Question

An analyst wants to determine the content of a 7% magnesium hydroxide bottle acquired in a drugstore, through the Volumetry method of neutralization.
8.995 g of magnesium hydroxide suspension were avolumados to 100ml of water, then,25 ml were added to the Erlenmeyer flask containing 50 ml of HCl 0.1 mol/L. The quantity of HCl 0.1 mol/L which did not react was titled with 36.30 ml of NaOH 0.1 mol/L. What percentage w/W of magnesium hydroxide in pharmaceutical preparation?

Answer 1

Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
HCl + NaOH ==> NaCl + H2O

millimols HCl initially added to sample = mL x M = 50*0.1 = 5
millimols HCl in excess (titrated with NaOH) = 36.30*0.1 = 3.63
millimols HCl used = 5-3.63 = 1.37
Convert 1.37 mmols HCl used to Mg(OH)2 = 1.37 x (1 mol Mg(OH)2/2 mols HCl) = 1.37/2 = 0.685 mmols or 6.85E-4 mols Mg(OH)2
grams Mg(OH)2 = mols x molar mass = ? and that is grams in the 25 mL. There are 4 times that in the 100 mL and that is the amount in 8.995 g of the suspension.
Then % Mg(OH)2 w/w = (g Mg(OH)2/8.995)*100 = ?

Post your work if you get stuck. .