Standardization of an NaOH Solution Experiment 1: Standardize an NaOH Solution Using Benzoic Acid as Primary Standard
Part 1: Prepare the NaOH Solution.
How many mL of water were used to prepare the NaOH solution?
200 mL
Data Analysis
Calculate the molarity of the NaOH solution. The molar mass of NaOH is 39.997 g/mol.
0.800g/39.99=0.0200015 moles
0.020 mol /0.200 L= 0.1M
Calculate the amount of benzoic acid to be neutralized by 20.00 mL NaOH solution, in both moles and grams. The molar mass of benzoic acid is 122.12 g/mol.
0.244 g
Part 2: Perform a Coarse Titration
Record the following data from your course titration in the table below.
mass of benzoic acid used (g) 0.244 g
pH of benzoic acid solution before titration 2.96
volume of NaOH in the burette before titration (mL) 50 mL
volume of NaOH in the burette after titration (mL) 17.86 mL
volume of NaOH dispensed in the titration (mL) 21.16 mL
Data Analysis
How do you expect your coarse titration volume to compare to your fine titration volumes?
Part 3: Perform Fine Titrations
Lab Results
Record the volume of NaOH solution dispensed in the 3 fine titrations.
20.05 mL, 20.05 mL, 19.95 mL
Data Analysis
Calculate the average concentration of the NaOH solution, using the average volume of NaOH solution dispensed in the 3 fine titrations.
(20.05+20.05+19.95)/3=20.02 mL
0.0200 mol/0.02002 L = 0.999 M
Experiment 2: Use the Standardized NaOH Solution to Determine the Concentration of an Acid
Part 1: Perform a Coarse Titration
Lab Results
What was the pH at the end point of the coarse titration?
11.25
Data Analysis
Based on your coarse titration volume, do you expect the acetic acid solution to have a higher or lower concentration than the NaOH solution?
Part 2: Perform Fine Titrations
Lab Results
For the 3 fine titrations of the acid of unknown concentration, fill in the following data.
Titration #1 Titration #2 Titration #3
volume of acid (mL) 34.05 mL 34.05 mL 34.05 mL
volume of NaOH dispensed (mL) 9.05 mL 9.05 mL 9.05 mL
Data Analysis
Calculate the following quantities and record the data in the table below.
average volume of NaOH solution dispensed (mL) 9.05 mL
average number of moles of NaOH dispensed 9.05 mL/39.997 = 0.226 mol
average concentration of the acid (mol/L) MaVa=MbVb
So: 0.226*9.05=2.05
2.05/25 mL = 0.082 mol/L
Conclusions
Phenolphthalein is pink over the range of pH 8 – 12. Why was it a useful indicator of when the equivalence point was reached?
It provides visual evidence for the equivalence point of the solution in which it is dispensed
Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized solution of NaOH. Given the data in the figure below, what is the concentration of the unknown acid?
volume of 0.125 M NaOH dispensed 24.68 mL
volume of acid solution 50.00 mL
24.68/39.997=0.6170
0.6170*24.68= 15.23
15.23/50 = 0.3046