ormat Arrange View Share Window Help Labe Report 27 Edited v 125% Insert Table C

Question

ormat Arrange View Share Window Help Labe Report 27 Edited v 125% Insert Table Chart Text Shape Media Comment Table 2 Titrations of Antacid Tablets 1 Data and Calculations Tablet 1 Brand Name Of Tablet Tums Active Ingredient Calcium Carbonate Milligrams of active ingredient 1000 mg 0.19 Mass of tablet beaker 100.37 g 98.15 g Mass of empty beaker 7 Net of mass tablet Molarity of HCL solution 3 M 9 Volume of HCL added to tablet 5 mL 10 Molarity of NaOH titrant 6 M 43.47 mL 11 Volume of NaoH Titrant Final buret reading 39.49 mL 12 Initial buret reading 13 Net volume of NaOH 3.98 mL 14 Millimoles of HCL solution added to tablet 3 MX 5 mL 15 mmil Millimoles of NaOH solution required to neutralize 6 M x 3.98 ml 23.88 mmol 15 the excess HCL Millimoles of HCL neutralized by the antacid tablet 17 Calculated mass, mcaol, of active ingredient Milligrams of active ingredient, mfg 18 percentage relative difference mcacl mmfg/ 20 Mass of inert binder in tablet Mass of inert binder

Explanation / Answer

The reaction that takes place in the solution-

CaCO3+2HCl----->CaCl2 +CO2 +H2O

HCl+NaOH----> NaCl +H2O(titration rxn)

the data provided must be incorrect in terms of the molarity of HCl or that of NaOH

The total mmol of acid (mmol reacted with CaCO3+mmol free acid) cannot be less than the mmol of base required to neutralize the excess acid as they react in 1:1 ratio)

possible molarity of HCl=6M

mmol of HCl added=6M*5ml=30mmol

--->So,mmol of HCl neutralized by the antacid=total mmol of HCl-mmol of excess HCl=total mmol of HCl-mmol of NaOH required to neutralize excess HCl=30-23.88=6.12 mmol

---->calculated mass of active ingredient =mmol of active ingredient reacting with HCl *molar mass of active ingredient

But, mmol of active ingredient reacting with HCl=1/2 *mmol of HCl [from the balanced chemical equation for the reaction]

So,mmol of active ingredient reacting with HCl=1/2 *mmol of HCl=1/2*6.12=3.06 mmol

Hence,calculated mass of active ingredient =mmol of active ingredient reacting with HCl *molar mass of active ingredient (CaCO3)=3.06 mmol*100.087 g/mol=306.266 mg

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