Question

Work out the volume in cm3 of 0.100 mol dm3 NaOH needed to neutralise each of 20.0cm3 of 0.200mol dm3 Hydrochloric acid?

could anyone show me the method to work out the above please.

Answer 1

Would it not be `40*cm^3`?

Explanation:

With all these problems, it is vital to write out a stoichiometric equation that informs you of the molar equivalence that occurs in the reaction.... And this one is a standard:

`underbrace(HCl(aq) + NaOH(aq))_"acid + base" rarr underbrace(NaCl(aq) + H_2O(l))_"SALT + water"`

And so here, conveniently we got 1:1 molar equivalence...and we work out the molar equivalence using the equation....

`n_"moles of solute"="volume"_"cubic decimetre"`

`xx"concentration"_"moles per cubic decimetre"`

`n_"HCl"=20.0*cm^3xx10^-3*dm^3*cm^-3xx0.200*mol*dm^-3` `-=4.0xx10^-3*mol`

And so we need an equivalent quantity of `NaOH` at `0.100*mol*dm^-3` concentration....

`-=(4.0xx10^-3*mol)/(0.100*mol*dm^3)xx1000*cm^3*dm^-3=40.0*cm^3`