Asked by Nina
100.0 mL Vinegar (d=1.06g/mL) contains 4% acetic acid.
a. How many moles of acid are in 100 mL of vinegar?
b. How many mL of 0.11 M NaOH are required to neutralize the vinegar?
a. How many moles of acid are in 100 mL of vinegar?
b. How many mL of 0.11 M NaOH are required to neutralize the vinegar?
Answers
Answered by
DrBob222
Is that 4% w/w. If so, that means 4g acetic acid/100 g soln.
4 g acetic acid = 4/molar mass = approximately 0.7 mols (but you need to do that exactly).
Convert 100 g soln to L, then M vinegar = mols/L.
For part b, mL NaOH x M NaOH = mL vinegar x ZM vinegar.
4 g acetic acid = 4/molar mass = approximately 0.7 mols (but you need to do that exactly).
Convert 100 g soln to L, then M vinegar = mols/L.
For part b, mL NaOH x M NaOH = mL vinegar x ZM vinegar.
Answered by
Nina
That is super helpful, thanks. I missed a two days of class when we learned acids and bases, so I was a bit stuck.
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