Calculate the number of milliliters of 0.636 M NaOH required to precipitate all of the Mg2+ ions in 187 mL of 0.718 M Mg(NO3)2

Question

Calculate the number of milliliters of 0.636 M NaOH required to precipitate all of the Mg2+ ions in 187 mL of 0.718 M Mg(NO3)2 solution as Mg(OH)2. The equation for the reaction is:
Mg(NO3)2(aq) + 2NaOH(aq) → Mg(OH)₂(s) + 2NaNO3(aq)

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