Answer
The concentration of that $Sr(OH)_2$ solution is equal to $ 1.6 \times 10^{-4} M$.
Work Step by Step
1. Calculate the hydroxide ion concentration:
pH + pOH = 14
10.5 + pOH = 14
pOH = 3.5
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 3.5}$
$[OH^-] = 3.2 \times 10^{- 4} M$
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2. Find the $Sr(OH)_2$ concentration:
- Each $Sr(OH)_2$ mol gives 2 $OH^-$ mol to the solution.
$3.2 \times 10^{-4} M (OH^-) \times \frac{1mol(Sr(OH)_2)}{2mol(OH^-)} = 1.6 \times 10^{-4} M (Sr(OH)_2)$
