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Analysis of Acid by Titration with Sodium Hydroxide

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Stephanie Thao
Chemistry 1151 Laboratory
Analysis of acid by titration with sodium hydroxide
Ms. Hoang
November 2012

Introduction: The purpose of this experiment is to demonstrate an example of how to determine the unknown molarity of hydrochloric acid by titration with a base (sodium hydroxide). Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte (wekipedia). The first step will be measuring and combining water and acid (Hydrochloric acid). An indicator anthocyanin will be added to the solution to change the color to pink. Anthocyanin is a water-soluble vacuolar pigment that may appear red, purple, or blue depending on the pH …show more content…

During this experiment after adding enough sodium hydroxide to the indicator solution; the solution changed color to blue, an indication of the stoichiometric end point. All three trials showed the proper color change. This proved that the color change indicated that the pH level had changed in solution and the amount of base added is chemically equivalent to the acid in the flask. The formula M1xV1=M2xV2 was used to calculate the unknown concentration of hydrochloric acid. The results from the trials showed that the concentration of hydrochloric acid was 0.2M. The concentration of the two acids should be relatively close in order to cancel each other out. It was complicated to compare the two concentrations since the given molarity of sodium hydroxide is in the one decimal place holder, whereas the calculated concentration for hydrochloric acid is in a two decimal place. From the lab it can be concluded that pH indicators can be an imprecise method to calculate the concentration of an unknown concentration.

Critical Thinking Questions
1. Trial 1
M1 = 0.1 (NaOH) V1 = 18.39mL (NaOH) M2=Unknown (HCl) V2=9.1mL (HCl)
0.1 x 18.39 = M2 x 9.1
1.839/9.1 = M2
M2 = 0.202M
Trial 2
M1 = 0.1 (NaOH) V1 = 18.40mL (NaOH) M2=Unknown (HCl) V2=9.2mL (HCl)
0.1 x 18.40 = M2 x 9.2

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