CHM-153 General Chemistry Principles ||Advance Study AssignmentName_Fboni 4larding2/12/21DateTITRATION OF AN UNKNOWN VINEGAR SOLUTIONHaving read the experiment, use the data below to carry out the calculations calledfor.DATA:Molarity of NaOH(phenophthalein)0.1132Volume of HC2H3O2Cacetic acid)25.00 ml.vinegar=TRIALACInitial Volume (ml.)0.150.430.31Final Volume (ml.)35.8134.0233.96CALCULATIONS:TRIALCVolume NaOH usedMoles NaOHMHC2 H3O2Rejected Molarities:Arithmetic Mean:

Given data,Molarity of NaOH=0.1132MVolume of acetic acid=25.0mL=0.025L

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

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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:

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