Ingrid CastilloPeriod 5 05/12/2013Larkin

Acid-Base Crime Scene Titration

Introduction:

The purpose of this lab was to find the concentration of the two unknown acids found in

the crime scene by using the titration technique, which involves the use of a base to neutralize

an acid to find the unknown concentration. A titration is a lab method used to determine to

exact concentration of an unknown acid or base. In a titration, the known concentration of base

is poured, from a burette, into a flask of unknown acid; eventually the solution will reach an

equivalence point causing the solution to neutralize. Many acidic and basic solutions are

colorless and odorless, which makes difficult to identify which is a base or acid.

Acids tend to lose hydrogen ions and bases tend to gain hydrogen ions. When a base

and acid react with each other they will neutralize because an acid donates H⁺ and bases

absorbs OH⁻ which produces water and what is left from the acid and base form a salt. Molarity

(M) or molar concentration is a common unit for expressing the concentration of solutions. In

order to find the molar concentration of an unknown acid or base, one has to divide number of

moles of solute by the liter of a solution.

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Procedures:

1. Goggles were put on and the following material were gathered:

100mL graduated cylinder. 50ml burette, stand, and clamp. 125mL Erlenmeyer flask. Phenolphthalein indicator. Glass funnel

2. The stopcock was turned at the bottom of the burette so that it was perpendicular to

the tube of the burette in a closed position.

3. The burette was carefully filled with 1.0 M NaOH base almost to the top. Then the initial

volume of NaOH was recorded.

4. 20.0mL of an acid sample was measured using a graduated cylinder. It was then poured

into the Erlenmeyer flask. The flask was positioned on the top of a white paper towel,

directly beneath the burette.

5. Three drops of phenolphthalein were added to the acid.

6. The unknown acid was titrated with the NaOH by opening the stopcock until there was a

slow stream of NaOH.

7. As the drops fell out of the burette, the flask was gently swirled to make the coloring

disappear. Then any observations were recorded.

8. When the solution became pinker faster and took a longer to swirl before it became

clear again, it was nearing neutralization.

9. When solution became turned clear again, the stopcock was turned very gently to be

“partly open” position. The stream of NaOH should been smaller and individuals drops

came down from the burette.

10. As soon as the coloring does not go away and the solution remains pink, the stopcock

was turned into a closed position.

11. The final volume of NaOH was recorded.

12. The burette was refilled with NaOH and the procedures were repeated for the other 3 examples.

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Results:

Table 1: Data of volume of HCl and NaOH and molarity of HCl found in the experiment.

Exact Volume of HCl (mL)

Initial Volume of NaOH (mL)

Final Volume of NaOH (mL)

Total Volume of

NaOH Used in titration

(mL)

Molarity of HCl (M)

Observations/ Notes

Suspect A Kirkpatrick HCl Acid

Trial 1

20 49.55 40 9.55 .4775 Pink faded after a minute

Trial 2

19.5 40 29.85 10.15 .5205

Suspect B LarkinHCL Acid

Trial 1

20.5 35.1 31.05 4.05 .1975 Barely Pink

Trial 2

21 31.5 26.55 4.45 .2119

Suspect C NguyenHCl Acid

Trial 1

20 45 29.2 15.8 .79

Trial 2

20.5 29.2 19.5 16.7 .8146

Crime Scene HCl Acid

Trial 1

19.5 30 19.5 10.5 .5384 Little too pink

Trial 2

20 43.2 33.5 9.5 .475

Calculations:

(See attached)

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Conclusion:

As the results of the investigation show, suspect A, Mrs. Kirkpatrick, was the murderer.

Mrs. Kirkpatrick’s HCL acid was proven to match the HCL acid of the crime scene. This was

concluded that the molarity of Mrs. Kirkpatrick’s acid was 0.4775 M of HCl which is similar to

the molarity of the crime scene’s acid of 0.475M of HCl. The titration technique was used to

determine the concentration of the unknown acid or base. When the burette was filled with 1.0

M of NaOH base, and then the NaOH was poured into a flask that contained HCl acid in order to

find its unknown concentration. In order to find the unknown concentration of the HCl acid, the

initial volume of NaOH must be recorded, which was 43.2 mL. After the titration of HCl, the final

volume of NaOH was recorded, this was 33.7mL. The total volume of NaOH that was used was

9.5mL, as Table 1 shows. The next step was to find the moles of NaOH used; in order to find the

moles of NaOH the molarity of NaOH was multiplied by the liters of NaOH, which was obtained

by dividing 9.5ml by 1,000ml because 1L equal 1,000ml the moles of NaOH was 0.0095; this was

used to titrate the HCl acid.

NaOH + HCl → NaCl + H₂O

After the moles of NaOH were calculated, the moles of HCl were found. In order to find

the moles of HCl, NaOH was needed to be converted to moles of HCl and this where the

balance equation was used to determine to find moles of HCl. The balance equation was used

for the neutralization reaction of NaOH and HCl. when acid and base neutralize each other, they

tend to produce a salt (NaCl) and water in the reaction. For every mole of NaOH, one mole of

HCl was needed to neutralize the solution, which means that 0.0095 moles of NaOH were

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needed to neutralize the acid; there must have been 0.0095 moles of HCl in the solution. The

final step was to calculate the molarity of HCL, which was done by dividing the moles of HCL by

the liters of HCL used, which resulted in a molarity of .0475M. This process was repeated for

each suspect.

Sources of Error:

As the solution was nearing neutralization, the NaOH was poured into the flask of HCl

very slowly. It could allow CO₂ in the air to react with water in the solution producing carbonic

acid. The presence of another acid would have affected the end result because more NaOH

would have been needed to fully neutralize and causing a higher molarity of HCl to be used in

the process of neutralization. The added carbonic acid than would actually been needed if only

the HCl was in the flask. Another error could have been in the measurement of the volume.

When the stopcock was closed when conducting titrations, there would be a drop of NaOH

hanging at the opening of the burette tube. This would cause lowering the volume inside, but

was not actually in the solution, and thus caused the volume of base used in the calculations to

be higher than the actual volume used. However, these errors would only have shifted the

results slightly.

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