\"Determination of the Composition of a TUMs Tablet\" DATAfile; Exact amount of
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"Determination of the Composition of a TUMs Tablet" DATAfile; Exact amount of NaOH used for the *1.00 Molar' solution: 18.81 grams of NaOH in a 0.250 ml volumetric flask Exact amount of 38% HCl used for the 1.062 Molar solution: 37.0 milliliters of HCl from bottle Mass of 1 tablet of TUMs: 1.364 grams Mass crushed up in mortar and used in titration: 1.334 grams First amount of. 1 Molar HCl solution' placed into TUMs beaker: 10.00 mL Second amoun(Total of 20 wL):10.00mL Titrations: Amount of '1 M NaOH into CONTROL beaker to turn phenolphthalein pink: 9.5 mL Amount of 1 M NaOH into TUMs beaker to turn phenolphthalein pink: 14.7 mL When a Turns tablet is dissolved into stomach acid, we find the following reaction: CaC03(s) + 2 HCI(aq) rightarrow CaCl2(aq) + H2)(l)+CO2(g) The difference between the first titration into the CONTROL beaker (with straight 1 M HCl acid) and the titration into the TUMs beaker can give us a handle on how much of the HCl was neutralized by the TUMs tablet. Determine the difference between mols of NaOH used in both titrations. This difference is equal to the mols of HCl that was neutralized by the TUMs. Usingthe equation of the reaction between the TUMs and the HCl given earlier, determine the mols of calcium carbonate from the TUMs tablet that must have reacted with the HCl From this, determine the % calcium carbonate per TUMs tablet. MY QUESTION IS HOW DO I CALCULATE THE NUMBER OF MOLS USED IN THIS TITRATION?Explanation / Answer
Amount of HCl neutralised due to TUMS tablets=14.7ml-9.5ml=5.2ml
No of moles of NaOH used in the first titration=molarityxvolume in L=1mol/Lx0.0095L=0.0095mol
No of moles of NaOH used in the second titration=1mol/Lx0.0147L=0.00147mol
difference between the no of moles=0.0052mol
no of moles of HCl consumed =difference between the no of moles NaOH in first and 2nd titration=0.0052mol
from the above equation 1 mol of CaCO3 reacts with 2 moles of HCl ,
So the number of moles of CaCO3 involved=0.0052molx1mol/2mol=0.0026mol
amount of CaCO3,g=no of molesxmolar mass=0.0026molx100g/mol=0.26g
% of CaCO3=0.26g/1.334gx100=19.5
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