CHE 101 7/29

4

Liu

CHE101

7/29

Lab19: Tritation

Pre-labQuestions

1.What is the difference between equivalence points and end points?

Themain difference between equivalence point and endpoint is thatequivalence point simply pertains to the point in titration that nomore reaction between a base and acid will take place despite moretitrant is dropped into the solution. The end point may indicate theequivalence point or other steps in the entire titration process by acolor change. This color change is due to the reaction of excesstitrant to an indicator. In other words, end points are a result ofthe reaction between an indicator and the titrant, while equivalencepoint is a result between the titrant and the chemical species beingtitrated – the analyte (US Department of Energy, 2014).

2.What would happen if you forgot to put the indicator in?

Ifthe analyst forgot to put the indicator, then the end point of thereaction mixture cannot be determined. This means that it will beimpossible to account how many molecules of titrant reacted with thetitrated chemical species at any particular time. This is true, ofcourse if the reaction between the analyte (the chemical speciesbeing titrated) and the titrant is not, in itself a coloredreaction). There are also reactions that require no indicator,because the reaction between the titrant and the analyte is a coloredreaction or a reaction that diminishes the initial color – considerthe reaction between boiled cabbage and bleach (Henrie, 2002).

3.If 2.078 g of Smarties requires 2.11 mL of 0.250 M Na2CO3toreach the endpoint, how many moles of acid is there in each gram ofthe candy?

Thereis one carbonate ion in a single molecule of Na2CO3.Thesecarbonate ions (CO32-)have a (-2) charge which means that it will take 3 moles of Na2CO3toneutralize 2 moles of Smarties.

Thechemical equation for this reaction is:

2HOC(CO2H)(CH2CO2H)2 + 3 Na2CO32 HOC(CO2H)(CH2CO2Na)2 + 6H2O

Molesof Acid:

0.00211LNa2CO3 (0.250 mols Na2CO3/LNa2CO3)= 5.275 x 10-04molesNa2CO3.

5.275x 10-04molesNa2CO3(2molesHOC(CO2H)(CH2CO2H)2/3 moles Na2CO3)= 3.52 X 10-04molesHOC(CO2H)(CH2CO2H)in 2.078 g of Smarties or

3.52X 10-04molesHOC(CO2H)(CH2CO2H)/2.078 g of Smarties = 1.69 X 10 10 -4moles acid /g of Smarties.

Therefore,there are 1.69 moles of acid molecules for every gram of Smarties.

Data

Table1:Titrationof grape juice indicator

Measurement

Grape Juice Trial 1

Grape Juice Trial 2

Molarity of Na2CO3

Initial syringe volume (mL)

Final syringe volume (mL)

Volume Na2CO3required (mL)

Average volume Na2CO3required (mL)

Table2:Titrationof candy

Measurement

SweeTarts Trial 1

SweeTarts Trial 2

Smartie

Trial 1

Smarties Trial 2

Mass of candy (g)

Concentration of Na2CO3, Mb(on the bottle)

Initial syringevolume (mL)

Final syringevolume (mL)

Volume Na2CO3solution required,Vb(mL)

Calculations

  1. Determine the volume of Na2CO3 used to titrate only the candy. HINT: Take the Volume of Na2CO3 used in Part 2 (candy and juice mixture) and subtract the volume required to titrate the grape juice alone (Part 1 average).

    1. SweeTarts Trial 1:

    2. SweeTarts Trial 2:

    3. Smarties Trial 1:

    4. Smarties Trial 2:

  2. Determine the amount of Na2CO3used per gram of candy (mL NaOH/g).HINT:Divide the volumes obtained above by the mass of each candy sample.

    1. SweeTarts Trial 1:

    2. SweeTarts Trial 2:

    3. Smarties Trial 1:

    4. Smarties Trial 2:

  1. Determine the average amount of base used per gram of candy for each set of trials (mL Na2CO3/g).

    1. SweeTarts Average:

    2. SweeTarts Average:

  1. Determine the molesacid per gram of candy.HINT:This is found by dividing the moles of acid found for each type of candy by the mass of the candy initially weighed out and recorded in the data table. Complete in the calculation in a stepwise process.

Molesof acid SweeTarts Trial 1:

    1. Write the balanced reaction equation.

    2. Determine the mol to mol ratio of the acid to base (What are the values a andb?).

    3. Apply the equation using your data in Table 1.(HINT: molesacid = a x Mbx Vb/ b . Be sure to convert mL to L.)

    4. Divide the number of moles of acid by the mass in grams of the candy used.

Molesof acid SweeTarts Trial 2:

RepeatSteps a through d for Trial 2.

Findthe average of the two trials: Average Moles acid = _

Molesof acid Smarties Trial 1:

RepeatSteps a through c for Smarties Trial 1.

Molesof acid Smarties Trial 2:

RepeatSteps a through c for Trial 2.

Findthe average of the two trials: Average Moles acid = _

Post-labQuestions

1.Which candy required more base per gram of candy? Is this what youexpected? Explain your answer.

Iam expecting that Smarties will require more base to neutralize. Thisexpectation is in accordance to the fact that Smarties has more acidgroups in its acid molecule than Sweet Tart does. There are threeacid groups in Smarties while there are only two acid groups in SweetTart. This means that it requires 1 mole of sodium bicarbonate toneutralize 0.67 mole of Smarties acid, while it takes 1 mole ofsodium bicarbonate to neutralize 1 mole of Sweet Tart acid (Henrie,2002).

2.Which candy contained more moles of acid per gram of candy? Is thiswhat you expected? Explain your answer.

Consideringthat Smarties has more acid groups in every molecule of the acid thanSweet Tart, then if each sample of both candies contained the sameamounts of acid molecules per grams, then it can be expected thatSmarties will have more moles of acids than Sweet Tarts. Note that inthe laboratory experiment sample calculations, Sweet Tart requiresfewer moles of NaOH to neutralize compared to Smarties (2:3). Thisratio means that Smarties has one mole of acid over Sweet Tart permolecule of both candies (Henrie, 2002).

3.How would your Na2CO3/gof candy reported change if there was a large air bubble in thestopcock when you started your titration?

Theair bubbles will result to positive error. Note that the air bubblewill not be involved in the acid base reaction but it does add to thereading of volume of the base used for titration. These means thatthe analyst will read higher volume of the base used compared to whatreally reacted. This means that there will be a higher reading ofNa2CO3 used than there should be. More Na2CO3 means that more of itreacted with the acid hence the report will show that there is moreacid molecules than there should be (Henrie, 2002).

References

Henrie,S. (2002). Greem Chemistry Manual. Tennessee: Union University.

USDepartment Of Energy (2014). Ask a Scientist. Retrieved from:&lthttp://www.newton.dep.anl.gov/askasci/chem03/chem03996.htm&gt.