DRY LAB B-VOLUMETRIC ANALYSIS DRY LAB WORKSHEETS Part 3 Analysis of CaCOs Conten

Question

DRY LAB B-VOLUMETRIC ANALYSIS DRY LAB WORKSHEETS Part 3 Analysis of CaCOs Content of Marble Using a Back Titration Reaction equation for the dissolution of calcium carbonate in hydrochloric acid acos (s) Reaction equation for the neutralization of the excess hydrochloric acid: ay) raa) Marble sample number 88 3u .a59 6:3474 no.a5g Mass Initial: 82a689 6.6138 Final: Marble Sample: . a736 Volume of HCI (titration 1): 5o Concentration of HCI: o.10o7mol-L- Concentration of NaOH:.O3O mol-L mL Remember, in order to dissolve the marble (rock) we require an excess of hydrochloric acid. To determine the actual amount of hydrochloric acid required for the dissolution of marble, we have to determine how much acid remains after the first step. To do this we titrate the solution with NaOH which reacts with the left-over HCI. HCI HCI reacts with CaCO3 Moles LLent Total Moles of HCI HCI Reacted

Explanation / Answer

Here, provided,

The mass of the marble sample = 0.2736 grams

The Volume of HCL is 50mL = 0.050L            

Concentration of HCl = 0.1007mol/L

The Volume of the NaOH = 23.86 mL

The Concentration of NaOH = 0.02510 mol/L

Solution

Moles NaOH = 0.02386 L x 0.02510 M = 0.000598

Moles HCl = 0.050 L x 0.1007 M = 0.005035

moles HCl consumed during sample titration = 0.005035 - 0.000598 = 0.004437

Moles CaCO3 = 0.004437 /2 =0.002218

Mass CaCO3 = 0.002218 mol x 100.0884 g/mol = 0.222 g

Therefore, percentage of CaCO3 = (0.222 x 100)/ 0.2736 = 81.15 %

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