Question

`0.013*dm^3` of `0.100*mol*L^-1` `NaOH` reacts completely with `0.010*dm^3` `HCl`; what is the concentration of `[HCl](aq)`?

Answer 1

The stoichiometric equation is...........

Explanation:

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

Sodium hydroxide and hydrogen chloride react with a 1:1 stoichiometry. And so we simply have to assess the molarity of the acid solution.

`"Moles of base"` `=` `0.100*mol*dm^-3xx0.013*dm^3=1.30xx10^-3*mol`.

Now this molar quantity was equimolar with the `0.010*dm^3` volume of `HCl(aq)`.

`"Concentration of HCl"=(1.30xx10^-3*mol)/(0.010*dm^-3)=0.130*mol*dm^-3`

Note that `1*dm^3=1xx(10^-1*m)^3=10^-3*m^3=1*L` as required. A `"cubic metre"` is a huge volume, and is equivalent to `1000*L`.