Question

derive the relation ph+poh=14

Answer 1

hi

We can write the equilibrium as follows:

2H2O⇌H3O++HO−

And, Kw = [H3O+][HO−]. This equilibrium has been carefully measured (by conductivity experiments) over a range of temperatures.[H2O] does not appear in the equilibrium because its concentration is effectively constant.

At 298⋅K we know that Kw=10−14. How do we know this? By careful and repeated measurements of the conductivity of the water solvent.

So Kw=[H3O+][HO−]=10−14.

Now this is an arithmetic expression, that we can divide, multiply, add to, subtract from, provided that we do it to both sides of the equation. We can take log10 of both sides to give:

log10Kw=log10[H3O+]+log10[HO−]=log10(10−14)

But, by definition of logarithms, if ab=c, logac=b, then log10(10−14)=−14, and thus,

log10Kw=log10[H3O+]+log10[HO−]=−14

But by definition, −log10[H3O+]=pH, and −log10[HO−]=pOH

pKw=pH+pOH=14 as required.

hope it helps u
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Answer 2

hey there here is the process

1) h2o→H+ + oh-

kw=10-14

-log 10-14 =14

now

14=-log ( h3o +) + -(log (oh -)

-log (h3o + ) poh=-log (oh -)

14=pH +poh

important:- log BC =log b + log c

which we have applied