An antacid tablet contains the weak base Mg(OH)2. After crushing the tablet and adding 20.00 mL of 1.00 M HCl(aq) to it, the remaining acid is titrated to the end point with 29.50 mL of 0.495 M NaOH. How many milligrams of Mg(OH)2 were present in the tablet?
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
mols HCl added initially = M x L = ?
mols HCl remaining = mols NaOH used = M NaOH x L NaOH = ?
mols HCl used by the tablet = initial mols HCl - mols NaOH = ?
mols Mg(OH)2 = mols HCl x 1/2 = ?
grams Mg(OH)2 = mols x molar mass = ? and convert to mg.
Post your work if you get stuck.
Isn't there supposed to be an equilibrium arrow in the chemical equation?
Why are we dividing mols HCl by 2?
1. ALL reactions are equilibrium reactions. Usually, when a base reacts with an acid the equilibrium is so far to the right that the reverse reaction is not significant so I didn't write the <==> arrow. If you prefer you may add it.
2. You divide by 2 because the reaction tells you that 2 mol HCl react with 1 mol Mg(OH)2. You have mols HCl and you want to convert to mols Mg(OH)2; therefore, mols HCl/2 = mols Mg(OH)2.
Okay, now I understand. So is my answer's correct?
Initially I questioned rounding to 0.0054 because of the number of significant figures (s.f.); however, that looks ok now. If your prof is a s.f. freak, s/he may prefer you to use 58.3 (3 s,f,) and not 58.33(4 s.f.) which gives an answer of 157 mg.
Ok Thanks!
To determine the number of milligrams of Mg(OH)2 present in the tablet, we need to perform stoichiometric calculations based on the given information.
First, let's write the balanced chemical equation for the reaction between Mg(OH)2 and HCl:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
From the equation, we can see that for each mole of Mg(OH)2, it reacts with 2 moles of HCl. Therefore, the number of moles of HCl reacted can be determined using the volume and molarity of HCl added.
Given:
Volume of HCl (V1) = 20.00 mL = 0.02000 L
Molarity of HCl (C1) = 1.00 M
Using the formula:
n = C * V
We can calculate the number of moles of HCl reacted:
n(HCl) = C1 * V1
= 1.00 M * 0.02000 L
= 0.02000 moles
According to the balanced equation, 1 mole of Mg(OH)2 reacts with 2 moles of HCl. Therefore, the number of moles of Mg(OH)2 present in the tablet is half the number of moles of HCl reacted:
n(Mg(OH)2) = 0.02000 moles / 2
= 0.01000 moles
Next, we need to determine the mass of Mg(OH)2 using its molar mass. The molar mass of Mg(OH)2 can be calculated by summing the atomic masses of its constituent elements:
Molar mass of Mg(OH)2 = (24.31 g/mol) + 2 * (1.01 g/mol + 16.00 g/mol)
= 58.33 g/mol
Finally, we can calculate the mass of Mg(OH)2 in milligrams:
Mass of Mg(OH)2 = n(Mg(OH)2) * Molar mass of Mg(OH)2
= 0.01000 moles * 58.33 g/mol
= 0.5833 grams
= 583.3 milligrams
Therefore, there were 583.3 milligrams of Mg(OH)2 present in the antacid tablet.
This is what I got:
20.00 mL / 1000 mL * 1.00 mol = 0.0200 mol HCl (initially added)
29.50 mL / 1000 mL * 0.495 mol + 0.0146 mol NaOH (used)
0.0200 mol HCl - 0.0146 mol NaOH = 0.0054 mol HCl (used)
0.0054 mol HCl / 2 mol HCl * 58.33 g * 1000 mg = 160 mg (answer)