How do you calculate [OH−] in [H3O+]=2.1×10−8 M at 25°C?

1 Answer
anor277
Mar 29, 2017

How do we calculate [HO−] IF [H3O+] =2.1×10−8⋅mol⋅L−1?

Well, [−OH]=4.78×10−7⋅mol⋅L−1.

Explanation:

We know that Kw=[H3O+][−OH]=10−14, in water at 298⋅K.

We take log10 of both SIDES............

log10Kw=log10[H3O+]+log10[HO−]=log1010−14

But log1010−14=−14 by definition............

And if we DEFINE,
−log10Kw=pKw, and −log10[H3O+]=pH, and −log10[HO−]=pOH

Then................

pH+pOH=pKw=14.

You should commit this last result to memory, as it makes the treatment of acidity/basicity straightforward.

So if [H3O+]=2.1×10−8⋅mol⋅L−1, pH=−(−log102.1×10−8)=−(−7.68)=+7.68

And thus pOH=14−7.68=6.32, and [HO−]=10−6.32=4.78×10−7⋅mol⋅L−1.

The solution is slightly BASIC....................

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