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CHM1311-Lab 4,Acid-Base titration
Principles of Chemistry (University of Ottawa)
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Experiment : 4
cid-Base Titrations
A
Chemistry 1311:Section Z08
Student Name : Aditi Biswas
Student Number:300091661
Lab Partner:Aidan Davies
Dr. Rashmi Vankateswaran ,TA:Emily Schaefer
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Introduction:
Acids and bases are defined by the Bronsted-Lowry theory,which states:when an acid
and a base react with each other, the acid forms its conjugate base and the base forms
its conjugate acid by exchange of a hydrogen cation (H+ ). These two substances are
used in calculating blood pH,body homeostasis, and neutralization reactions. Acids
increase acidity,therefore decreasing pH and bases inversely increase pH.When they
dissolve in water, strong acids dissociate to H+ ions and bases dissociate to form OH-
ions.
In this experiment, different concentrations of known and unknown acids were titrated
against a standardized base (NaOH).The equation to determine the concentrations of
solutions is:
HCl+NaOH=NaCl+H2O
This formula will give the amount of moles per litre of a substance in a solution.
In this experiment, A monoprotic acid gives off 1 hydrogen ion in
solution(diprotic=2,triprotic=3).
Ref:Nguyen, Kao, & Kurtz,2011
The goal of this experiment was to determine the concentrations of an unknown acid
sample and determine the percentage by mass of acid.This was done by titrating these
unknown solutions with a standardized solution of NaOH.
Procedure : As outlined in the lab manual.
Observation:
Table 1 :
NaOH solution :
Concentration of distilled water 250 ml
Volume of NaOH 4.1 ml
Table 1 shows the initial volume of NaOH(4.1 ml), when 250 mL of water was added.
Before and after the addition of NaOH to the beaker, the solutions were both clear and
colourless.
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Table 2 : Standardization of the diluted NaOH :
Trial 1 Trial 2 Trial 3
Concentration of HCl 0.1 M 0.1 M 0.1 M
Volume of HCl 10 ml 10 ml 10 ml
pH of HCl 2.25 2.34 2.37
Final volume of NaOH 11.4 ml 11.2 ml 13.3 ml
pH of NaOH 11.4 11.2 11.7
Table 2 shows the initial concentration and volume of HCl acid and the final volume of
NaOH and their pH in 3 trials. HCl was a clear colourless solution. 2 drops of
phenolphthalein were added to the acidic solution before the titration. The colour at the
end point was a bright pink.
Table 3 : Titration of Unknown Acid :
Trial 1 Trial 2
Volume of unknown acid 10 ml 10 ml
Concentration of unknown acid 0.1 M 0.1 M
Final volume of NaOH 10.8 ml 11.7 ml
Volume of water in the beaker 99.2 ml 108.3 ml
Total volume in the beaker 120 ml 120 ml
pH of the solution 11.2 11.5
Table 3 shows the volume and pH of the unknown acid and the volume of the NaOH
solution. It also present the final pH of the solution.The unknown acid was a clear
colourless solution that turned pink with the addition of 2 drops of phenolphthalein when
the end point was reached.
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Graphs :
Fig-1:Titration of HCl with NaOH(trial 1)
Fig 2:Titration of HCl with NaOH(trial 2)
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Fig 3:Titration of HCl with NaOH(trial 3)
Graphs for titration of unknown acid:
Fig 4: Titration of unknown acid with NaOH(trial 1)
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Fig 5: Titration of unknown acid with NaOH(trial 2)
Calculation :
Concentration of NaOH solution (approximately):
C1V1=C2V2
C2=(6 M x 0.041L)/0.25 L
C2=0.984 M
Concentration of NaOH :
Trial 1 :
CNaOH VNaOH = CHCl VHCl
CNaOH = (0.1 M x 0.01 L)/0.0114 L
CNaOH = 0.0877 M
Trial 2 :
CNaOH VNaOH = CHCl VHCl
CNaOH = (0.1 M x 0.01 L)/0.0112 L
CNaOH = 0.0892 M
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Trial 3 :
CNaOH VNaOH = CHCl VHCl
CNaOH = (0.1 M x 0.01 L)/0.0133 L
CNaOH = 0.0752 M
Average concentration of NaOH = (0.0877 M +0.0892 M +0.0752 M)/3
=0.084 M
Concentration of unknown acid :(For diprotic acid )
Trial 1 :
CNaOH VNaOH = CAcid VAcid
CAcid = ½(0.1 M )(10 ml)/(10.8 ml)
= 0.04630 M
Trial 2 :
CNaOH VNaOH = CAcid VAcid
CAcid = ½(0.1 M )(10 ml)/(11.7 ml)
= 0.04274 M
Average concentration of the unknown acid = (0.04630 M + 0.0427 M)/2
=0.0445 M
The unknown acid is Maleic acid (C4H4O4).
Mass percentage of acid = [(0.0445 M )(116.1 g/mol)/1.59 g/cm3 ]x 100%
=0.3249%
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Discussion :
After the results of this experiment were calculated, it was determined that the
concentration of the unknown acid was 0.0445 M and the percentage of the acid is
1.2%.As the actual values are not available,it is impossible to know what the percent
error actually was. However,there are a few possible error.
1.Firstly, the same beaker was used for all trials.The beaker was washed after every
trial,however there could be still traces of acid/base in the beaker, which could cause
the substance to reach its equivalence point faster/slower than it otherwise would.
2.Secondly, the drop counter accurately counting drops,the drop counter could have
missed any number of drops which would lead to an incorrect amount of NaOH being
used in the titration.
All of these errors will cause inaccurate titration curves as well as inaccurate calculated
equivalence points.
The Values observed in the table were very similar to those calculated by Logger Pro,
they only differed by decimal numbers.
Conclusion :
The concentration of an unknown acid and the concentration and the percentage of an
acid in an unknown sample was determined by titrating an unknown acid with a
known(NaOH) base.Lastly,it was found that the concentration of the unknown acid was
0.0445 M with a percentage of 0.3249%.
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