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

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Answer 1 · 2017-05-18T14:27:37

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

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.

A 45.33*mL45.33*mL volume of Sr(OH)_2(aq)Sr(OH)_2(aq) was titrated with a 19.81*mL19.81*mL volume of 0.44*mol*L^-10.44*mol*L^-1 concentration. What is [Sr(OH)_2][Sr(OH)_2]?