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]#?

1 Answer
May 18, 2017

Answer:

#[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.