Question

How do you calculate `[OH^-]` in `[H_3O^+] = 2.1 times 10^-8` `M` at 25°C?

Answer 1

How do we calculate `[HO^-]` IF `[H_3O^+]` `=2.1xx10^-8*mol*L^-1`?

Well, `[""^(-)OH]=4.78xx10^-7*mol*L^-1.`

Explanation:

We know that `K_w=[H_3O^+][""^(-)OH]=10^-14`, in water at `298*K`.

We take `log_10` of both SIDES............

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14`

But `log_(10)10^-14=-14` by definition............

And if we DEFINE,
`-log_10K_w=pK_w`, and `-log_10[H_3O^+]=pH`, and `-log_10[HO^-]=pOH`

Then................

`pH+pOH=pK_w=14`.

You should commit this last result to memory, as it makes the treatment of acidity/basicity straightforward.

So if `[H_3O^+]=2.1xx10^-8*mol*L^-1`, `pH=-(-log_(10)2.1xx10^-8)=-(-7.68)=+7.68`

And thus `pOH=14-7.68=6.32`, and `[HO^-]=10^(-6.32)=4.78xx10^-7*mol*L^-1.`

The solution is slightly BASIC....................