LAB 20C

Feb. 19, 2015 Liangting Lin 1-4 Lab Partner: Sunny Zhao
20C: Acid-Base Titration
Purpose/Objective:
1. To titrate a hydrochloric acid solution of unknown concentration with 0.50M sodium hydroxide, and determine the molarity of the hydrochloric acid
2. To titrate an acetic acid solution with 0.50M sodium hydroxide, and determine the molarity and percentage composition of the vinegar.

Apparatus and Materials:
Please refer to P.236 of Health Chemistry Laboratory Experiments.

Procedure:
Part I Determination of molarity of Hydrochloric Acid Solution

Part II Determination of Percentage Composition of Vinegar

Data and Observations
Part I Determination of Molarity of Hydrochloric Acid Solution

Table 1 Volume of NaOH Needed to Neutralize 10.00 ml of Unknown HCl
Molarity of NaOH=
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Initial volume of NaOH (ml)

Final volume of NaOH (ml)

Volume of NaOH used

Average volume of NaOH

Part II Determination of Percentage Composition of Vinegar

Table 2 Volume of NaOH Needed to Neutralize 10.00 ml of Vinegar
Molarity of NaOH=
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Initial volume of NaOH (ml)

Final volume of NaOH (ml)

Volume of NaOH used

Average volume of NaOH

Calculations
1. Moles of NaOH = 0.500 mol/L X 0.010525 L = 5.26 x 10-3 mol

2. NaOH + HCl --> H2O + NaCl
NaOH: HCl 1:1 ratio
Moles of HCl = Moles of NaOH = 5.26 X 10-3 mol

3. Molarity of HCl = 5.26 x 10-3 mol / 0.010 L = 0.526 M

4. Moles of NaOH = 0.500 mol/L x 0.016275 L = 8.14 X 10-3 mol

5. CH3COOH + NaOH --> H2O + CH3COONa
CH3COOH: NaOH 1:1 ratio Moles of CH3COOH = Moles of NaOH = 8.14 X 10-3 mol

6. Molarity of CH3COOH = 8.14 x 10-3 mol/ 0.010 L = 0.814 M

7. Mass of CH3COOH in 1L solution = 0.814 mol/L X 60.0g/mol = 48.8g

8. Mass solution = 1.00g/ml x 1000ml = 1000g
Percentage of acetic acid = 48.8g/ 1000g x 100%