PH of Salts

When we talk about the pH of salts, we mean the pH of aqueous solutions of soluble salts. Such salts dissociate in solution according to Eq

$$\mathrm{BA} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{+} + \mathrm{A}^{-} + \mathrm{H}_2\mathrm {O},$$

where $$\mathrm{BA}$$ is a salt of the acid $$\mathrm{HA}$$ and the base $$\mathrm{B(OH)},$$ which is formed e.g. by neutralization according to Eq

$$ \mathrm{HA} + \mathrm{BOH} \rightarrow \mathrm{BA} + \mathrm{H}_2\mathrm{O}.$$

pH of salt of strong acid and strong base
In the case of a salt of a strong acid and a strong base, we consider that
 * cation $$\mathrm{B}^{+}$$ — because it is strong — will remain dissociated: $$\mathrm{B}^{+} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{+} + \mathrm{H}_2\mathrm{O}$$

So none of the salt ions will react with water molecules and the only source of $$\mathrm{H}_3\mathrm{O}^{+}$$ and \mathrm{OH}^{-}< /math> will be autoprotolysis of water itself. I mean
 * anion $$\mathrm{A}^{-}$$ — because it is strong — will also remain dissociated: $$\mathrm{A}^{-} + \mathrm{H}_2\mathrm{O } \rightarrow \mathrm{A}^{-} + \mathrm{H}_2\mathrm{O}$$

$$\mathrm{pH} = -\log\ [\mathrm{H}_3\mathrm{O}^{+}] = -\log\ [\mathrm{OH}^{-}] = -\log \sqrt{K_w} = \frac{1}{2} \mathrm{p}K_w$$

and at 25 °C the pH will be equal to 7.

pH of salt of weak acid and strong base
In the case of a salt of a weak acid and a strong base, we consider that
 * cation $$\mathrm{B}^{+}$$ — because it is strong — will remain dissociated: $$\mathrm{B}^{+} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{+} + \mathrm{H}_2\mathrm{O}$$


 * anion $$\mathrm{A}^{-}$$ — because it is weak — will react with water, i.e. undergo so-called hydrolysis, according to the equation: $$\mathrm{A}^{-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HA} + \mathrm{OH}^{-}$$


 * will hydrolyze very few anions, the amount of hydrolyzed anions will be negligible, i.e. $$c_{\mathrm{BA}} - [\mathrm{A}^{-}] \approx 0, or else

The above hydrolysis reaction has an equilibrium constant
 * the only source of is the aforementioned hydrolysis reaction, we neglect other sources of hydroxide anions and oxonium cations, i.e. the amount of \mathrm{HA} and \mathrm{OH}^{-} $$ will be the same according to her equation.

$$K = \frac{[\mathrm{HA}] \cdot [\mathrm{OH}^{-}]}{[\mathrm{A}^{-}] \cdot [\mathrm{H}_2 \mathrm{O}]}.$$

We will consider the concentration of water in water as constant and introduce a new constant, the so-called hydrolytic constant:

$$K_{h,A} = [\mathrm{H}_2\mathrm{O}] \cdot K = \frac{[\mathrm{HA}] \cdot [\mathrm{OH}^{-}] }{[\mathrm{A}^{-}]}$$

If we adjust the (slightly imprecise) notation of water in chemical equations, it will be easier to see that the hydrolysis equation is de facto just the opposite equation to dissociation:
 * dissociation: $$\mathrm{HA} \rightarrow \mathrm{H}^{+} + \mathrm{A}^{-},$$

It is therefore natural that the hydrolytic constant $$K_{h,A}$$ and the dissociation constant $$K_A$$ will be related:
 * hydrolysis: $$\mathrm{A}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{HA}.$$

$$K_{h,A} \cdot K_A = \frac{[\mathrm{HA}] \cdot [\mathrm{OH}^{-}]}{[\mathrm{A}^{-}]} \cdot \frac{[\mathrm{A}^{-}] \cdot [\mathrm{H}_3\mathrm{O}^{+}]}{[\mathrm{HA}]} = [\mathrm{ OH}^{-}] \cdot [\mathrm{H}_3\mathrm{O}^{+}] = K_w$$

So we get the formula for the constant $$K_{h,A}$$

$$K_{h,A} = \frac{K_w}{K_A}.$$

If we substitute $$[\mathrm{HA}]$$ in the definition of the hydrolytic constant $$K_{h,A}$$ according to our assumptions $$[\mathrm{OH}^{-} ],$$ and $$c_{\mathrm{BA}}$$ for $$[\mathrm{A}^{-}]$$ we get

$$K_{h,A} \cdot [\mathrm{A}^{-}] = K_{h,A} \cdot c_{\mathrm{BA}} = [\mathrm{OH}^{-} ]^{2}.$$

