Question

Is a solution with a pOH of 12.1 acidic, basic, or neither?

Answer 1

This value specifies a highly acidic solution...............`pH=1.9`

Explanation:

For the acid-base equilibrium that operates in water.....

`2H_2O rightleftharpoons H_3O^+ + HO^-`

We know that at under standard conditions, `298*K, "1 atmosphere"`.............

`K_w=[H_3O^+][HO^-]=10^-14`

This defines the acid-base equilibrium, the autoprotolysis of water.

We can take `log_10` of BOTH sides........

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]`

And on rearrangement,

`underbrace(-log_10[H_3O^+])_(pH)-underbrace(log_10[HO^-])_(pOH)=underbrace(-log_10K_w)_(pK_w)`

Given that `K_w=10^-14`, and `pH=-log_10[H_3O^+]`, then BY DEFINITION, `underbrace(-log_10K_w)_(pK_w)=-log_(10)10^(-14)=-(-14)=14`, and the defining relationship, which you may not have to derive, but WILL have to remember,

`14=pH +pOH`

And given that `pOH=12.1`, this means that `[HO^-]=10^(-12.1)*mol*L^-1`, and `pH=14-12.1=1.9`, and (FINALLY) `[H_3O^+]=10^(-1.9)*mol*L^-1=0.0126*mol*L^-1.`

Now this might seem a lot of work, but only because I derived the equation. You must be able to use the relationship...........

`14=pH +pOH`

These logarithmic terms were introduced back in the day, before the advent of electronic calculators. Log tables, printed values of `log_10` and `log_e` were widely used by scientists, engineers, and by students of course.