Question

What volume of a 1.0 M HCI is required to completely neutralize 25.0 ml of a 1.0 M KOH?

Answer 1

Why, an equal volume, i.e. a `25.0*mL` volume............

Explanation:

We interrogate the reaction:

`KOH(aq) + HCl(aq) rarr KCl(aq) + H_2O(l)`

When we quote solution concentration, we refer to the quotient:

`"Concentration"` `=` `"Moles of solute"/"Volume of solution"`, i.e.

`C=n/V`, and thus `"concentration"` has units of `mol*L^-1`

And you will habitually use this quotient to calculate to calculate solution concentrations, moles and mass of solute, and volume of solution needed.

It should not take a great deal of insight to appreciate that the given volume of `NaOH(aq)` solution contains THE SAME AMOUNT OF SOLUTE that is in the given volume of `HCl(aq)`.

i.e. `"moles of KCl"` `=`

`"concentration"xx"volume"=1.0*mol*L^-1xx25.0xx10^-3*L`

`=25xx10^-3*mol`.

What is the final concentration of `KCl(aq)`? Why is it different from the starting concentrations of `[KOH]`, and `[HCl]`?