When we express the dependence of the concentration of oxonium cations on the concentration of hydroxide anions from the definition of the ion product of water $$K_w$$, we obtain

$$[\mathrm{H}_3\mathrm{O}]^{2} = \frac{K_w^2}{[\mathrm{OH}^{-}]^{2}} = \frac{K_w^2}{K_{h,A} \cdot c_{\mathrm{BA}}} = \frac{K_w^2}{\frac{K_w}{K_A} \cdot c_{\mathrm{BA}}} = \frac{K_w^2 \cdot K_A}{K_w \cdot c_{\mathrm{BA}}} = \frac{K_w \cdot K_A}{c_{\mathrm{BA}}}$$

Odmocníme (koncentrace jsou vždy kladné), zlogaritmujeme a vynásobíme −1:

$$-\log [\mathrm{H}_3\mathrm{O}^{+}] = -\log \sqrt{\frac{K_w \cdot K_A}{c_{\mathrm{BA}}}} = -\log \frac{K_w^{\frac{1}{2}} \cdot K_A^{\frac{1}{2}}}{c_{\mathrm{BA}}^{\frac{1}{2}}} = \frac{1}{2} \log c_{\mathrm{BA}} - \frac{1}{2} \log K_w - \frac{1}{2} \log K_A = \frac{1}{2} \log c_{\mathrm{BA}} + \frac{1}{2} \mathrm{p}K_w + \frac{1}{2} \mathrm{p}K_A$$

Při 25 °C dostáváme vzorec

$$\mathrm{pH} = 7 + \frac{1}{2} \log c_{\mathrm{BA}} + \frac{1}{2} \mathrm{p}K_A$$

The resulting pH will be alkaline. This is due to the fact that the anion of the acid draws hydrons from the system.

pH of salt of strong acid and weak base
In the case of a salt of a strong acid and a weak base, we consider that \mathrm{B}^{+} + 2\ \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BOH} + \mathrm{H}_3\mathrm{O}^{+}< /math>
 * cation $$\mathrm{B}^{+}$$ — because it is weak — will hydrolyze according to the reaction


 * anion $$\mathrm{A}^{-}$$ — because it is strong — will not hydrolyze, i.e.

$$\mathrm{A}^{-} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{A}^{-} + \mathrm{H}_2\mathrm{O}$$
 * will hydrolyze 'very few cations, and the amount of hydrolyzed cations will be negligible, i.e. c_{\mathrm{BA}} - [\mathrm{B}^{+}] \approx 0, or $$c_{\mathrm{BA}} \approx [\mathrm{B}^{+}]$$

For hydrolysis, we introduce the hydrolytic constant $$K_{h,B} = K \cdot [\mathrm{H}_2\mathrm{O}]^{2}$$ as
 * hydrolysis of cations is the only source of oxonium cations, we neglect other sources, so according to the hydrolysis equation $$[\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{BOH}]$$

$$K_{h,B} = \frac{[\mathrm{H}_3\mathrm{O}^{+}] \cdot [\mathrm{BOH}] }{ [\mathrm{B}^{+ }] },$$

for which it can be proved again that $$K_{h,B} \cdot K_B = K_w.$$

Podle průženů dosadíme do hydrolytické konstanty $$c_{\mathrm{BA}}$$ za $$B^{+}$$ a $$[\mathrm{H}_3\mathrm{O} ^{+}]$$ za $$[\mathrm{BOH}]$$:

$$[\mathrm{B}^{+}] \cdot K_{h,B} = c_{\mathrm{BA}} \cdot K_{h,B} = [\mathrm{H}_3\mathrm{ O}^{+}]^{2}$$

Odmocníme (concentrace jsou váze kladná numára), zlogaritmuje a vyőlőníme −1:

$$-\log\ [\mathrm{H}_3\mathrm{O}^{+}] = \frac{1}{2} \left(-\log c_{\mathrm{BA}} - \log K_{h,B} \right) = \frac{1}{2} \left(-\log c_{\mathrm{BA}} - \log \frac{K_w}{K_B} \right)$$

$$-\log\ [\mathrm{H}_3\mathrm{O}^{+}] = \frac{1}{2} \left(- \log c_{\mathrm{BA}} + \log K_B - \log K_w \right) = \frac{1}{2} \left(\mathrm{p}K_w - \mathrm{p}K_B - \log c_{\mathrm{BA}} \right)$$

At 25 °C, we get the formula:

$$\mathrm{pH} = 7 - \frac{1}{2} \mathrm{p}K_B - \frac{1}{2} \log c_{\mathrm{BA}}$$

The resulting pH will be acidic. This is because the base cation adds hydrons to the system.

pH of salt of weak acid and weak base
In the case of a salt of a weak acid and a weak base, we consider that \mathrm{B}^{+} + 2\ \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BOH} + \mathrm{H}_3\mathrm{O}^{+}< /math>
 * cation $$\mathrm{B}^{+}$$ will hydrolyze according to Eq


 * anion $$\mathrm{A}^{-}$$ will react with the resulting oxonium cations and then possibly further hydrolyze according to the equation

$$\mathrm{A}^{-} + \mathrm{H}_3\mathrm{O}^{+} \rightleftharpoons \mathrm{HA} + \mathrm{H}_2\mathrm{O}$$ $$c_{\mathrm{BA}} \approx [\mathrm{B}^{+}] \approx [\mathrm{A}^{-}]$$
 * both ions will hydrolyze in negligible amounts, i.e
 * there is no other source of hydroxide anions and oxonium cations in the system, therefore


