If 1.12 g of Mg(OH) 2 precipitate forms from mixing 100 mL of an NaOH solution withexcess magnesium ions, what was the sodium hydroxide solution concentration?
- 1 month ago
Mg2+ +2OH- = Mg(OH)2
moles of precipitate = mass / molar mass = 1.12 g / 58.32 g/mol = 0.0192 moles
moles of OH- in the reaction = 2 * 0.0192 = 0.0384 moles
these are in 100 ml so concentration of the solution = moles / liter
= 0.0384 / 0.1 L = 0.384M ( 3 significant figures)