posted by Casey on .
My mind is bleeding again. I am trying to figure out the molarity of acetic acid in Vinegar. We used a standard solution of .613 Molars of NaOH. We added 33.75mL of NaOH to completely titrate/neutralize 25mL of Vinegar and we're supposed to be able to figure out the molarity of acidic acid in the vinegar, but I cannot figure this out. Can anybody help me?
moles NaOH = M x L = 0.613 x 0.03375= ??
So moles acidity in the solution (the vinegar) is the same as moos NaOH since the reaction is a 1 mole to 1 mole ratio.
Since moles vinegar is in 25 mL, then
M = moles/L and M vinegar = moles from above/0.025 = ??
OR, if you don't want to go through all the explanations, just use this neat formula (good when 1:1 mole ratio exists as it does in this problem).
mL NaOH x M NaOH = mL vinegar x M vinegar.
Solve for M vinegar.