What is the #pH# of a solution for which #[H_3O^+]=6.31xx10^-4molL^-1#?

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Answer 1 · 2016-07-30T22:19:13

#pH=3.20#

#pH=-log_10[H_3O^+]#

#=-log_10(6.31xx10^-4)=#

#-log_10{6.31xx10^-4}=-(-3.20)=3.20#

What is the #pH# of a solution for which #[H_3O^+]=6.31xx10^-4*mol*L^-1#?