A 4.0 M solution of Mg(OH)2 is completely neutralized 160 mL of a 1.0 M solution of HCl. How many milliliters of magnesium hydroxide were required?

Feb 26, 2018

The question is not legitimate. The solubility of magnesium hydroxide in water is approx. $6 \cdot \text{ppm}$ at room temperature...

Explanation:

Of course we can interrogate the molar quantity with respect to hydrochloric acid....

We gots...

$160 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 1.0 \cdot m o l \cdot {L}^{-} 1 = 0.160 \cdot m o l$ with respect to $H C l$

And this molar quantity will react with HALF an EQUIVALENT with respect to magnesium hydroxide according to the following equation...

$\frac{1}{2} \times 0.160 \cdot m o l \times 58.32 \cdot g \cdot m o {l}^{-} 1 = 4.67 \cdot g$

Note that I answered the question I wanted to, and not the question you asked, but magnesium hydroxide is not so soluble in aqueous solution. The question should be reformed....

Feb 26, 2018

20 milliliters

Explanation:

The first step is to find out how many moles of $H C l$ reacted with the Hydroxide.

Using the concentration formula:

$c = \frac{n}{v}$

c= concentration in $m o l {\mathrm{dm}}^{-} 3$
n= number of moles
v= volume in liters.

Thus we find out that the $H C l$ solution:
$1 = \frac{n}{0.16}$
n=0.16 mol
So we used 0.16 mol of $H C l$ in the titration.
As this is a reaction with a strong base and a strong acid (presumably as a hydroxide can be considered strong) The reaction goes to completion.

We know that ${H}^{+}$ ions react with $O {H}^{-}$ ions to form water, aka neutralizes the solution. However, $M g {\left(O H\right)}_{2}$ is a dihydroxide, meaning 1 mol of it will release 2 mol of $O {H}^{-}$ ions. Hence 0.16 mol of $H C l$ will react with 0.08 mol of $M g {\left(O H\right)}_{2}$ to completely neutralize.

Going back to the original equation we can thus find the volume of the Magnesium hydroxide, as we know the concentration from the question.

$c = \frac{n}{v}$

$4 = \frac{0.08}{v}$

$v = 0.02$ litres, or 20 millilitres.