# Question #b267a

Dec 9, 2014

The answer is pH = $13.26$.

$N a O H$ is a strong base, which means that it dissociates completely to $N {a}^{+}$ and $O {H}^{-}$when placed in water. This also gives us an estimate of its pH, since strong basic solutions usually have a very high $p H$.

$N a O H \left(a q\right) \iff N {a}^{+} \left(a q\right) + O {H}^{-} \left(a q\right)$

Complete dissociation means that the concentrations of$N {a}^{+}$ cations and $O {H}^{-}$ anions will be equal to the initial concentration of $N a O H$. We can determine that by calculating the number of $N a O H$ moles found in $8.6 g$.

${n}_{N a O H} = \frac{m}{m o l a r m a s s} = \frac{8.6 g}{40 \frac{g}{m o l}} = 022$ moles

Therefore, $N a O H$ molarity is

$C = \frac{n}{V} _ \left(s o l u t i o n\right) = \frac{0.22 m o l e s}{1.2 L} = 0.18 M$

We know that $\left[O {H}^{-}\right]$ (the concentration of $O {H}^{-}$) is equal to this value, so $\left[O {H}^{-}\right]$ = $0.18 M$.

We can now calculate $p O H$ by

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(0.18\right) = 0.74$

Therefore, the solution's $p H$ is

$p H = 14 - p O H = 14 - 0.74 = 13.26$, which matches our initial estimate of a very high $p H$.