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

1 Answer
Apr 25, 2017

Answer:

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.