What volume of 0.12 M HCl is needed to neutralize a solution containing 3.56 g of Mg(OH)2 dissolved in enough water to make a 300 mL solution.?

Answer 1

Short answer: about 1.0 L of the 0.12 M HCl (1017 mL before rounding).

The balanced equation is where this one is won or lost.

Mg(OH)2 + 2 HCl -> MgCl2 + 2 H2O

Magnesium hydroxide carries two hydroxides, so each mole of it consumes two moles of HCl. Forgetting that 2 is far and away the most common way this problem goes wrong.

Step 1 - moles of Mg(OH)2.
Molar mass = 24.31 + 2(16.00) + 2(1.008) = 58.32 g/mol
n = 3.56 g / 58.32 g/mol = 0.0610 mol

Step 2 - moles of HCl required.
n(HCl) = 2 x 0.0610 mol = 0.122 mol

Step 3 - volume of acid.
V = n / M = 0.122 mol / 0.12 mol/L = 1.02 L

Rounded to the two significant figures that 0.12 M allows: 1.0 L, or about 1.0 x 10^3 mL.

The 300 mL is a distractor, and that is deliberate. How much acid you need is set by how much Mg(OH)2 is present, not by how much water it happens to be sitting in. Dilute that base to 600 mL or squeeze it into 150 mL and the answer does not move by a drop. When a problem hands you a volume you never used, confirm it is genuinely irrelevant instead of assuming you skipped a step - here it would only matter if you were asked for the concentration of the base (0.0610 mol / 0.300 L = 0.203 M) or for the pH at some point in the titration.

One honest footnote. Mg(OH)2 is only very slightly soluble - milk of magnesia is a suspension, not a solution - so a true 0.203 M solution of it does not really exist. The problem is idealizing so the stoichiometry stays clean, and the arithmetic above is still exactly what the acid neutralizes when you tip all 3.56 g of the solid and its water into the flask, which is how it runs in the lab.

A quick self-check: you need 0.122 mol of HCl out of a bottle holding 0.12 mol in every liter, so the volume had to land right around 1 L. If you got something near 500 mL, you almost certainly dropped the 2.