# Question #d8f04

Dec 5, 2014

The answer is $5.5 \cdot {10}^{- 14}$.

$N a O H$ is a strong base, which means it dissasociates completely into $N {a}^{+}$ and $O {H}^{-}$.

$N a O H \left(a q\right) + {H}_{2} O \left(l\right) \iff N {a}^{+} \left(a q\right) + O {H}^{-} \left(a q\right)$

Starting from the given mass of $8.6$ grams of $N a O H$ and knowing that its molar mass is $40 \frac{g}{m o l e}$, we get

${n}_{N a O H} = \frac{8.6 g}{40 \frac{g}{m o l e}} = 0.215$ moles.

The molar concentration of $N a O H$ is

$C = \frac{n}{V} = \frac{0.215 m o l e s}{1.2 L} = 0.18 M$

Now, a strong base (and a strong acid for that matter) dissasociates completely, which means that the concentrations of $N {a}^{+}$ and $O {H}^{-}$ in aqueous solution are equal to the initial concentration of $N a O H$, $0.18 M$.

Therefore, we can calculate the $p O H$ using

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

We can determine the $p H$ by using

pH $= 14 - p O H = 14 - 0.74 = 13.26$

This means that the $\left[{H}^{+}\right]$ is equal to

$\left[{H}^{+}\right] = {10}^{- p H} = {10}^{- 13.26} = 5.5 \cdot {10}^{- 14}$