Answer to Question #108883 in Physical Chemistry for Yugneswari Muniandy

Question

A 375ml bottle of antacid was analysed and found to contain 30g of Mg(OH)2 and 30g of Al(OH)3. The instructions on the bottle recommend a dose of 10ml of the antacid. Calculate the quantity of stomach acid that could be neutralized by taking the recommended dosage of antacid. Stomach acid has a pH of 0.900

Answer 1

The neutraliation reactions are

2 H⁺ + Mg(OH)₂ = Mg²⁺ + 2 H₂O

3 H⁺ + Al(OH)₃ = Al³⁺ + 3 H₂O

Concentration of [H+] = 10^(-pH) = 10^(-0.9) = 0.126 M

The weight of Mg(OH)2 and Al(OH)3 in the dose are equal to 10/375*30 = 0.8 g,

the respective amounts are: Mg(OH)2 = 0.8/58.31 = 0.0137 mole and Al(OH)3 = 0.8/78 = 0.0103 mole.

The total amount of neutralized H⁺ ions is n = 2*0.0137 + 3*0.0103 = 0.0583 mole.

The volume of neutralized stomach acid is then: V = [H⁺]/n = 0.126 M/0.0583 mole = 2.16 L.

Answer: the volume of stomach acid that could be neutralized is 2.16 L