Question

Calculate the [H³O+] with POH 4.5

Answer 1

 pH and pOH are only meaningful when applied to aqueous (water-based) solutions. When water dissociates it yields a hydrogen ion and a hydroxide.

H2O ⇆ H+ + OH-

When calculating pOH, remember that [] refers to molarity, M.

Kw = [H+][OH-] = 1x10-14 at 25°C
for pure water [H+] = [OH-] = 1x10-7
Acidic Solution: [H+] > 1x10-7
Basic Solution: [H+] < 1x10-7

How to Find pOH Using Calculations

There are a few different formulas you can use to calculate pOH, the hydroxide ion concentration, or the pH (if you know pOH):

pOH = -log10[OH-]
[OH-] = 10-pOH
pOH + pH = 14 for any aqueous solution

pOH Example Problems

Find the [OH-] given the pH or pOH. You are given that the pH = 4.5.

pOH + pH =14
pOH + 4.5 = 14
pOH = 14 - 4.5
pOH = 9.5

[OH-] = 10-pOH
[OH-] = 10-9.5
[OH-] = 3.2 x 10-10 M

Find the hydroxide ion concentration of a solution with a pOH of 5.90.

pOH = -log[OH-]
5.90 = -log[OH-]
Because you're working with log, you can rewrite the equation to solve for the hydroxide ion concentration:

[OH-] = 10-5.90
To solve this, use a scientific calculator and enter 5.90 and use the +/- button to make it negative and then press the 10x key. On some calculators, you can simply take the inverse log of -5.90.

[OH-] = 1.25 x 10-6 M

Find the pOH of a chemical solution if the hydroxide ion concentration is 4.22 x 10-5M.

pOH = -log[OH-]
pOH = -log[4.22 x 10-5]

To find this on a scientific calculator, enter 4.22 x 5 (make it negative using the +/- key), press the 10x key, and press equal to get the number in scientific notation. Now press log. Remember your answer is the negative value (-) of this number.
pOH = - (-4.37)
pOH = 4.37

Understand Why pH + pOH = 14

Water, whether it's on its own or part of an aqueous solution, undergoes self-ionization which can be represented by the equation:

2 H2O ⇆ H3O+ + OH-

Equilibrium forms between the unionized water and the hydronium (H3O+) and hydroxide (OH-) ions. The expression for the equilibrium constant Kw is:

Kw = [H3O+][OH-]

Strictly speaking, this relationship is only valid for aqueous solutions at 25°C because that is when the value of Kw is 1 x 10-14. If you take the log of both side of the equation:

log (1 x 10-14) = log [H3O+] + log [OH-]

(Remember, when numbers are multiplied, their logs are added.)

log (1 x 10-14) = - 14
- 14 = log[H3O+] + log [OH-]

Multiplying both sides of the equation by -1:

14 = - log [H3O+] - log [OH-]

pH is defined as - log [H3O+] and pOH is defined as -log [OH-], so the relation becomes:

14 = pH - (-pOH)
14 = pH + pOH