If the #pH=9# for a solution of #KOH(aq)#, what is the concentration of #HO^-#?

Question

#A.# #9molL^-1#;
#B.# #5molL^-1#;
#C.# #10^-5molL^-1#;
#D.# #"Cannot be determined."#

Noah Aldrich · Accepted Answer

#"Option C"#

It is a known fact that in standard conditions and an aqueous solution,

#pH+pOH# #=# #14#

If #pH# #=# #9#, then #pOH=14-9=5#

And thus, given the definition of #pOH#, #[HO^-]# #=# #10^(-5)*mol*L^-1#.

To further clarify, let's revisit the process of water autoprotolysis:

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

We can now definitely write this rxn's equilibrium expression:

#([H_3O^+][HO^-])/([H_2O]^2)# #=# #K_w#

At #298# #K# and #1*atm#, #K_w# has been precisely and accurately measured, and the #[H_2O]# term can be removed because it is so large it is effectively constant. And thus,

#[H_3O^+][HO^-]# #=# #K_w# #=# #10^(-14)#

Now this is a mathematical expression, which we can divide, mulitply, etc. PROVIDED that we do it to both sides of the equation. One thing that can be done (and was done routinely in the days before electronic calculators) was take #log_10# of both sides:

#log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14#

OR

#-log_(10)10^-14=-log_10[H_3O^+]-log_10[HO^-]#

But, #-log_(10)10^-14=-(-14)=14#, and #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]# #=# #pOH#. All of this is by definition.

And thus #pH+pOH=14#.

This final one is definitely worth memorizing. In the case of an undergraduate, memorize the entire derivation.

Ellie Andrews · Answer

undefined $$ \text{pOH} = 14 - \text{pH} $$
$$ [\text{OH}^-] = 10^{-\text{pOH}} $$

If the #pH=9# for a solution of #KOH(aq)#, what is the concentration of #HO^-#?

#A.# #9*mol*L^-1#; #B.# #5*mol*L^-1#; #C.# #10^-5*mol*L^-1#; #D.# #"Cannot be determined."#

#A.# #9molL^-1#;
#B.# #5molL^-1#;
#C.# #10^-5molL^-1#;
#D.# #"Cannot be determined."#