The molarity of NaOH = 0.5M, 30mL. Please i need help with these. The following

Question

The molarity of NaOH = 0.5M, 30mL. Please i need help with these. The following readings where gotten during a titration between NaOH and white vinegar. Trial 1=7.6mL, trial 2= 8.0mL. Average volume of NaOH used =7.8mL. question 1. What is the concentration of CH3CooH in vinegar(mol/L). 2. %CH3COOH in vinegar. 3. If the manufacturer used 5.0% acetic acid, what would the percent error between your result and the manufacturer's statement be?. 4. If a 7.0mL,sample of vinegar was titrated to the stoichiometric equivalence point with 7.5mL of 1.5M NaOH, what is the mass percent of CH3COOH in the vinegar sample?. 5. Why is it important to do multiple trials of titration, instead of only one trial. 6. If a student vinegar in the titrator instead of NaOH and put NaOH in the beaker and added phenolpthalein to the solution in the beaker, what would he expect to see happen for results?

Explanation / Answer

1. What is the concentration of CH3CooH in vinegar(mol/L).

The reaction between NaOH and CH3COOH is as follows:

NaOH + CH3COOH = CH3COONa + H2O

Reaction occurred in 1:1 ratio thus the concentration of vinegar:

M1V1= M2V2

M2 = M1V1/V2

= 0.5 * 7.8 ml/ 5.00 ml

= 0.78 M

2. %CH3COOH in vinegar.

Assume that the density of vinegar is 1 g/ ml, so the mass of vinegar is 5 g.

Now calculate the moles of NaOH ,then mole of acetic acid in vinegar:

Moles of NaOH: 0.5* 7.8*10^-3

= 3.9*10^-3 moles NaOH or CH3COOH

Mass of 3.9*10^-3 moles CH3COOH

= 3.9*10^-3 moles * 60.05 g/ moles

= 0.234 g

% of CH3COOH in vinegar = amount of acetic / amount of vinegar *100

0.234 g/ 5 g*100

4.68 %

3. If the manufacturer used 5.0% acetic acid, what would the percent error between your result and the manufacturer's statement be?

% error = absolute value / the theoretical value * 100

= 4.68/5.0 *100

= 93.6%

4. If a 7.0mL,sample of vinegar was titrated to the stoichiometric equivalence point with 7.5mL of 1.5M NaOH, what is the mass percent of CH3COOH in the vinegar sample?.

Concentration of vinegar :

M1V1=M2V2

7.5*1.5= M2*7.0

M2 = 1.61 M concentration of vinegar

Assume that the density of vinegar is 1 g/ ml, so the mass of vinegar is 7 g.

Now calculate the moles of NaOH ,then mole of acetic acid in vinegar:

Moles of NaOH: 1.5* 7.5*10^-3

= 0.01125 moles NaOH or CH3COOH

Mass of 0.01125 moles CH3COOH

= 0.01125 moles * 60.05 g/ moles

= 0.6755 g

% of CH3COOH in vinegar = amount of acetic / amount of vinegar *100

0.6755 g/ 7 g*100

9.65 %

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