Question

What is `pH` if `(i)` `[H_3O^+]=5.6xx10^-5*mol*L^-1`; and `(ii)` if `[HO^-]=10^-7*mol*L^-1`?

Answer 1

`(i)` `pH=4.25.................`

Explanation:

By definition, `pH=-log_10[H_3O^+]`, and thus for (i)....

`pH=-log_10(5.6xx10^-5)=-(-4.25)=4.25` (if I have counted up the zeroes properly........)

But since `[H_3O^+]xx[HO^-]=10^-14`, we can take `log_10` of both sides, and gets........

`log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14=-14`

Equivalently, multiplying each side by `-1`............

`underbrace(-log_10[H_3O^+])_color(red)(pH)underbrace(-log_10[HO^-])_color(blue)(pOH)=+14`

And thus our defining relationship......

`color(red)(pH)+color(blue)(pOH)= 14`

And so (ii) if `[HO^-]=10^-7*mol*L^-1`, `pOH=7` (why?), and `pH=7`. Agreed?