Question

The pH of a solution changes fro 300 to 6.00. By what factor does the `[H_3O^+]` change?

Answer 1

By a factor of `1000`.

Explanation:

`pH=-log_10[H_3O^+]`

If `pH_1=3.00`, then `[H_3O^+]=10^-3*mol*L^-1.`

If `pH_2=6.00`, then `[H_3O^+]=10^-6*mol*L^-1.`

`(pH_1)/(pH_2)=(10^-3*mol*L^-1)/(10^-6*mol*L^-1)=10^3`

The latter concentration is less concentrated with respect to `H_3O^+` than the former by a factor of `1000`.