## Chemistry 9th Edition

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