Question

What volume of 0.500 M `NaOH` is needed to neutralize 45.0 mL of 0.400 M `HCl`?

Answer 1

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

Explanation:

Clearly there is a 1:1 equivalence, and as a first step we calculate the number of moles of hydrochloric acid:

`45.0xx10^-3*Lxx0.400*mol*L^-1` `=` `1.80xx10^-2*mol" hydrochloric acid"`.

We find an equivalent molar quantity of sodium hydroxide:

`(1.80xx10^-2*mol)/(0.500*mol*L^-1)xx10^3*mL*L^-1` `=` `36.0*mL`.

This volume is reasonable, in that it is LESS than the volume of the less concentrated acid.