Question

Experiment 10 Prelaboratory Assignment Vinegar Analysis Lolo Sec. Name I. Asunning the to of a 59. acetic acid by mass sclution is 10 Desk No. 25.0 ml. of density 0.10 M NaoH Also record desemnine volume of aceic acid" dhis calculation your Report Sheet A chemist often uses a white card with a black mark to aid in reading the meniscus of a technique make the reading more clear liquid. How does this b. A wait 10-15 seconds afier dispensi a volume of titrant before ing is made. Explain why the wait is good laboratory technique. ng a The color change at the endpoint should persist for 30 seconds, Explain why the time lapse is a good titration technique. 3 Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due solely to citric acid, that citric acid is a monoprotic acid, and that the density of lemon juice is 1.0 g/mL, then the citric acid concentration calculates to 0.5% by mass. Estimate the volume of 0.0100 M NaoH required to neutralize a 3.1-g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol. Experiment 10 1

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Answer 1

1) We have 5% acetic acid solution, i.e, 100 gm vinegar solution contains 5 gm acetic acid.

Therefore, 1000 gm vinegar solution will have (5*1000)/100 gm = 50 gm acetic acid.

Now, density of the solution is 1 gm/mL; therefore, volume of 1000 gm solution is (1000 gm)/(1 gm/mL) = 1000 Ml = 1 L.

Thus 1 L vinegar solution contains 50 gm acetic acid.

Molar mass of acetic acid = 60.05 gm/mol.

Therefore, moles of acetic acid in 50 gm sample = (50 gm)*(1 mole/60.05 gm) = 0.8326 mole.

Molarity of the supplied vinegar/acetic acid solution = 0.8326 mole/1 L = 0.8326 M (since the volume of the solution is 1 L).

The reaction between acetic acid and NaOH is

CH₃COOH + NaOH ---------> CH₃COO^-Na⁺ + H₂O

The reaction is a 1:1 reaction. Therefore, using the law of molarity, we have

(25 mL)*(0.10 M) = V*(0.8326 M)

===> V = (25*0.10)/0.8326 = 3.002 ≈ 3.0

The volume of acetic acid required = 3.0 mL (ans)

2a) It is easy to identify a white mark or reading against a black background rather than reading a white mark against a white background. Hence, a white card with a black mark is used to make the reading easier and more prominent.

b) The titre (the solution in the burette) can stick to the walls of the container in which the titration is being carried out and thus the wait is ideal to make sure that the titrant droplets have fallen into the solution and are not sticking to the walls. Such sticking to the walls will give an erroneous result.

c) A temporary color change occurs possibly due to a local reaction between the titre (solution in the burette) and the titrant (solution in the beaker) to afford a complex. Most titration reactions are pH dependent; hence if the correct pH is not attained, the said complex will decompose. At the equivalence point, the solution has the correct pH to stabilize the said complex. The complex will no longer decompose at the end point and the color change will persist. Thus, the 30 second wait is required to ascertain the correct equivalence point.

3) The citric acid concentration is 0.5% by mass, i.e, 100 gm lemon juice contains 0.5 gm citric acid.

Therefore, 3.71 gm lemon juice contains (3.71 gm)*(0.5 gm/100 gm) = 0.01855 gm citric acid.

The density of lemon juice is 1gm/mL; therefore, volume of 3.71 gm lemon juice = 3.71 gm/(1.0 gm/mL) = 3.71 mL.

So, 3.71 mL lemon juice contains 0.01855 gm citric acid.

Molar mass of citric acid = 190.12 gm/mol. Therefore, moles of citric acid in 0.01855 gm citric acid = (0.01855 gm)*( 1 mole/190.12 gm) = 9.757*10^-5 mole.

Now, the reaction between NaOH and citric acid is HA + NaOH ------> NaA + H₂O, i.e, 1:1 ratio. Hence, 9.757*10^-5 mole citric acid requires 9.757*10^-5 mole NaOH for neutralization.

The molarity of the given NaOH solution is 0.0100 M (= 0.0100 mole/L).Let the volume required be V mL.

Thus,

V*(1 L/1000 mL)*(0.0100 mole/L) = 9.757*10^-5 mole

===> V = (9.757*10^-5*1000)/0.0100 mL = 9.757 mL ≈ 9.8 mL

The volume of NaOH required = 9.8 mL (ans)

5a) (B) (3) Volume of NaOH used = (final reading – initial reading) = (25.40 – 3.70) mL = 21.70 mL.

(5) Moles of NaOH added = (volume in L)*(concentration in moles/L) = 21.70 mL*(1 L/1000 mL)*(0.0940 mol/L) = 2.0398*10^-3 mole ≈ 2.04*10^-3 mol

(6) Moles of acetic acid in vinegar = 2.04*10^-3 mol (since the reaction between acetic acid and NaOH is a 1:1 reaction).

(7) Mass of CH₃COOH in vinegar = (moles of CH₃COOH)*(molar mass of CH₃COOH) = (2.04*10^-3 mol)*(60.05 gm/1 mol) = 0.1224 gm

(8) Percent by mass of CH₃COOH in vinegar = (0.1224 gm)/(3.06 gm)*100 = 4.00 ≈

5b) The three values are: 4.00%, 5.01% and 4.66%. The average value is (4.00 + 5.01 + 4.66)/3 % = 4.556% ≈ 4.56%