Question

1. Write the chemical equation for the reaction of an antacid containing magnesium hydroxide, Mg(OH)₂, with stomach acid (hydrochloric acid, HCl) and then balance the equation.

2. If a tablet of the antacid neutralized 41.4 mL of 0.150 M hydrochloric acid then how many grams of magnesium hydroxide was in the tablet?

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3.Concentrated gastric acid is about 0.16 molar in concentration (0.16 M HCl), corresponding to a pH of 0.8. In the stomach, however, the solution becomes diluted to a pH between 1 and 2. Using the information from Q.5 calculate how many milliliters of gastric acid can be neutralized by 1 tablet of antacid?

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Answer 1

The balanced reaction of Mg(OH)2 and HCl is

Mg(OH)2+2HCl -------->MgCl2+2H2O

molar masses : Mg(OH)2= 58 g/mole, HCl= 36.5 and MgCl2= 95 and H2O= 18

moles of HCl in 41.4ml of 0.15M= Molarity* Volume (L)=0.15*41.4/1000 =0.00621

The reaction suggets 2 moles of HCl requires 1 mole of Mg(OH)2 for complete neutralization

hence 0.00621moles of HCl requires 0.00621/2= 0.003105

Mass of Mg(OH)2= Moles* Molar mass of Mg(OH)2= 0.003105*58=0.18 gm

3. Basis :1 Liter of solution of molarity 0.16M. Moles of HCl =Molarity* Volume L=0.16

pH=-log [H+], when pH= 1, [H+]= 10^-1=0.1 moles/L

since acid is diluted while moles of HCl remains at 0.16, Volume of solutopn = 0.16/0.1= 1.6L

A single dose of ant acid contains 250 mg of Mg(OH)2

Moles of Mg(OH)2= Mass/molar mass =250*10^-3/ 58=0.0043 moles

From the reaction, moles of HCl required = 2*0.0043= 0.0086 moles

Volume of HCl required ( for a pH of 1)= 0.0086/0.1= 0.086 L = 86 ml

and volume of HCl required for a pH of 2= 0.0086/0.01 ( pH=2 means [H+] =10-2)= 0.86L= 860 ml