Codie Trent
New member
...volume of acetic acid? the equation of the titration is:
NaOH + CH3COOH --> NaCH3COOH + H2O
the volume of NaOH used is .0232L and the concentration is .10M
the number of mole works out to be .00232 mol
a 3.01g sample of acetic acid is used with a molar mass of 60.06g
there is a 1:1 ratio so the number of mole of acetic acid is also .00232mol
i have worked out the grams of acetic acid by multiplying the number of moles by the molar mass (.00232 x 60.06 = .0139g acetic acid)
from this i worked out the percentage by mass of acetic acid by dividing the grams by the mass of sample used (.139g / 3.01g = .046179 i then multiplied by 100 to get a percentage .046179 x 100 = 4.62%)
however im having trouble working out the concentration of acetic acid as the volume used is unknown!
id really appreciate it if someone could help me out
thanks
NaOH + CH3COOH --> NaCH3COOH + H2O
the volume of NaOH used is .0232L and the concentration is .10M
the number of mole works out to be .00232 mol
a 3.01g sample of acetic acid is used with a molar mass of 60.06g
there is a 1:1 ratio so the number of mole of acetic acid is also .00232mol
i have worked out the grams of acetic acid by multiplying the number of moles by the molar mass (.00232 x 60.06 = .0139g acetic acid)
from this i worked out the percentage by mass of acetic acid by dividing the grams by the mass of sample used (.139g / 3.01g = .046179 i then multiplied by 100 to get a percentage .046179 x 100 = 4.62%)
however im having trouble working out the concentration of acetic acid as the volume used is unknown!
id really appreciate it if someone could help me out
thanks