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

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