Molarity of NaOH (phenophthalein) 0.1132 Volume of HC2H3O2 (acetic acid) 25.00 ml. vinegar %3D TRIAL A C Initial Volume (ml.) 0.15 0.43 0.31 Final Volume (ml.) 35.81 34.02 33.96 CULATIONS: TRIAL A. C Volume NaOH used Moles NaOH MHC2 H3O2 Rejected Molarities: Arithmetic Mean: II
Molarity of NaOH (phenophthalein) 0.1132 Volume of HC2H3O2 (acetic acid) 25.00 ml. vinegar %3D TRIAL A C Initial Volume (ml.) 0.15 0.43 0.31 Final Volume (ml.) 35.81 34.02 33.96 CULATIONS: TRIAL A. C Volume NaOH used Moles NaOH MHC2 H3O2 Rejected Molarities: Arithmetic Mean: II
Chapter21: Potentiometry
Section: Chapter Questions
Problem 21.22QAP
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![CHM-153 General Chemistry Principles ||
Advance Study Assignment
Name_Fboni 4larding
2/12/21
Date
TITRATION OF AN UNKNOWN VINEGAR SOLUTION
Having read the experiment, use the data below to carry out the calculations called
for.
DATA:
Molarity of NaOH
(phenophthalein)
0.1132
Volume of HC2H3O2
Cacetic acid)
25.00 ml.
vinegar
=
TRIAL
A
C
Initial Volume (ml.)
0.15
0.43
0.31
Final Volume (ml.)
35.81
34.02
33.96
CALCULATIONS:
TRIAL
C
Volume NaOH used
Moles NaOH
MHC2 H3O2
Rejected Molarities:
Arithmetic Mean:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb61f129-4f59-4af9-aedb-1d7e1dad227c%2F10757c30-ee3e-4d0a-b2f0-fae41adab592%2Fxvs0bdj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:CHM-153 General Chemistry Principles ||
Advance Study Assignment
Name_Fboni 4larding
2/12/21
Date
TITRATION OF AN UNKNOWN VINEGAR SOLUTION
Having read the experiment, use the data below to carry out the calculations called
for.
DATA:
Molarity of NaOH
(phenophthalein)
0.1132
Volume of HC2H3O2
Cacetic acid)
25.00 ml.
vinegar
=
TRIAL
A
C
Initial Volume (ml.)
0.15
0.43
0.31
Final Volume (ml.)
35.81
34.02
33.96
CALCULATIONS:
TRIAL
C
Volume NaOH used
Moles NaOH
MHC2 H3O2
Rejected Molarities:
Arithmetic Mean:
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