How do you calculate [OH] in [H3O+]=2.1×108 M at 25°C?

1 Answer
Mar 29, 2017

How do we calculate [HO] IF [H3O+] =2.1×108molL1?

Well, [OH]=4.78×107molL1.

Explanation:

We know that Kw=[H3O+][OH]=1014, in water at 298K.

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

log10Kw=log10[H3O+]+log10[HO]=log101014

But log101014=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+]=2.1×108molL1, pH=(log102.1×108)=(7.68)=+7.68

And thus pOH=147.68=6.32, and [HO]=106.32=4.78×107molL1.

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