 * oxonium cations 'formed by hydrolysis of the base cation $$\mathrm{B}^{+}$$ are denoted by $$[\mathrm{H}_3\mathrm{O}

^{+}]_{\mathrm{B}}$$ and according to the chemical equation of hydrolysis, $$[\mathrm{H}_3\mathrm{O}^{+}]_{\mathrm{ B}} = [\mathrm{GOD}]$$


 * oxonium cations 'depleted by hydrolysis of the acid anion $$[\mathrm{A}^{-}]$$ are denoted by $$[\mathrm{H}_3\mathrm{O}^{ +}]_{\mathrm{A}}$$ and according to the chemical equation of hydrolysis, $$[\mathrm{H}_3\mathrm{O}^{+}]_{\mathrm{A} } = [\mathrm{HA}]$$


 * 'we calculate the equilibrium concentration of oxonium cations as the difference between the concentration of oxonium cations formed by the hydrolysis of the cation of the base and the concentration of the oxonium cations consumed by the hydrolysis of the anion of the acid, i.e.
 * $$\mathbf [\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{H}_3\mathrm{O}^{+}]_{\mathrm{B}} - [\mathrm{H}_3\mathrm{O}^{+}]_{\mathrm{A}}$$

For the hydrolysis of the cation, we have the hydrolysis constant:

$$K_{h,B} = \frac{K_w}{K_B} = \frac{[\mathrm{H}_3\mathrm{O}^{+}] \cdot [\mathrm{BOH}]}{ [\mathrm{B}^{+}]}$$

and for the hydrolysis of the anion it is better to express the concentration of oxonium cations using another constant describing the equilibrium, namely the dissociation constant:

K_A = \frac{[\mathrm{H}_3\mathrm{O}^{+}] \cdot [\mathrm{A}^{-}]}{[\mathrm{HA}]}

By adding the penultimate assumption to the last assumption, we get

$$[\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{BOH}] - [\mathrm{HA}].$$

We express $$[\mathrm{BOH}]$$ and $$[\mathrm{HA}]$$ from the equations for the constants describing the equilibria:

[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w \cdot [\mathrm{B}^{+}]}{K_B \cdot [\mathrm{H}_3 \mathrm{O}^{+}]} - \frac{[\mathrm{H}_3\mathrm{O}^{+}] \cdot [\mathrm{A}^{-}] }{K_A}< /math>

Upravíme:

$$[\mathrm{H}_3\mathrm{O}^{+}] + \frac{[\mathrm{H}_3\mathrm{O}^{+}] \cdot [\mathrm{A}^ {-}] }{K_A} = [\mathrm{H}_3\mathrm{O}^{+}] \left(1 + \frac{[\mathrm{A}^{-}]}{K_A} \ right) = \frac{K_w \cdot [\mathrm{B}^{+}]}{K_B \cdot [\mathrm{H}_3\mathrm{O}^{+}]} $$

Vyjádříme koncentrisi oxoniových kationtů:

$$[\mathrm{H}_3\mathrm{O}^{+}]^{2} = \frac{ \frac{K_w \cdot [\mathrm{B}^{+}]}{K_B} } { 1 + \frac{ [\mathrm{A}^{-}] }{ K_A } } = \frac{ \frac{K_w \cdot [\mathrm{B}^{+}]}{K_B} }{ \ frac{ K_A + [\mathrm{A}^{-}] }{ K_A } } = \frac{ K_w \cdot [\mathrm{B}^{+}] \cdot K_A }{ K_B \cdot (K_A + [ \mathrm{A}^{-}]) }$$

Since $$[\mathrm{A}^{-}] \gg K_A,$$ we can neglect $$K_A$$ from the denominator and approximate the formula to

$$[\mathrm{H}_3\mathrm{O}^{+}]^{2} = \frac{ K_w \cdot [\mathrm{B}^{+}] \cdot K_A }{ K_B \cdot [\mathrm{A}^{-}] }$$

Adding $$c_{\mathrm{BA}}$$ according to assumptions, we get

$$[\mathrm{H}_3\mathrm{O}^{+}]^{2} = \frac{ K_w \cdot c_{\mathrm{BA}} \cdot K_A }{ K_B \cdot c_{\ mathrm{BA}} } = \frac{ K_W \cdot K_A }{ K_B }$$

We take the square root (these are positive constants), take the logarithm, multiply by −1 and get

$$ -\log\ [\mathrm{H}_\mathrm{O}^{+}] = -\frac{1}{2} \log K_w -\frac{1}{2} \log K_A + \frac{1}{2} \log K_B = \frac{1}{2} (\mathrm{p}K_w + \mathrm{p}K_A - \mathrm{p}K_B)$$

The pH'  of salts of weak acids and weak bases therefore (after approximation)  does not depend on the concentration of the salt.

At 25 °C we get

$$ \mathrm{pH} = 7 + \frac{1}{2} \mathrm{p}K_A - \frac{1}{2} \mathrm{p}K_B$$

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