All work must be shown with units cancellation and proper format: 10.0mLs of vin
ID: 536132 • Letter: A
Question
All work must be shown with units cancellation and proper format:
10.0mLs of vinegar is pipetted into a 200mL volumetric flask and diluted to volume with water. (Note- vinegar is just a specific concentration of acetic acid in water.) 10.0mls of this diluted solution is then transferred to a 100mL Erlenmeyer flask, 25mLs of water is added as is 2 drops of indicator. The solution is then titrated to an end point (complete reaction) with 15.0mLs of 0.025M NaOH.
a. What is the concentration of the acetic acid in the diluted solution?
b . What is the concentration of the acetic acid in the vinegar solution?
Explanation / Answer
The answers are
a) Concentration of Acetic acid in diluted solution = 0.0375M
b) Concentration of Acetic acid in vinegar solution = 0.75M
Explanation
The reaction of acetic acid with NaOH is
CH3COOH + NaOH -------> CH3COO- + NaCl
1 mole of CH3COOH react 1 mole of NaOH
10ml of Vinegar solution is diluted to 200ml
Dilution = 200ml/10ml= 20 time
Volume of diluted vinegar solution for titration,V1 = 10ml
Molarity of Vinegar solution,M1 = ?
Volume of NaOH solution consumed ,V2 = 15ml
Molarity of NaOH solution , M2= 0.025M
Since , 1 mole of NaOH react with 1mole of CH3COOH
V1M1 = V2M2
M1= V2 M2/V1
= 15 ml × 0.025M/10ml
= 0.0375M
Therefore,
Molarity of CH3COOH in diluted solution = 0.0375MM
Since vinegar solution 20time diluted
Molarity of CH3COOH in Vinegar solution= 0.0375M× 20 = 0.75M