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?
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. .
To determine the percentage w/W of magnesium hydroxide in the pharmaceutical preparation, we need to calculate the amount of magnesium hydroxide reacted with the HCl and the amount of unreacted HCl.
Let's break down the steps to solve the problem:
Step 1: Calculate the moles of HCl used.
For this, we need to use the formula:
moles = concentration (mol/L) * volume (L)
moles of HCl = 0.1 mol/L * 0.05 L (50 mL/1000 mL) = 0.005 mol
Step 2: Calculate the moles of NaOH used.
The reaction between HCl and NaOH is a 1:1 ratio. So the number of moles of NaOH used is equal to the moles of HCl reacted.
moles of NaOH = moles of HCl = 0.005 mol
Step 3: Calculate the moles of Mg(OH)2 reacted.
The reaction between HCl and Mg(OH)2 is a 2:2 ratio. So the number of moles of Mg(OH)2 reacted is half the moles of NaOH used.
moles of Mg(OH)2 reacted = 0.005 mol / 2 = 0.0025 mol
Step 4: Calculate the mass of Mg(OH)2 reacted.
For this, we need to use the formula:
mass = moles * molar mass
The molar mass of Mg(OH)2 is approximately 58.33 g/mol.
mass of Mg(OH)2 reacted = 0.0025 mol * 58.33 g/mol = 0.1458 g
Step 5: Calculate the percentage w/W of Mg(OH)2.
For this, we need to use the formula:
percentage w/W = (mass of Mg(OH)2 reacted / mass of sample) * 100
The mass of the sample is 8.995 g.
percentage w/W = (0.1458 g / 8.995 g) * 100 = 1.62%
Therefore, the percentage w/W of magnesium hydroxide in the pharmaceutical preparation is approximately 1.62%.