Question

What is the pH and pOH of a 1.2 x 10-3 [H+] solution?

What is the pH and pOH of a 1.2 x 10-3 [H+] solution?

Answer 1

Well, `pH-=-log_10[H_3O^+]`

Explanation:

And so if `[H_3O^+]=1.2xx10^-3*mol*L^-1`, then `pH=-(log_10{1.2xx10^-3})=-(-2.92)=2.92`...

The logarithmic scale is a throwback before the advent of electronic calculators.. And so background in water WE KNOW the protonolysis reaction:

`2H_2OrightleftharpoonsH_3O^+ + HO^-`

Under standard conditions, this expression has been quantified...

`K_w=underbrace([H_3O^+])_("equivalently, "H^+)[HO^-]=10^-14`

We take logs of both sides to get....

`log_10[H_3O^+]+log_10[HO^-]=underbrace(log_10(10^-14))_(-=-14" by definition")`

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

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

And so (finally!)..`14=pH+pOH`...and you will have to know how and when to use the expression....

And so `pOH=14-pH=14-2.92=11.08`