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?

Answer 1

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.

Answer 2

Isn't there supposed to be an equilibrium arrow in the chemical equation?

Answer 3

Why are we dividing mols HCl by 2?

Answer 4

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)

Answer 5

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.

Answer 6

Okay, now I understand. So is my answer's correct?

Answer 7

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.

Answer 8

Ok Thanks!