Question

What is the [OH-] of a solution whose [H3O+] is 2.50 x 10-4 ?

Answer 1

Approx....`4xx10^-11*mol*L^-1`

Explanation:

We use the old autoprotolysis equation:

`2H_2O(l) rightleftharpoonsH_3O^+ +HO^-`

And this equilibrium has been carefully measured....

`K_w=[H_3O^+][HO^-]=10^-14`...

And given such an equation we may manipulate it provided that we maintain the equality by doing to the manipulation to both sides.... One thing we can do is to take `log_10` of each side....

`logK_w=underbrace(log_10[H_3O^+])_(-pH)+underbrace(log_10[HO^-])_(-pOH)=underbrace(log_10(10^-14))_(-14)`

...And so `pH+pOH=14`

We are given `[H_3O^+]=2.50xx10^-4`

...and thus `pH=-log_10(2.50xx10^-4-(3.60)=+3.60`...

...so `pOH=14-3.60=10.4`

And `[HO^-]=-log_10(10^(-10.4))=3.98xx10^-11*mol*L^-1`