I need help calculating [H3O] from pH and the pOH scale
using my calculator. I have a TI-30XIIS
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 10x 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.
To calculate [H3O+] from pH, you can use the equation:
[H3O+] = 10^(-pH)
To calculate [H3O+] from pOH, you can use the equation:
[H3O+] = 10^(14 - pOH)
Now, let's calculate [H3O+] using your calculator, the TI-30XIIS:
1. Turn on your calculator.
2. Make sure it is in the "normal" mode.
3. Start by calculating [H3O+] from pH.
a. Enter the pH value.
b. Press the key for the exponent symbol "^".
c. Enter -1 (since the exponent is negative).
d. Press the "=" (equals) key.
The result displayed on the screen is the value of [H3O+].
4. To calculate [H3O+] from pOH:
a. Enter the pOH value.
b. Subtract the entered pOH value from 14.
c. Press the key for the exponent symbol "^".
d. Enter -1 (since the exponent is negative).
e. Press the "=" (equals) key.
The result displayed on the screen is the value of [H3O+].
Remember to check the mode and settings on your calculator beforehand, as different versions may have slightly different key functions, but the general steps should remain the same.