I need help calculating [H3O] from pH and the pOH scale

Question

using my calculator. I have a TI-30XIIS

Answer 1

Given pH = 4.67, pH = ?
pH = -log(H^+)
4.67 = -log(H^+)
-4.67 = log(H^+)
(H^+) = 2.29E-5

Here is how you do it on the calculator.
Punch in 4.67. Hit the change sign to it becomes -. (If you don't have a change sign, punch in -4.67 at the start.)
Now locate and hit the 10^x key. That should return 2.29086E-5 which can be rounded.
For pOH, the easy way, if you have pH, is
pH + pOH = pKw = 14
So 14=pH = pOH.

If you have OH^- already, then
pOH = -log(OH^-) and it works just like the pH does. By the way, if you need pKa, it works just like pH and pOH, pKa = -log Ka. pKb = -log Kb.