bleakarcher
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- Jul 8, 2011
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- HSC
- 2013
re: HSC Chemistry Marathon Archive
pH=-log[H+]
pOH=-log[OH-]
At 25 degrees C, [H+][OH-]=10^(-14)
Taking the common log of both sides:
log[[H+][OH-]]=log[10^(-14)]
log[H+]+log[OH-]=-14
-pH-pOH=-14
pH=14-pOH
By definition,I'm a little unsure with the bolded part and Sy do you make these questions up? I could use a few for practice lol.
pH=-log[H+]
pOH=-log[OH-]
At 25 degrees C, [H+][OH-]=10^(-14)
Taking the common log of both sides:
log[[H+][OH-]]=log[10^(-14)]
log[H+]+log[OH-]=-14
-pH-pOH=-14
pH=14-pOH