Ex 9 Evaluating Antacids PRELAB QUESTIONS broken in half one of the pieces weigh

Question

Ex 9 Evaluating Antacids PRELAB QUESTIONS broken in half one of the pieces weighs o.56g. This 1. An antacid ta from brandA is burets are set up. The piece is stirred with water and 4 drops of Two buret buret contains 0.2063M hydrochloric acid solution. The base the acid sodium hydroxide solution. The initial readings burets are: is buret 1.4 the base buret 0,85mL. Acid added the titration flask, the mixture boiled for two minutes, cooled, and then titrated with the base solution until the mixture is a faint pink. The final bu are: acid buret 28.44mL base buret 17.25mL. What volume of hydrochloric acid was added to the flask? 26.44 mL How many moles of hydrochloric acid was added to the flask? 02 lol L ss 12 moles What volume of sodium hydroxide solution was added to the flask? How many moles of sodium hydroxide was added to the flask? How many moles of the hydrochloric acid did the sodium hydroxide neutralize? O. o olue 53 moku 5 HCi How many moles of hydrochloric acid were required to neutralize the base in the antacid tablet? ODS 12 What was the neutralizing power of the antacid tablet expressed in moles of hydrochloric acid neutralized per gram of antacid tablet? A tablet of Brand A antacid weighs 2.11g, what is the neutralizing power of this antacid in moles of HCl per tablet? o ml Hol per tablet

Explanation / Answer

HCl was added= final buret reading-initial buret reading=28.44-1.43 ml=27.01 ml

Moles of HCl added =27.01*.2063 /1000 mol=0.0057 mol

Added NaOH=final buret reading-initial buret reading=(17.25-0.85) ml=16.40 ml

Moles of NaOH added to flask=16.40*0.1008/1000 =0.0016

Moles of HCl required to neutralise NaOH=0.0016

Moles of HCl which neutralised antacid=total moles of HCl used-moles of HCl used for NaOH=(0.0057-0.0016) =0.0041

Neutralising power of antacid =0.0041/0.56 moles HCl/g of antacid=0.0073 moles/g of antacid

Neutralising power of brand A antacid which is weighing 2.11 g=0.0073*2.11=0.0154 moles/g

1 dollar is the cost of 12/0.59 tablets.

So neutralising power of these tablets=0.0154*12/0.59=0.313 moles/g

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