In this exercise the amount of calcium carbonate (CaCO_3) in both an antacid tab

Question

In this exercise the amount of calcium carbonate (CaCO_3) in both an antacid tablet, and an unknown sample must be determined and reported. The chemical analysis method used will be a titration of Ca^2+ with the titrant EDTA. EDTA is a ligand that will from a complex ion with Ca^2+. EBT is a dye that will be used to indicate the equivalence point of the titration. EBT is also a ligand, and therefore will form a different complex ion with Ca^2+ than that of EDTA. In solution the two ligands, EDTA and EBT, compete for the Ca^2+ ion. EBT has the property that when it is bonded to Ca^2+ it forms a RED complex. When it is not bonded to Ca^2+ (or is free from Ca^2+) it is BLUE in color. As EDTA is added to the dissolved CaCO_3 solution, Ca^2+ ions will break the bonds they have initially formed with the EBT dye and make new bonds with the EDTA to form a more stable complex. This process "frees up" EBT molecules in the solution. Each molecule of EDTA makes___ coordinate covalent bonds to one Ca^2+ cation. Enter an integer as an answer. Do not include units. Enter Your Answer: 1 mmol of Ca^2+ bonds with ___ mmol of EDTA. Enter an integer as an answer. Do not include units. Enter Your Answer: When you titrate a solution that contains CaCO_3 obtained by dissolving part of an antacid tablet, you want to optimize the accuracy of the buret by dissolving a mass that will consume approximately 25 mL of the titrant EDTA. If the molarity of the EDTA solution is 0.04820, how many mmols of EDTA are there in 25.0 mL? Enter a numeric answer, to the correct number of significant figures. Do not include units. Enter Your Answer: How many mmols of Ca^2+ from an antacid tablet will react with the number of mmols of EDTA from Question 3? Refer back to Question 2, if necessary. Enter a numeric answer, to the correct number of significant figures. Do no include units. Enter Your Answer: The amount of Ca^2+ you reported in Question 4 all came from CaCO_3, the active ingredient in the antacid tablet. How many mmols of CaCO_3 correspond to your answer for Question 4 ? Enter a numeric answer, to the correct number of significant figures. Do not include units. Enter Your Answer: If the molarity of the EDTA solution is 0.04820, how many mmols of EDTA are there in 25.0 mL ? Enter a numeric answer, to the correct number of significant figures. Do not include units. Enter Your Answer: How many mmols of Ca^2+ from an antacid tablet will react with the number of mmols of EDTA from Question 3 ? Refer back to Question 2, if necessary. Enter a numeric answer, to the correct number of significant figures. Do no include units. Enter Your Answer: The amount of Ca^2+ you reported in Question 4 all came from CaCO_3, the active ingredient in the antacid tablet. How many mmols of CaCO_3 correspond to your answer for Question 4 ? Enter a numeric answer, to the correct number of significant figures. Do not include units. Enter Your Answer: How many mg of CaCO_3 correspond to your answer for question 5 ? Use 100.1 g/mol as the molar mass of CaCO_3. Enter a numeric answer to THREE significant figures. Do not include units. Enter Your Answer: If one antacid tablet contains 500 mg of CaCO_3, what percentage of the tablet will provide the mass of Question 6? See Slides 19 and 20 of the posted lecture notes. The answer should be around 25%. Meaning that 1/4 of the antacid tablet contains enough Ca^2+ to consume 25 mL of EDTA. Enter a numeric answer only, to TWO significant figures. Do not include the percent sign. Enter Your Answer: The solution described above is now titrated with EDTA. At the equivalence point you will see it change to a pure ___ color. Fill in the bank with a one word answer. Enter Your Answer: You find that your mass of 0.1983 g of powder requires 22.38 mL of EDTA to reach the equivalence point. How many mmol of CaCO_3 are in your powder? The molarity of the EDTA is given above in Question 3 Enter a numeric answer only, to the correct number of significant figures. Do not include units. Enter Your Answer: How many grams of CaCO_3 are in your powder? Enter a numeric answer only, to the correct number of significant figures. Do not include units. Enter Your Answer: What is the mass percent of CaCO_3 m your powder? See Slide 28 of the posted lecture notes. This is, of course, also the mass percent of CaCO_3 in the entire antacid tablet. Enter a numeric answer only, to the correct number of significant figures. Do not include units. Enter Your Answer: How many mg of CaCO_3 were in the entire solid antacid tablet before grinding and separating? See slide 28 of the posted lecture notes. Mass of tablet is given in Question 8. Enter a numeric answer only, to the correct number of significant figures. Do not include units. Enter Your Answer: You weigh out 119.4 mg of your sample and find it requires 23.28 mL of EDTA to reach the EQ. PT. What is the mass percent of CaCO_3 in your sample ? The molarity of the EDTA solution is given in Question 3. Enter a numeric answer only to the correct number of significant figures. Do not include units. Enter Your Answer:

Explanation / Answer

Question 6:

Number of moles of CaCO3 is 1.205 mmol

Molar mass of CaCO3 is 100.1 g mol-1

Mass of 1.205 mmoles of CaCO3 is

= 1.205 x 10-3 mol x 100.1 g mol-1

= 0.1206 g

Question 7:

Number of moles of EDTA is 1.205 mmol

EDTA and Ca2+ reacts in 1:1 ratio

500 mg of CaCO3 has (500 x 10-3 g / 100.1 g mol-1) 0.00499 moles

Percentage of CaCO3 is (1.205 x 10-3 / 0.00499) x 100

=24.124%

Question 11:

We know that molarity of EDTA 0.04820M

Number of moles of EDTA in 22.38mL is

= 0.04820 mmol /mL x 22.38 mL

= 1.078 mmol.

EDTA and Ca2+ reacts in 1:1 ratio

So the mmol of CaCO3 in our powder is 1.078

Question 12:

Grams of CaCO3 is (1.078 x 10-3 mol x 100.1 g mol-1) = 0.1079g

Question 13:

Mass percentage of CaCO3 is (0.1079 g / 0.1983 g)x 100 = 54.45%

Question 15:

We know that EDTA and Ca2+ reacts in 1:1 ratio

Number of moles of EDTA in 22.38 mL of solution is 0.04820 mmol /mL x 22.38 mL

= 1.078 mmol.

Mass of 1.078mmol of CaCO3 is (1.078 x 10-3 mol x 100.1 g mol-1) 0.1079g

Mass percentage of CaCO3 is (0.1079 g / 119.4 mg) x 100 is 90.43%

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