Question

Question #52060

Answer 1

I have given a full answer here:
https://socratic.org/questions/ph-of-10-8m-naoh?source=search

Explanation:

I assume you mean `sf(10^-8color(white)(x)M)` NaOH solution.

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

`sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2))` at 298K

`:.` `sf([H^+]=10^(-14)/[[OH^-]]=10^(-14)/10^(-8)=10^(-6)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log(10^-6)=6)`

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

`sf(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 `sf(H^+)` formed from the auto - ionisation of water. This is done in the answer which I have given the link to.

Answer 2

The `pH` is indeed `6.0`.

Explanation:

You can find the `pH` from a given `[OH^-]` using an equation like this:

`pH = 14.0-(-log([OH^-]))`

`pH = 14.0-(-log(10^-8 M))`

`pH = 14.0-(8.0)= **6.0**`

`NaOH` is a base, but the solution is basic only if the `[OH^-]` is greater than `1 xx 10^-7 M`, i.e. greater than `[H^+]`.