The equilibrium obtained following the ionization of water makes it possible to explain the behaviour of acids and bases in aqueous solution, as well as the resulting concentrations of |H^{+}| and |OH^{-}| ions.
The pH scale is a way of expressing |H^+| ion concentration in an aqueous solution. This scale expresses low concentration values in a more practical way. Therefore, |\text{pH}| measurements correspond to different values of |H^+| ion concentrations:
| pH | Solution | [|H^+|] |
| pH < 7 | Acid | [|H^+|] > |1\times 10^{-7}| mol/L |
| pH = 7 | Neutral | [|H^+|] = |1\times 10^{-7}| mol/L |
| pH > 7 | Basic | [|H^+|] < |1\times 10^{-7}| mol/L |
We can, therefore, express pH as follows:
|pH = -log [H^+]|
|[H^{+}] = 10^{-pH}|
As for |\text{pOH}|, it can be expressed as follows:
|pOH = -log [OH^-]|
|[OH^{-}] = 10^{-pOH}|
Finally, it is important to remember that the sum of pH and pOH is always equal to 14 in standard conditions:
|pH + pOH = 14|
The ionization constant of water applies to all aqueous solutions. Since it is not affected by the concentration of ions in solution, the ionization constant of water is always the same for a given temperature. Thus, it can be used to calculate the concentration of one of the ions in solution (hydronium or hydroxide), as long as one of the two concentrations or the pH of the solution is known. This calculation is possible whether or not there is an acid or a base in solution.
The following example shows how to use the ionization constant of water to find out the molar concentrations of the present ions when the pH is known.
A solution of phosphoric acid |(H_{3}PO_{4})| has a |\text{pH}| of |3.7|. What is its |OH^{-}| ion concentration?
1. Calculation of |H^+| ions:
|[H^{+}] = 10^{-pH}|
|[H^{+}] = 10^{-3.7}|
|[H^{+}] = 2\times 10^{-4} M|
2. Calculation of |OH^-| ions:
We will use the ionization constant of water to perform this calculation.
|K_{water} = [H^{+}]\cdot[OH^{-}] = 1\times 10^{-14}|
|[OH^{-}] = \displaystyle \frac{1\times 10^{-14}}{[H^+]}|
|[OH^{-}] = \displaystyle \frac{1\times 10^{-14}}{2\times 10^{-4}}|
|[OH^{-}] = 5\times 10^{-11} M|
The concentration of |OH^{-}| ions is |5\times 10^{-11}\ \text{mol/L}|.
The following example shows how to use the ionization constant of water to find out the molar concentration of one ion present when the concentration of the other ion is known.
At 25 ºC, an aqueous solution is prepared by adding |7.3\ \text{g}| of hydrochloric acid |(\text{HCl})| in a container that will hold a total volume of |10\ text{L}| of this solution. Determine the |OH^{-}_{(aq)}| ion concentration present.
Solution :
1. Molar concentration of |HCl| in this resulting solution
|7.3\ \text{g de HCl}| corresponds to:
|\displaystyle \frac{7.3\ \text{g HCl}}{36.5\ \text{g/mol HCl}}\ =\ 0.2\ \text{moles of HCl}|
|HCl| dissociates completely according to the following equation:
|HCl_{(aq)} \rightleftharpoons H^{+}_{^(aq)} + Cl^{-}_{(aq)}|
I : 0.2 mol 0 mol
E : 0 mol 0.2 mol
The total volume being |10\ \text{L}| , the molar concentration becomes:
|\displaystyle \frac{0.2\ \text{mol}}{10\ \text{L}} = 0.02\ \text{mol/L or}\ 0.02\ \text{M}|
So the final |[H^{+}_{(aq)}]| will be |2\times10^{-2}\ \text{mol/L}| because the concentration of |H^+_{(aq)}| already present in the water is not significant.
Consequently:
|K_{H_{2}O}| = |[H^+_{(aq)}]\times[OH^-_{(aq)}]|
|1\times10^{-14}\ =\ {2\times10^{-2}}\times[OH^-_{(aq)}]|
|[OH^-_{(aq)}]\ = \displaystyle \frac{1\times10^{-14}}{2\times10^{-2}}|
|[OH^-_{(aq)}]\ = 5\times10^{-13}\ \text{mol/L}|
The concentration of ions |[OH^-_{(aq)}]| is of |5\times10^{-13}\ \text{mol/L}|.