# Question #9ebcd

Mar 30, 2017

Explanation below.

#### Explanation:

Strontium hydroxide or $S r O {H}_{2}$ is a strong base and will completely ionize. This means that virtually all $S r O {H}_{2}$ will dissociate in water into metal cations and hydroxide ions.

${\text{Sr"("OH")_2(s) -> "Sr"^{2+}+2"OH}}^{-}$

$p O H$ is the negative log of the concentration of hydroxide molecules. Since we know that $S r O {H}_{2}$ will completely ionize in solution, we can use the given concentration to solve for $p O H$.

$p O H = - \log \left[O {H}^{-}\right]$
$p O H = - \log \left[2.0 \cdot {10}^{-} 1 M\right]$
$p O H = 0.70$

The question asks for $p H$ and not $p O H$. Using the formula below, we can solve for $p H$.

$p H + p O H = 14$
$p H + 0.70 = 14$
$p H = 14 - 0.70 = 13.3$
$p H = 13.3$

The $p H$ is 13.3 at $\text{25 degrees Celsius}$