determine the molar concentration of acetic acid solution with a mass fraction of acid in the solution of 9.12%. the density of the solution is 1g/ml.
Answers
We're asked to find the molar concentration of the
CH
3
COOH
solution, with some known titration measurements.
Let's first write the chemical equation for this neutralization reaction:
NaOH
(
a
q
)
+
CH
3
COOH
(
a
q
)
→
CH
3
COONa
(
a
q
)
+
H
2
O
(
l
)
Let's find the number of moles of
NaOH
using the given molarity and volume (which we convert to liters):
mol NaOH
=
(
0.1047
mol
L
)
(
0.02865
L
)
=
0.003000
mol NaOH
Using the coefficients of the equation, let's now find the relative number of moles of
CH
3
COOH
used up:
0.003000
mol NaOH
(
1
l
mol CH
3
COOH
1
mol NaOH
)
=
0.003000
mol CH
3
COOH
Lastly, using the given volume of acetic acid solution (agin converting to liters), let's find the molarity of the acetic acid solution:
molarity
=
mol solute
L soln
=
0.003000
l
mol CH
3
COOH
0.025
l
L soln
=
0.12
M
rounded to
2
significant figures.
thank you