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]?

May 18, 2017

$\left[S r {\left(O H\right)}_{2}\right] \cong 0.1 \cdot m o {L}^{-} 1$

Explanation:

We assess the stoichiometric reaction:

$S r {\left(O H\right)}_{2} \left(s\right) + 2 H C l \left(a q\right) \rightarrow S r C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

$\text{Moles of HCl} \equiv 19.81 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.44 \cdot m o l \cdot {L}^{-} 1 = 8.72 \times {10}^{-} 3 \cdot m o l$.

And thus there were $\frac{8.72 \times {10}^{-} 3 \cdot m o l}{2}$ WITH RESPECT TO $S r {\left(O H\right)}_{2}$

$\left[S r {\left(O H\right)}_{2}\right] = \frac{\frac{8.72 \times {10}^{-} 3 \cdot m o l}{2}}{45.33 \times {10}^{-} 3 \cdot L} = 0.0961 \cdot m o l \cdot {L}^{-} 1$.

I don't think this problem is too realistic because I doubt that you could prepare a $S r {\left(O H\right)}_{2}$ solution this concentrated.