Question

A `45.33*mL` volume of `Sr(OH)_2(aq)` was titrated with a `19.81*mL` volume of `0.44*mol*L^-1` concentration. What is `[Sr(OH)_2]`?

Answer 1

`[Sr(OH)_2]~=0.1*moL^-1`

Explanation:

We assess the stoichiometric reaction:

`Sr(OH)_2(s) +2HCl(aq) rarr SrCl_2(aq) + 2H_2O(l)`

`"Moles of HCl"-=19.81*mLxx10^-3*L*mL^-1xx0.44*mol*L^-1=8.72xx10^-3*mol`.

And thus there were `(8.72xx10^-3*mol)/2` WITH RESPECT TO `Sr(OH)_2`

`[Sr(OH)_2]=((8.72xx10^-3*mol)/2)/(45.33xx10^-3*L)=0.0961*mol*L^-1`.

I don't think this problem is too realistic because I doubt that you could prepare a `Sr(OH)_2` solution this concentrated.