# Question #52060

May 18, 2017

#### Explanation:

I assume you mean $\textsf{{10}^{-} 8 \textcolor{w h i t e}{x} M}$ NaOH solution.

The pH cannot be 6 because, as you correctly state, NaOH is a base or alkali. I guess you reasoned:

$\textsf{{K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = {10}^{- 14} \textcolor{w h i t e}{x} {\text{mol"^2."l}}^{- 2}}$ at 298K

$\therefore$$\textsf{\left[{H}^{+}\right] = {10}^{- 14} / \left[\left[O {H}^{-}\right]\right] = {10}^{- 14} / {10}^{- 8} = {10}^{- 6} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left({10}^{-} 6\right) = 6}$

Clearly, this cannot be right as such a solution would be weakly acidic.

$\textsf{{10}^{-} 8}$M is a fantastically dilute solution so you would expect the pH to fall towards 7 as the solution is diluted. It cannot fall below 7.

At these low concentrations you need to take into account the $\textsf{{H}^{+}}$ formed from the auto - ionisation of water. This is done in the answer which I have given the link to.

May 18, 2017

The $p H$ is indeed $6.0$.

#### Explanation:

You can find the $p H$ from a given $\left[O {H}^{-}\right]$ using an equation like this:

$p H = 14.0 - \left(- \log \left(\left[O {H}^{-}\right]\right)\right)$

$p H = 14.0 - \left(- \log \left({10}^{-} 8 M\right)\right)$

$p H = 14.0 - \left(8.0\right) = \ast 6.0 \ast$

$N a O H$ is a base, but the solution is basic only if the $\left[O {H}^{-}\right]$ is greater than $1 \times {10}^{-} 7 M$, i.e. greater than $\left[{H}^{+}\right]$.