Chemistry Question?

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?

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  • 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) 

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