Question

To what volume must a `250*mL` volume of `10*mol*L^-1` `HCl` be diluted to give a `0.050*mol*L^-1` concentration?

Answer 1

Before we start remember the CARDINAL RULE:
`"Acid to water not water to acid....."` I get a final volume of `50*L` of dilute acid.

Explanation:

`"May her rest be long and placid, she added water to the acid."`

`"The other girl did as she oughter, she added acid to the water."`

A small drop of water placed in a strong acid might spit as it heats up upon hydration (another old saw: `"if you spit in acid it spits back at you"`).

For your problem, we use the relationship.....

`"Concentration"="Moles of solute (moles)"/"Volume of solution (L)"`

We have a starting volume of `250*mL` of `10.0*mol*L^-1` `HCl` (the stuff you get out of the bottle is `10.6*mol*L^-1`).

`"Moles of HCl"=250*mLxx10^-3*L*mL^-1xx10*mol*L^-1`

`=2.50*mol` with respect to `HCl`.

Now this molar quantity is diluted up to give a `0.050*mol*L^-1` (and remember acid to water!).

We solve for `"volume"` in `"concentration"="moles of HCl"/"volume of solution"`

i.e. `V=(2.50*mol)/(0.050*mol*L^-1)=50*L`, why? Because `1/L^-1=1/(1/L)=L` as required.

AND WHEN YOU WORK WITH CONC. ACIDS YOU MUST WEAR (i) SAFETY SPECTACLES to protect your mince pies; and (ii) a LAB-COAT to protect your clothing.