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What is #[H^+]# if #pH=4.18#?

1 Answer

Jul 20, 2017

#[H^+]=10^(-4.18)*mol*L^-1#

Explanation:

By definition, #pH=-log_10[H^+]#; equivalently, #pH=-log_10[H_3O^+]#.

We take antilogs..........

#[H_3O^+]=10^(-4.18)*mol*L^-1=6.61xx10^-5*mol*L^-1#.

See https://socratic.org/questions/does-flat-coke-go-ph-neutral for background......and also here.

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