Question

Assume that you are working with an Antacid tablet that contains 550mg CaCO₃ and 110mg Mg(OH)₂. Each tablet weighs 745mg.

Part A step 1. Determine the amount of tablet powder that would need to be weighed to give you 1000mg of CaCO₃.

Part A Step 3. How much CaCO₃ is in the 20 ml portion?

Part A Step 4. How much EDTA is needed for one titration? Given that each mL of EDTA is equivalent to 5.004 mg of CaCO₃.

If the tablets (tablet powder) actually contain both CaCO₃ and Mg(OH)₂ – How much Mg(OH)₂ was weighed in Part A Step 1 along with the CaCO₃.

Part B Step1: In each ml of filtrate, how much (mg) of each, CaCO₃ and Mg(OH)₂, exist?

So how much filtrate would be equal to 120mg of CaCO3 and Mg(OH)₂?

Part B Step 3: How much EDTA is needed for one titration? (Remember for the Ca present: each mL of EDTA is equivalent to 5.004 mg of CaCO₃ and for the Mg present: each mL of EDTA is equivalent to 2.916 mg of Mg(OH)₂.)

Answer 1

1: Part A step 1: Given the weight of each tablet = 745 mg

Also each tablet contains 550 mg CaCO3 i.e.

550 mg of CaCO3 is present in 745 mg of the tablet power

Hence 1000 mg of CaCO3 that would be present in the mass of tablet power

= (745 mg tablet power / 550 mg CaCO3) x 1000 mg CaCO3 = 1355 mg tablet power (answer)

2: Part A Step 2 is needed to solve the problem in Part A Step 3 and step - 4, as volume term is given here that is not mentioned anywhere else.

3:  If the tablets (tablet powder) actually contain both CaCO3 and Mg(OH)2 – How much Mg(OH)2was weighed in Part A Step 1 along with the CaCO3.

Given the weight of each tablet = 745 mg

Also each tablet contains 110 mg Mg(OH)2 i.e.

110 mg Mg(OH)2 is present in 745 mg of the tablet power

Hence the mass of Mg(OH)2 that would be present in 1355 mg of tablet power

= [110 mg Mg(OH)2 / 745 mg tablet power) x 1355 mg tablet power = 200 mg Mg(OH)2 (answer)

Part B Step1: Volume of the titrant prepeared is needed to solve this problem and also to solve PartB step-2