It was found that 20.55mL of alkali was needed to neutralise the acid. What is the molarity of the acetic acid in the vinegar?
2007-04-11 13:51:23 · 4 answers · asked by angeybutt 1 in Science & Mathematics ➔ Chemistry
Neutralization occurs when the reaction
NaOH+HOAc = NaOAc + H2O is complete
For titration, (VbMb) = (VaMa) where b refers to the base and a refers to the acid. BTW: don't refer to NaOH as "alkali", it can be confusing.
You have Va,Vb and Mb. So solve for Ma.
2007-04-11 13:58:49 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
MaVa=MbVb Ma*25.00=0.1*20.55 Ma=0.0822M
2007-04-12 16:29:00 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The formula is CH3COOH + NaOH = Na(CH3COO) + H2O
[OH-] = Moles / Litres (NOTE [ ] means concentration)
Rearrange for moles = [OH-]*Litres
= 0.1*20.55/1000
= 2.06*10^(-3)
So 1 mole of NaOH will neutralise 1mole of CH3COOH
So the number of moles CH3COOH in the solution is also 2.06*10^(-3)
Using the equation above
[CH3COOH] = 2.06*10^(-3)/(25.00/1000)
= 8.24*10^(-2)
2007-04-11 21:06:54 · answer #3 · answered by ktrna69 6 · 0⤊ 0⤋
0.0822 M
2007-04-11 21:01:19 · answer #4 · answered by misoma5 7 · 0⤊ 0⤋