Question

What volume (in mL) of a 3.50M `LiF` stock solution would you use to make 1.5L of 0.500M `LiF` solution?

Answer 1

A volume a bit over `214*mL` is required.

Explanation:

`"Moles of lithium fluoride"` `=` `1.5*Lxx0.500*mol*L^-1=0.75*mol`.

And thus we need a volume of the stock solution that contains `0.75*mol` lithium fluoride.

`(0.75*cancel(mol))/(3.50*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1)=214.3*mL`

This solution would be slightly basic, i.e. `pH>7`. Why?