Description: Vinegar is a solution of acetic acid, CH 3 COOH, which is a weak ac
ID: 994151 • Letter: D
Question
Description: Vinegar is a solution of acetic acid, CH3COOH, which is a weak acid. In this experiment you will determine the actual amount of acetic acid in commercial vinegar to 4 sf. This will be done by using the titration technique from the previous NaOH lab. Household vinegar contains between 5-6% m/m acetic acid. You will be provided with a ~0.1M solution of NaOH, which you need to standardize to 4 sf with KHP (that is, the 0.1 M is an approximation). You will then need to prepare a solution of about 0.1M KHP; the exact concentration to 4sf will be calculated when the exact mass of KHP is obtained in lab.
1) What is the molarity of a household vinegar labeled as containing 5.0% m/m acetic acid? (The density of the solution is 1.056 g/mL.)
2) How much of the ~0.1 M NaOH solution would you need to titrate 25.00 mL of the above solution to the endpoint?
3) Obviously, the household vinegar must be diluted in order to titrate with the ~0.1 M NaOH. A student took 10 mL of this vinegar with a volumetric pipet and diluted it with water in a 100.0 mL volumetric flask. Then 25.00 mL of the diluted vinegar was titrated with the standardized NaOH, which was 0.1106 M. The volume of NaOH used was 18.94 mL. What was the concentration of the diluted vinegar?
4) What is the %m/m of the undiluted vinegar?
Please show the calculations and explaination.
Explanation / Answer
1)
molarity = mass % x 10 x density / molar mass
= 5 x 10 x 1.056 / 60
= 0.88 M
2)
at end point mimmoles of acid = millimoles of base
0.88 x 25.00 = 0.1 x V
V = 220 mL
volum of NaOH needed = 220 mL
3)
in dilution : M1 V1 = M2 V2
10 x 0.88 = M2 x 100
M2 = 0.088 M
molarity of diluted vinegar = 0.088M
4)
molarity = mass % x 10 x d / molar mass
0.088 = mass % x 10 x 1.056 / 60
mass % = 0.5 %