Answer to Question #142252 in General Chemistry for Aida

Question

Suppose you are analyzing an antacid that contains Mg(OH)2 and Al(OH)3
in a 1:1 molar ratio. The antacid was neutralized with an excess of 0.500 M HCl, and the excess HCl was then back titrated with 0.500 M NaOH. The following data was obtained:


Volume of 0.500 M HCl utilized: 38.25 mL
Volume of 0.500 M NaOH utilized: 3.60 mL

Calculate the masses of Mg(OH)2 and Al(OH)3 contained in the antacid.

Answer 1

To the mixture solution total volume of 0.500 M HCl utilized: 38.25 mL

Let, S = 0.500 M

V = 38.25 mL

So, total amount of HCl = S×V

= 0.500 M×38.25 mL

= 19.125 (mol.L-1).mL

= 19.125 mol.(1000ml)-1.ml

= 19.125 × 10^–3 mol.ml-1.ml

= 19.125 × 10^–3 mol

To titrate the excess amount of HCl volume of 0.500 M NaOH utilized: 3.60 mL

Let, S' = 0.500 M

V' = 3.60 mL

So, amount of excess HCl = S'V'

= 0.500 M×3.60 mL

= 1.8 (mol.L-1).mL

= 1.8 mol.(1000ml)-1.ml

= 1.8× 10^–3 mol.ml-1.ml

= 1.8× 10^–3 mol

Hence,

HCl consumption by the mixture of Mg(OH)2 and Al(OH)3 in solution

= Total amount of HCl – amount of excess HCl

= (19.125 – 1.8)×10^–3 mol

= 17.325×10^–3 mol

Let,

Moles of Mg(OH)2 in the solution

= Moles of Al(OH)3 in the solution

= A mole

1 mole of Mg(OH)2 produce 2 mole of OH–

And

1 mole of Al(OH)3 produce 3 Mole of OH–

So, total number of moles of OH– in the solution

= (A×2 + A×3) mole

= 5A mole

total number of moles of OH– in the solution

= HCl consumption by the mixture of Mg(OH)2 and Al(OH)3 in solution

So, 5A = 17.325×10^–3

Or, A =(17.325×10^–3)/5

Or, A = 3.465×10^–3

So, in solution 3.465×10^–3 mole of Al(OH)3 and 3.465×10^–3 mole of Mg(OH)2 present

Molar mass of Al(OH)3 = 78 g

So, mass of 3.465×10^–3 mole of Al(OH)3

= 3.465×10^–3×78 grams

=0.2703 grams

Molar mass of Mg(OH)2 = 58.32 grams

So, mass of 3.465×10^–3 mole of Mg(OH)2

= 3.465×10^–3×58.32 grams

=0.2021 grams

Hence,

0.2021 grams of Mg(OH)2 and 0.2703 grams of Al(OH)3 contained in the antacid