# Question #7c9bd

Dec 13, 2014

The answer is pH = $11.97$.

The balanced chemical equation is

$K O {H}_{\left(s\right)} + {H}_{2} {O}_{\left(l\right)} \to {K}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

Since $K O H$ is a strong base, it will completely dissociate into ${K}^{+}$ and $O {H}^{-}$ ions; this means that the concentration of both ${K}^{+}$ and $O {H}^{_}$ will be equal to the starting concentration of $9.28 \cdot {10}^{-} 3 M$.

We know that pH = $14 - p O H$, which means that we need to determine the pOH first

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(9.28 \cdot {10}^{-} 3\right) = 2.03$

Therefore, the pH is equal to $p H = 14 - 2.03 = 11.97$