Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

1. A Phillips antacid tablet, weighing 0.5729 g, was dissolved and diluted to 25

ID: 484236 • Letter: 1

Question

1. A Phillips antacid tablet, weighing 0.5729 g, was dissolved and diluted to 250.00 mL using the procedures described on p. 66 of your lab manual (see item 1). The resulting solution was observed to have an orange coloration (see Figure 2). Three 20.00 mL aliquots of the antacid tablet solution were prepared (see p.67, item 4) and titrated to the Calmagite endpoint shown in Figure 3. Calculate the quantities necessary to complete Table 2.

Molecular Weight Na2EDTA•2H2O = 372.24 g/mole

Molecular Weight Mg(OH)2 = 58.326 g/mole

Table 2: Antacid Tablet Analyses Results

Trial 1

Trial 2

Trial 3

Burette Start (mL):

0.00

0.00

0.00

Burette End (mL):

39.55

38.60

38.61

Na2EDTA Added (mL):

39.55

38.60

38.61

Moles Na2EDTA:

Conc. Mg2+ (moles/L):

Average Conc. Mg2+ (moles/L):

Total Moles Mg2+ in Antacid Tablet:

Total Moles Mg(OH)2 in Antacid Tablet:

Total Grams Mg(OH)2 in Antacid Tablet:

Total Milligrams Mg(OH)2 in Antacid Tablet:

Difference from Manufacturer Label (%):

Trial 1

Trial 2

Trial 3

Burette Start (mL):

0.00

0.00

0.00

Burette End (mL):

39.55

38.60

38.61

Na2EDTA Added (mL):

39.55

38.60

38.61

Moles Na2EDTA:

Conc. Mg2+ (moles/L):

Average Conc. Mg2+ (moles/L):

Total Moles Mg2+ in Antacid Tablet:

Total Moles Mg(OH)2 in Antacid Tablet:

Total Grams Mg(OH)2 in Antacid Tablet:

Total Milligrams Mg(OH)2 in Antacid Tablet:

Difference from Manufacturer Label (%):

Explanation / Answer

0.5729 g tablet dissolved in 250 ml water

molarity of Mg(OH)2 = (0.5729/58.326) / 0.250 = 0.039 M according to the label

now,

reaction equation.

3, 20 ml solutions of this were neutralized by Na2EDTA

Mg2+ + 2 (OH)2- + 2 Na2EDTA = (MgEDTA)2- + 4 NaOH

we have done the calculations using an assumption that 0.25 mol/l of the solution in the burette is taken. please correlate if the concentration used in burette is different.

formulas applied are as follows:

moles of EDTA = (0.25*volume in trials)/1000

Trial 1 Trial 2 Trial 3 Na2 EDTA 39.55 38.6 38.61 moles of Na 2EDTA 0.009888 0.00965 0.009653 moles of Mg2+ 0.004944 0.004825 0.004826 Concentration 0.247188 0.24125 0.241313 Molesof Mg2+ in 250 ml 0.001978 0.00193 0.001931 Moles of MgOH2 0.005933 0.00579 0.005792 g of MgOH2 0.346019 0.337708 0.337795 mg of MgOH2 346.019 337.7075 337.795 Difference(%) 60.3978 58.94703 58.9623