Question

How many milliliters of 0.250M `NaOH` are required to neutralize 30.4 mL of 0.152 M `HCl`?

Answer 1

Approx. `19*mL`. We know that

`"Concentration"="Amount of substance in moles"/"Volume of solution"`

Explanation:

`"Moles of HCl"` `=` `30.4xx10^-3cancelLxx0.152*mol*cancel(L^-1)` `=` `4.62xx10^-3*mol`.

We need an equivalent quantity of sodium hydroxide; `0.250*mol*L^-1` sodium hydroxide is available.

So, `(4.62xx10^-3*cancel(mol))/(0.250*cancel(mol)*cancel(L^-1))xx10^3*mL*cancel(L^-1` `=` `??mL`

What I should have done at the beginning is to write the stoichiometric equation:

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

Why should I write this first? Because it establishes the stoichiometry; the 1:1 molar equivalence. One equiv of sodium hydroxide reacts with one equiv hydrochloric acid. Thus if I know the quantity of acid, I also know the quantity of base. There will be reactions where a different stoichiometry pertains.