What is the pH of a #1.0 x 10^-5# M solution of NaOH?

1 Answer
anor277
Dec 2, 2016

In water #pH+pOH=14#

#pH=14-pOH=14-5=9#

Explanation:

#pOH=-log_10[HO^-]# by definition.

Thus #pOH=-log_10(1xx10^-5)=-(-5)=5#

#pH+pOH=14#

#pH=14-pOH=14-5=9#

When we write #log_ab=c#, we ask to what power we raise the base, #a#, to get #b#; here #a^c=b#. The normal bases are #10#, #"(common logarithms)"#, and #e#, #"(natural logarithms)"#.

Thus when we write #log_10(10^-5)#, we are asking to what power we raise #10# to get #10^-5#. Now clearly the answer is #-5#, i.e. #log_10(10^-5)=-5#, alternatively #log_10(10^5)=+5#.

What are #log_10(100), log_10(1000), log_10(1000000)??#

You shouldn't need a calculator, but use one if you don't see it straight off.

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