Question

-09 244m Wao Estimating the Acid Content of a Soft Drink Lab Worksheet Section: Name: 1. Complete the Data Table below for each of your titrations. Titration #3 Practice Titration Titration #1 Titration #2 Molarity of NaOH used Starting Volume (mL) Final Volume (mL) Net Volume of N OH (mL) Net Volume of NAOH (L) 9744 097490974a .2 .3 2s.7 34,7 S4.2 2. Using the data for Titration #1, show how to calculate the: a) moles of NaOH needed to neutralize the citric acid in the 100-mL sample of Mountain Dew. b) moles of citric acid in the 100-mL sample of Mountain Dew c) grams of citric acid contained in 100-mL of Mountain Dew. 3. A 12-ounce can of Mountain Dew contains approximately 355 mL. Based on your calculations in question 2, how many grams of citric acid are ingested by drinking one can of Mountain Dew ? 4. If you performed more titrations, what is the average mass of citric acid found in 100 mL of Mountain Dew? (Or, you could list the results from several other students.) i. 5. Look back at step 9 of the procedure. Why do we use a faint but persistent change in the color of the solution to mark the completion of the neutralization reaction? Why is it wrong to add NaOH until the reaction acquires a deep and intense color? urns Pik and n6ans se 4-00 much was adlac on the Wa want to Knou 6X0044wha neutralention happans wn a Spacae amount irsc 164

Need help figuring out net volume of Naoh in data table (am I just subtracting starting volume and final volume?) and questions, but need to know if I answered number 5 correctly, please.

Answer 1

Here Net Volume of NaOH calculation (Q1.) , Q2. (with all 3 sub-parts) and Q 5. has been answered. Kindly post other questions separately.

Net Volume of NaOH = Final volume reading - Initial volume reading

The Table in Q1. can be completed:

	Trial1	Trial 2	Trial 3
Starting volume reading (in ml)	0.2	0.3	0.1
Final Volume reading (in ml)	34.2	34.1	35.7
Net volume (in ml)	34	33.8	35.6
Net Volume in (L)	0.034	0.0338	0.0356

Note that 1 L = 1000 ml

Q2. For Titartiom # 1, or Trial 1,

Here, Assuming that this addition of NaOH is done to a 100 ml sample of mountain dew. (Experiment #1 mentiooned in Q6)

a) moles of NaOH used = (Voume of NaOH in litre) x (Molarity of NaOH i.e. mol per litre )

                                     = (0.034 ) x (0.09749)

                                    = 0.0033   or 3.3 x 10^-3

b) Note that C₆H₈O₇ i.e. citric acid has 3 acidic hydrogens. Therefore 3 moles of NaOH are required to neutralize 1 mole of citric acid completely.

That is, if 1 mole of NaOH was used up, (1/3) mole of citric acid would have been present.

For  3.3 x 10^-3 mole of NaOH , (1/3)x 3.3 x 10^-3 = 1.1 x 10^-3 mole of citric acid is present.       

c) grams of citric acid = moles present x molar mass

= (1.1 x 10^-3 mole) x (192 g/mol) = 0.211 g

Q5. The Answer you have written is correct. It should be further explained.

The end-point is detected in citric acid and NaOH titration using Phenolphthalein as indicator. It changes from colorless to pink to deep red as pH increases on addition of NaOH.

The pink color starts to appear even at the start of titration, since citric acid is a weak acid, and absorbs OH- ions slowly. But this pink color disappears on swirling the solution. The end point is detected when the pink color persists for at least a minute. If we add even a single drop of NaOH more, then the pH rise is extremely sharp and the color changes to intense red, which does not indicate the correct end-point. It will lead to erroneous calculations in Q2, solved above.

Kindly submit feedback if explanation is satisfactory.