Question

A sample of stomach acid has a pOH of 12.60. What is the pH of this solution?

Answer 1

Well...`pOH+pH=14` in aqueous solution under standard conditions....

Explanation:

And so `pH=14-12.60=1.40`...the solution is QUITE acidic...

And why? Well we interrogate the following equilibrium...

`2H_2O(l) rightleftharpoonsH_3O^+ +HO^(-)`

For which `K_w=10^(-14)=[H_3O^+][HO^-]`

And we takes `log_10` OF BOTH SIDES....

`log_10K_w=underbrace(log_(10)10^(-14))_(-14)=log_10[H_3O^+]+log_10[HO^-]`

And so `14=-log_10[H_3O^+]-log_10[HO^-]`

..but by definition....

`14=underbrace(-log_10[H_3O^+])_"pH"underbrace(-log_10[HO^-])_"+ pOH"`

And so `"pH+pOH=14"`....and you will end up using this relationship a lot....and likewise for a given acid, and conjugate base under these conditions....`pK_a+pK_b=14"`...