 1.46 / 36.0729 = .0404735965 x .050 = .0020236798
V = .0020236798 / .141 = .0143523392 L = 14.3523392 mL

[Yggdrasil]

b)  You have the wrong formula for barium hydroxide.
c)  You have the wrong molar mass for NaOH is wrong.

[kentucky_girl]

Ok, so for b, the formula is Ba(OH)2, but I don't understand why that would affect my calculations? Would it affect my number of moles?

and for c, I redid it and got:

1.36/39.996 = .0365018251 x .050 = .0018250913 mol
.0018250913/.141 = .0129439096L = 12.9439096 mL

[enahs]

Ok, so for b, the formula is Ba(OH)2, but I don't understand why that would affect my calculations? Would it affect my number of moles?

Seeing is how you do not actually say what solution B is, and you are getting it wrong; I would say yes it is effecting your calculations.

[Yggdrasil]

Ok, so for b, the formula is Ba(OH)2, but I don't understand why that would affect my calculations? Would it affect my number of moles?

Writing out a balanced chemical equation for the reaction between barium hydroxide and hydrochloric acid would be helpful here.

[kentucky_girl]

I'm sorry... I carelessly left out the actual problem for b. :\ I edited it... thanks for pointing that out.