Question

You have 505 mL of a 0.125 M `HCl` solution and you want to dilute it to exactly 0.100 M. How much water should you add?

Assume volumes are additive.

Answer 1

Approx. `100*mL` of water should be added.

Explanation:

`"Moles of HCl"=505*mLxx10^-3L*mL^-1xx0.125*mol*L^-1=0.0631*mol.`

We require a `"concentration"` of `0.100*mol*L^-1`.

But `"Concentration"="Moles of solute"/"Volume of solution"`

Thus `"Volume of solution"` `=` `"Moles of solute"/"Concentration"`

`=(0.0631*mol)/(0.100*mol*L^-1)=0.631*L,` or `631*mL`.

And thus we dilute the original `505*mL` volume to `631*mL`.

`"IMPORTANT EXPERIMENTAL NOTE:"`

Ordinarily, we would never add water to an acid. Why not? Well, because if you spit in conc. acid, it spits back at you. Because here we deal with fairly dilute acids (i.e. `~0.1*mol*L^-1`), we can relax this rule a bit and add water to the acid.