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

0.202g of Mg(OH)₂ and 0.270g of Al(OH)₃

Step-by-step explanation:

The OH⁻ ions of the Mg(OH)2 and Al(OH)3 reacts with the HCl. The HCl in excess reacts with the NaOH. To solve this question we must find the moles of HCl that react in the beginning with the antiacid and as the ratio is 1:1 we can find the moles of Mg(OH)2 and Al(OH)3. The reactions are:

OH⁻ + HCl → Cl⁻ + H₂O

HCl + NaOH → NaCl + H₂O

Moles HCl added:

0.03825L * (0.500mol / L) = 0.019125 moles

Moles NaOH utilized:

0.0036L * (0.500mol / L) = 0.0018 moles

Moles HCl that react = Moles OH⁻ in the antiacid

0.019125 moles - 0.0018moles =

0.017325 moles OH⁻ in the antiacid

As the ratio is 1:1. the moles Mg(OH)2 = Moles Al(OH)3

And these 5 hydroxides produce 0.017325 moles:

5X = 0.017325 moles

X = 3.465x10⁻³ moles

Where X are moles of both Mg(OH)2 and Moles Al(OH)3

Mass -Molar mass: 58.3197 g/mol-:

3.465x10⁻³ moles * (58.3197g/mol) =

0.202g of Mg(OH)₂

And

Mass -Molar mass: 78g/mol-:

3.465x10⁻³ moles * (78g/mol) =

0.270g of Al(OH)₃