How ph depends on concentration?

1 Answer
May 8, 2018

Well, by definitions, #pH=-log_10[H_3O^+]#

Explanation:

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

Just to note that in aqueous solution under standard conditions, the ion product...

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

And we can take #log_10# of both sides to give....

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

And thus.... #-14=log_(10)[H_3O^+]+log_(10)[HO^-]#

Or.....

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

#14=underbrace(-log_10[H^+])_(pH)underbrace(-log_10[OH^-])_(pOH)#

#14=pH+pOH#
By definition, #-log_10[H^+]=pH#, #-log_10[HO^-]=pOH#