Calculate either [h3o ] or [oh–] for the solutions below at 25 c. [oh-]=1.83x10^-7 m [h3o+]=? m

Answer is: concentration of hydronium ions is 5,5·10⁻⁸ M.
[OH⁻] = 1,83·10⁻⁷ M; equilibrium concentration.
The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 1·10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = Kw ÷ [OH⁻].
[H₃O⁺] = 1·10⁻¹⁴ M² ÷ 1,83·10⁻⁷ M.
[H₃O⁺] = 5,5·10⁻⁸ M.

This problem requires using the concept of pH and pOH. The “p” stands for the negative log so pH and pOH represent the negative log of the concentration of hydrogen or hydroxide ions.

Here is the solution:

pOH = -log [OH-]

pOH = -log [1.83x10^-7 M]

pOH = 6.74

 

pH + pOH = 14

pH = 14 - 6.74

pH = 7.26

 

pH = -log [H3O+]

7.26 = -log[H3O+]

[H3O+] = 5.46 x 10^-8 M

It will appear as o2(g) as a reactant 

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