Question

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?

Answer 1

The question is not legitimate. The solubility of magnesium hydroxide in water is approx. `6*"ppm"` at room temperature...

Explanation:

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

We gots...

`160*mLxx10^-3*L*mL^-1xx1.0*mol*L^-1=0.160*mol` with respect to `HCl`

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

`1/2xx0.160*molxx58.32*g*mol^-1=4.67*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....

Answer 2

20 milliliters

Explanation:

The first step is to find out how many moles of `HCl` reacted with the Hydroxide.

Using the concentration formula:

`c=(n)/(v)`

c= concentration in `mol dm^-3`
n= number of moles
v= volume in liters.

Thus we find out that the `HCl` solution:
`1=(n)/0.16`
n=0.16 mol
So we used 0.16 mol of `HCl` 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 `OH^-` ions to form water, aka neutralizes the solution. However, `Mg(OH)_2` is a dihydroxide, meaning 1 mol of it will release 2 mol of `OH^-` ions. Hence 0.16 mol of `HCl` will react with 0.08 mol of `Mg(OH)_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=(n)/(v)`

`4=(0.08)/(v)`

`v=0.02` litres, or 20 millilitres.