What is #pH# for a solution whose #[H_3O^+]=6.7xx10^-5*mol*L^-1#?

1 Answer
anor277
Aug 19, 2016

#pH=4.17#

Explanation:

#pH=log_10[H_3O^+]# (equivalently #pH=log_10[H^+]#).

Thus #pH=log_10(6.7xx10^-5)# #=# #-(-4.17)#

Note that when we write #log_ab=c#, we are looking for the power to which we raise the base #a# to get #b#; here #a^c=b#. What are #log_(10)100#, #log_(10)1000000#, and #log_(10)0.001#?

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