Problem 3 Show that pH pOH 14 Provide a detailed solution sh

Problem 3 Show that pH+ pOH 14. Provide a detailed solution showing each step.

Solution

We can write the equilibrium as follows:

2H₂OH₃O⁺+HO

And, K_w = [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 298K we know that K_w=10¹⁴. How do we know this? By careful and repeated measurements of the conductivity of the water solvent.

So Kw=[H₃O⁺][HO]=10¹⁴.

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:

log10K_w=log10[H₃O⁺]+log10[HO]=log10(10¹⁴)

But, by definition of logarithms, if ab=c, logac=b, then log10(10¹⁴)=14, and thus,

log10K_w=log10[H₃O⁺]+log10[HO]=14

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

pKw=pH+pOH=14 as required.