Analysis of Acid by Titration with Sodium Hydroxide Essay

1169 Words Nov 9th, 2012 5 Pages
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
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The titration of the acid was repeated in three trials. All data and calculations were recorded.

Data Tables:
Solution Table Acid Bottle Code: #9 | Trial # 1 | Trial #2 | Trial #3 | Final buret reading acid (mL) | 9.1 | 9.2 | 10.10 | Volume of acid used (mL) | 9.1 | 9.2 | 10.10 | Final buret reading NaOH (mL) | 26.30 | 44.70 | 45.50 | Initial buret reading NaOH (mL) | 7.91 | 26.30 | 26.50 | Volume of NaOH (mL) | 18.39 | 18.40 | 19.00 | Concentration of NaOH (M) | 0.1 | 0.1 | 0.1 | Molarity of the acid (HCl) | 0.202M | 0.199M | 0.182M | Ratio of volume of NaOH used (mL) to volume of acid used (mL) | | | |

Calculations:
As previously stated in introduction, formula M1xV1=M2xV2 found on page 93 of Survey of Chemistry laboratory manual was used to calculate the concentration of hydrochloric acid (HCl).
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
1.839/9.2 = M2
M2 = 0.199M
Trial 3
M1 = 0.1 (NaOH) V1 = 19.00mL (NaOH) M2=Unknown (HCl) V2=10.10mL (HCl)
0.1 x 18.39=M2 x 10.10
1.839/10.10 = M2
M2 = 0.182M
Average of acid
0.202 + 0.199 + 0.182/3 = 0.194M = M average
Discussion and conclusion:
This lab was successful in the proper color change needed to represent an end point when acid

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