What is the [H3O+] of a solution with the PH of 4.98?

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Answer 1 · 2018-04-23T12:29:13

#[H_3O^+]=1.05xx10^-5*mol*L^-1#...

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

And given that if #log_ay=z# if follows that #a^z=y#

And so if #pH=4.98#, #[H_3O^+]=10^(-4.98)*mol*L^-1=??*mol*L^-1#..

Answer 2 · 2018-04-23T12:29:26

#10^-4.98=1.05 times 10^-5 mol dm^-3#

As

#pH=-log_10[H_3O^(+)]#

Hence, using log laws:

#10^-(pH)=[H_3O^(+)]#

Therefore if #pH=4.98#

Then

#10^(-4.98)=[H_3O^(+)]=1.05 times 10^(-5) mol dm^-3#