PH of strong acids and bases

When calculating pH, it is always necessary to consider what is the source of oxonium cations in a given environment.

Strong monosaturated acids
For strong monosaturated acids the dissociation follows the equation

$$\mathrm{HA} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{A}^{-} + \mathrm{H}_3\mathrm{O}^{+}.$$

For the calculation we assume: Let's deduce:
 * the substance quantity of $$\mathrm{H}_3\mathrm{O}^{+}$$ according to the above equation will be the same as $$\mathrm{A}^{-}$$, which, given an identical volume, is also true for the concentration, i.e. $$[\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{A}^{-}]$$;
 * all acid - because it is a strong acid - is converted into $$\mathrm{A}^{-}$$ a $$\mathrm{H}_3\mathrm{O}^{+}$$, therefore we will mark $$[\mathrm{A}^{-}]$$ its concentration, i.e. $$[\mathrm{A}^{-}] = c_{\mathrm{HA}}$$

$$\mathrm{pH} = -\log\ [\mathrm{H}_3\mathrm{O}^{+}] = -\log\ [\mathrm{A}^{-}] = -\log\ c_{\mathrm{HA}},$$

and to calculate the pH we get the formula $$\mathrm{pH} = -\log\ c_{\mathrm{HA}}.$$

Strong monosaturated bases
For strong monosaturated bases the dissociation follows the equation

$$\mathrm{BOH} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{+} + \mathrm{OH}^{-} + \mathrm{H}_2\mathrm{O}$$

We assume, as in the case of strong monosaturates, that: The calculation is therefore analogous, we just have to remember that unlike acids, the base is not a source of oxonium cations, but takes oxonium cations from the environment (see the theory of acids and bases), so we add from the equation for the ionic product of water: and from these assumptions, we deduce
 * the amount, or concentration, of hydroxide ions and the resulting $$\mathrm{B}^{+}$$ is the same according to the above chemical equation, i.e. $$[\mathrm{OH}^{-}] = [\mathrm{B}^{+}]$$;
 * dissociation occurs completely, i.e. $$[\mathrm{B}^{+}] = c_{\mathrm{BOH}}.$$
 * $$[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}$$

$$\mathrm{pH} = -\log\ [\mathrm{H}_3\mathrm{O}^{+}] = -\log\ \frac{K_w}{[\mathrm{OH}^{-}]} = \log\ [\mathrm{OH}^{-}] - \log K_w = \log\ [\mathrm{B}^{+}] - \log K_w = \log c_{\mathrm{BOH}} -\log K_w.$$

Calculate the pH at 25 °C using the formula $$\mathrm{pH} = 14 + \log\ c_{\mathrm{BOH}}.$$

Strong dibasic acids
Strong dibasic acids dissociate according to the equation

$$\mathrm{H}_2\mathrm{A} + 2\ \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{A}^{2-} + 2\ \mathrm{H}_3\mathrm{O}^{+},$$

we assume, then: From this we derive $$\mathrm{pH} = -\log\ [\mathrm{H}_3\mathrm{O}^{+}] = -\log\ [\mathrm{A}^{2-}] = -\log (2 \cdot c_{\mathrm{H}_2\mathrm{A}}) = -\log 2 - \log c_{\mathrm{H}_2\mathrm{A}}$$
 * complete dissociation, i.e. $$c_{\mathrm{H}_2\mathrm{A}} = [\mathrm{A}^{2-}];$$
 * however, the amount of oxonium cations and the amount of formed $$\mathrm{A}^{2-}$$ is - in contrast to monosaturated acids - in a ratio of 1:2, i.e. $$[\mathrm{H}_3\mathrm{O}^{+}] = 2 \cdot c_{\mathrm{H}_2\mathrm{A}}.$$

and the pH is calculated according to the formula $$ \mathrm{pH} = - \log\ c_{\mathrm{H}_2\mathrm{A}} - \log\ 2.$$

Strong dibasic bases
Strong dibasic bases dissociate according to the equation

$$\mathrm{B(OH)}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{2+} + 2\ \mathrm{OH}^{-} + \mathrm{H}_2\mathrm{O}$$

as we assume for monosaturated bases and dibasic acids: Then we derive
 * complete dissociation, i.e. $$c_{\mathrm{B(OH)}_2} = [\mathrm{B}^{2+}];$$
 * concentration of the formed $$\mathrm{B}^{2+}$$ and the concentration of hydroxide anions is in the ratio 1:2, i.e.$$[\mathrm{OH}^{-}] = 2 \cdot [\mathrm{B}^{2+}]$$, in addition, according to the previous assumption $$[\mathrm{OH}^{-}] = 2 \cdot c_{\mathrm{B(OH)}_2}$$
 * hydroxide anions drain oxonium cations from the environment, $$[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}$$.

$$- \log [\mathrm{H}_3\mathrm{O}^{+}] =- \log \frac{K_w}{[\mathrm{OH}^{-}]} = \log\ [\mathrm{OH}^{-}] - \log K_w = \log (2 \cdot c_{\mathrm{B(OH)}_2}) - \log K_w = \log 2 + \log\ c_{\mathrm{B(OH)}_2} - \log K_w$$

and the pH at 25 °C is calculated according to the formula $$\mathrm{pH} = 14 + \log 2 + \log\ c_{\mathrm{B(OH)}_2}.$$

Related articles

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