Question

What is `[HCl]` if a `17.04*mL` volume of `0.468*mol*L^-1` concentration of `[Ca(OH)_2]` was titrated with `13.7*mL` of said acid?

Answer 1

Approx. `1* mol*L^-1`

Explanation:

We need (i) the balanced chemical equation:

`2HCl(aq) + Ca(OH)_2(aq) rarr CaCl_2(aq) + 2H_2O(l)`

And (ii) the amount in moles of calcium hydroxide.

`"Moles of calcium hydroxide"` `=` `17.04xx10^-3Lxx0.468*mol*L^-1` `=` `7.98xx10^-3*mol`

Twice this molar quantity of `HCl` was present in the `13.7*mL` `HCl(aq)`.

Thus `[HCl]=(2xx7.98xx10^-3*mol)/(13.7xx10^-3*L)` `=` `??mol*L^-1`

Note that the concentration of `Ca(OH)_2` is unreasonable....