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According to the experiment procedure, given that the NaOH solution has a concentration of 0.100M, and that the vinegar (acetic...Asked by Pauline
According to the experiment
procedure, given that the NaOH
solution has a concentration of
0.100M, and that the vinegar
(acetic acid) has a concentration of
5.0%m/m (mass percent), what
volume(mL) of titrant would be
required to reach the "equivalence
point", if you place 2.500g of
vinegar in the flask (show
calculations) ?
procedure, given that the NaOH
solution has a concentration of
0.100M, and that the vinegar
(acetic acid) has a concentration of
5.0%m/m (mass percent), what
volume(mL) of titrant would be
required to reach the "equivalence
point", if you place 2.500g of
vinegar in the flask (show
calculations) ?
Answers
Answered by
DrBob222
Vinegar is acetic acid (CH3COOH) but I will shorten that to HAc.
5.0% w/w means 5g HAc in 100 g solution.
g HAc in 2.5g vinegar is 0.05 x 2.5 = about 0.125
mols HAc = g/molar mass = ?
mols KOH = mols HAc
M KOH needed = mols KOH/L KOH. You know mols and M, solve for L and convert to mL.
5.0% w/w means 5g HAc in 100 g solution.
g HAc in 2.5g vinegar is 0.05 x 2.5 = about 0.125
mols HAc = g/molar mass = ?
mols KOH = mols HAc
M KOH needed = mols KOH/L KOH. You know mols and M, solve for L and convert to mL.
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