Question

Question #55028

Answer 1

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

Start with the balanced chemical equation for this double replacement reaction

`MgCl_(2(aq)) + color(red)(2)NaOH_((aq)) -> Mg(OH)_(2(s)) darr + 2NaCl_text((aq])`

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:color(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 `color(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 = n/V => n = C * V`

`n_(MgCl_2) = 0.0500"mol"/cancel("L") * 1.0 * 10^(-6)cancel("L") = 0.0500 * 10^(-6)"moles"` `MgCl_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"` `NaOH`

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

`C = n/V => V = 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 * 10^(-6)cancel("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

`Mg_((aq))^(2+) + 2OH_((aq))^(-) -> Mg(OH)_(2(s)) darr`