# Question 55028

May 17, 2015

You'd need 0.500 ML in order to precipitate that much magnesium from your solution.

$M g C {l}_{2 \left(a q\right)} + \textcolor{red}{2} N a O {H}_{\left(a q\right)} \to M g {\left(O H\right)}_{2 \left(s\right)} \downarrow + 2 N a C {l}_{\textrm{\left(a q\right]}}$

This reaction will form sodium chloride, a soluble salt, and magnesium hydroxide, an insoluble solid that will precipitate out of the solution.

Notice the $1 : \textcolor{red}{2}$ mole ratio that exists between magnesium chloride and sodium hydroxide. This tells you that, for every 1 mole of magnesium chloride, you need $\textcolor{red}{2}$ moles of sodium hydroxide in order for all of the magnesium to be removed from the solution.

So, you know that you're dealing with a 1.00-ML, 0.0500-mol/L solution of magnesium chloride.

Use the solution's molarity to determine how many moles of magnesium chloride you have

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{M g C {l}_{2}} = 0.0500 \text{mol"/cancel("L") * 1.0 * 10^(-6)cancel("L") = 0.0500 * 10^(-6)"moles}$ $M g C {l}_{2}$

Use the aforementioned mole ratio to determine how many moles of sodium hydroxide you'd need

0.0500 * 10^(-6)cancel("moles"MgCl_2) * (color(red)(2)"moles"NaOH)/(1cancel("mole"MgCl_2)) = 0.100 * 10^(-6)"moles" $N a O H$

Since you know the molarity of the sodium hydroxide solution, you can determine the volume you'd need by

$C = \frac{n}{V} \implies V = \frac{n}{C}$

V_(NaOH) = (0.100 * 10^(-6)"moles")/(0.200cancel("mol")/"L") = 0.5 * 10^(-6)"L"#

This is equivalent to having

$0.5 \cdot {10}^{- 6} \cancel{\text{L") * "1 ML"/(10^(-6)cancel("L")) = color(green)("0.5 ML}}$

SIDE NOTE Here's how the net ionic equation looks like for your reaction

$M {g}_{\left(a q\right)}^{2 +} + 2 O {H}_{\left(a q\right)}^{-} \to M g {\left(O H\right)}_{2 \left(s\right)} \downarrow$