How does #10^(–4.80) = 10^(log[H3O+])# become #10^(–4.80) = [H_3O^+]#? Please see image below, under the heading #Method 2: 10^x Function#.

Fundamentals of Chemistry by Kate Rowen

1 Answer
anor277
Feb 9, 2018

You are asking us how to define #pH#...

Explanation:

Back in the day, (the which I can BARELY remember), students, scientists, and engineers, used to use sets of log tables, or slide rules, for calculations the which a modern calculator would make very short work of.... The #pH# scale utilizes the logarithmic scale...and it has stuck with us...

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

And this is the autoprotolysis reaction in aqueous solution under standard conditions, but we can take #log_10# of both sides...

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

#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#

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

Now your question actually tells you what to do...you still have to use your calculator effectively and competently...good luck,

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