Question

How many milliliters of 0.20N sodium hydroxide must be added to 75 mL of 0.050 N hydrochloric acid to make a neutral solution?

Answer 1

Approx. `19*mL`

Explanation:

We need (i) a stoichiometric equation in order to inform our calculations. Now with one equiv of simple acids and bases, this is trivial:

`NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)`

`"Moles of HCl"=75xx10^-3cancelLxx0.050*mol*cancel(L^-1)`

`3.75xx10^-3*mol` with respect to `HCl`.

And thus we need an equiv of `NaOH`, and we have `0.20*mol*L^-1` available, so..........we divide the molar quantity of `HCl` by the `NaOH` concentration to get the req'd volume

`(3.75xx10^-3*mol)/(0.20*mol*L^-1)=0.01875*L=18.8*mL`