Question

3. Tilration of antacid tablets Tablet 2 Philips MgCOH)2 311 Data and calculations Brand name of tablet Active ingredient Milligrams of active ingredient (from the manufacturer's label) Mass of tablet + beaker Mass of empty beaker Net mass of tablet Molarity of HCl solution Volume of HCI added to tablet Molarity of NaOH titrant Volume of NaOH titrant Initial buret reading Final buret reading Tablet 1 CaCo3 Cacos 500 26-391 25-051 8 1.3408 3100 M 500 ml 6.00 M 24.086) 23.056 0 550 3:00 600 6.00 mL 0.00 0.89 0.80 150 0.74 0.74 150 ml mL mmol mmol Net volume of NaOH Millimoles of HCl solution added to tablet Millimoles of NaOH solution required to neutralize the excess HCI Millimoles of HCl neutralized by the antacid tablet 4.8 mmol mmol 10.2 10.6 mmol mmol Write the balanced chemical equation(s) for the reaction of HCI with the active ingredient in the tablet(s). Calculate from your titration data the millimoles and mass (in milligrams) of active ingredient in the tablet(s). Compare your results with the values given on the manufacturer's label. Tablet 1: Equation Tablet 2: Equation 28-11

Answer 1

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Tablets phillips Mg (OH)2 3 11 mg 24.08674 23.056 g yo. 580 g ANSWER Data and Tablet e - Calculations Brand name CaCO₃ of tablet Milligrams of CaCO3 achie ingredient 500mg Mats of tablet theates 26.391 g Mass of emphy beakes 25.051 g Net mass of tablet. Molanty of HCI sdubiony! 1.340 y Volume of HCI added 6.00 ML to tablet Molarity of Nach titrant 600 M Net Volume of Na on titraut 0.80 me Millimoles of Hai Solection 15.0mmol added to tablet Millimoles of NaOH solehan 4.8mmo) required to neutralize the excesuch Millimoles of HCl neutralize 10.2 mmol by the antacid tablet Y 300M 3:00M SoomL 6.com 0.74 mL 15.0 mmoj 4.4 mmel 10.6 mmol

x CaCl2 + CO2 + H₂O as Tablet I: Equation . CaCO3 + 2HCl(aq) Tablet 2: Equation Mg COMI scan) + 2HClan - Mg Cl2 egg + 2H20ls

2 Tablet 7 Mass calcealed Meale, of achive o ingredient = mole of CaCO₃ x molecular weight of CaCO3 Note (i) Mole of CaCO3 is fx mmd of HCl = 10.2 mmol = Solomon 0 Note from reachin stiochiometric we see that - 1 male of CaCO₂ take 2 mole of HCl, hence 1 mmol of cacoz is half of moud of the Mcale, of Calg = 5.1 mmol x 100. 0869 g/mol Mcale, of Ca CO₂ = 510.44 my Mmfg i 500 mg Percentage relatiue difference - Maifiye Meate Micalc- Mnfy) X1.00 Mapy -1510.44mg - Soomy ) *.00 L'ercentage gelative = 10.414 x3.00 2.088% difference of laco Percentage velatine difference of Cacą = 2.0887.) mass of inert binder in tablet 1 = not given overaly 500mg 500 tablet mass Mass calculated Meale, of active ingredient = mole of Mgom, * Edo. Moleculey weight of Mg(OH) Note Tyrole of my 6mis įx mmol of 141 = 70.6 = 5-3 ramos - 2 0 Reason Imde of Mycoth give 2 m do of one, hence amole of one take zinde of hcl . 5.3 mmel x 58. 3197 = 309.04441 mg Mcale = Mmfy 311 g i Mercentage relative difterence - meals - mmfy) xter =/303.09441m3). 9.69441my vreo - -711ng mufry [31lmy a Percentage relative ctitorence = -0.61273% mass of inert binder in tablet 2 not given mass of overall ablet