PH of strong acids and bases: Difference between revisions

Revision as of 23:56, 17 December 2022

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:

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. $[\mathrm {A} ^{-}]=c_{\mathrm {HA} }$

Let's deduce:

$\mathrm {pH} =-\log \ [\mathrm {H} _{3}\mathrm {O} ^{+}]=-\log \ [\mathrm {A} ^{-}]=-\log \ c_{\mathrm {HA} },$

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 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. $[\mathrm {B} ^{+}]=c_{\mathrm {BOH} }.$

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:

$[\mathrm {H} _{3}\mathrm {O} ^{+}]={\frac {K_{w}}{[\mathrm {OH} ^{-}]}}$

and from these assumptions, we deduce

$\mathrm {pH} =-\log \ [\mathrm {H} _{3}\mathrm {O} ^{+}]=-\log \ {\frac {K_{w}}{[\mathrm {OH} ^{-}]}}=\log \ [\mathrm {OH} ^{-}]-\log K_{w}=\log \ [\mathrm {B} ^{+}]-\log K_{w}=\log c_{\mathrm {BOH} }-\log K_{w}.$

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:

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. $[\mathrm {H} _{3}\mathrm {O} ^{+}]=2\cdot c_{\mathrm {H} _{2}\mathrm {A} }.$

From this we derive $\mathrm {pH} =-\log \ [\mathrm {H} _{3}\mathrm {O} ^{+}]=-\log \ [\mathrm {A} ^{2-}]=-\log(2\cdot c_{\mathrm {H} _{2}\mathrm {A} })=-\log 2-\log c_{\mathrm {H} _{2}\mathrm {A} }$

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:

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, $[\mathrm {H} _{3}\mathrm {O} ^{+}]={\frac {K_{w}}{[\mathrm {OH} ^{-}]}}$ .

Then we derive

$-\log[\mathrm {H} _{3}\mathrm {O} ^{+}]=-\log {\frac {K_{w}}{[\mathrm {OH} ^{-}]}}=\log \ [\mathrm {OH} ^{-}]-\log K_{w}=\log(2\cdot c_{\mathrm {B(OH)} _{2}})-\log K_{w}=\log 2+\log \ c_{\mathrm {B(OH)} _{2}}-\log K_{w}$

and the pH at 25 °C is calculated according to the formula

 $\mathrm {pH} =14+\log 2+\log \ c_{\mathrm {B(OH)} _{2}}.$

@@ Line 1: / Line 1: @@
-When calculating pH, it is always necessary to consider what is the source of oxonium cations in a given environment.
+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
+For '''strong monosaturated acids''' the dissociation follows the equation
 <math>\mathrm{HA} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{A}^{-} + \mathrm{H}_3\mathrm{O}^{+}.</math>
 For the calculation we assume:
-* [[látkové množství]] <math>\mathrm{H}_3\mathrm{O}^{+}</math> bude podle výše uvedené rovnice stejné jako <math>\mathrm{A}^{-}</math>, což vzhledem k totožnému objemu platí i pro koncentraci, tedy <math>[\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{A}^{-}]</math>;
+* [[the substance quantity]] of  <math>\mathrm{H}_3\mathrm{O}^{+}</math> according to the above equation will be the same as <math>\mathrm{A}^{-}</math>, which, given an identical volume, is also true for the concentration, i.e. <math>[\mathrm{H}_3\mathrm{O}^{+}] = [\mathrm{A}^{-}]</math>;
-* všechna kyselina se – neboť je kyselinou silnou – přemění na <math>\mathrm{A}^{-}</math> a <math>\mathrm{H}_3\mathrm{O}^{+}</math>, proto označíme <math>[\mathrm{A}^{-}]</math> její koncentraci, tedy <math>[\mathrm{A}^{-}] = c_{\mathrm{HA}}</math>
+* all acid - because it is a strong acid - is converted into <math>\mathrm{A}^{-}</math> a <math>\mathrm{H}_3\mathrm{O}^{+}</math>, therefore we will mark <math>[\mathrm{A}^{-}]</math> its concentration, i.e. <math>[\mathrm{A}^{-}] = c_{\mathrm{HA}}</math>
-Odvodíme tedy:
+Let's deduce:
 <math>\mathrm{pH} = -\log\ [\mathrm{H}_3\mathrm{O}^{+}] = -\log\ [\mathrm{A}^{-}] = -\log\ c_{\mathrm{HA}},</math>
-a pro výpočet pH dostáváme vzorec
+and to calculate the pH we get the formula
   <math>\mathrm{pH} = -\log\ c_{\mathrm{HA}}.</math>
-===Silné jednosytné zásady===
+===Strong monosaturated bases===
-U '''silných jednosytných zásad''' probíhá disociace podle rovnice
+For '''strong monosaturated bases''' the dissociation follows the equation
 <math>\mathrm{BOH} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{+} + \mathrm{OH}^{-} + \mathrm{H}_2\mathrm{O}</math>
-Předpokládáme, stejně jako v případě silných jednosytných kyselin, že:
+We assume, as in the case of strong monosaturates, that:
-*látkové množství, resp. koncentrace, hydroxidových iontů a vzniklého <math>\mathrm{B}^{+}</math> je podle výše uvedené chemické rovnice stejné, tedy <math>[\mathrm{OH}^{-}] = [\mathrm{B}^{+}]</math>;
+*the amount, or concentration, of hydroxide ions and the resulting <math>\mathrm{B}^{+}</math> is the same according to the above chemical equation, i.e. <math>[\mathrm{OH}^{-}] = [\mathrm{B}^{+}]</math>;
-*disociace probíhá úplně, tedy <math>[\mathrm{B}^{+}] = c_{\mathrm{BOH}}.</math>
+*dissociation occurs completely, i.e. <math>[\mathrm{B}^{+}] = c_{\mathrm{BOH}}.</math>
-Výpočet je tedy analogický, musíme si jen uvědomit, že na rozdíl od kyselin není zásada zdrojem oxoniových kationtů, ale oxoniové kationty z prostředí odebírá (viz [[teorie kyselin a zásad]]), proto dosadíme z rovnice pro iontový součin [[voda|vody]]:
+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]]:
 *<math>[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}</math>
-a z těchto předpokladů odvodíme:
+and from these assumptions, we deduce
 <math>\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.</math>
-Při 25&nbsp;°C pH spočítáme podle vzorce
+Calculate the pH at 25 °C using the formula
   <math>\mathrm{pH} = 14 + \log\ c_{\mathrm{BOH}}.</math>
-===Silné dvousytné kyseliny===
+===Strong dibasic acids===
-'''Silné dvousytné kyseliny''' disociují podle rovnice
+'''Strong dibasic acids''' dissociate according to the equation
 <math>\mathrm{H}_2\mathrm{A} + 2\ \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{A}^{2-} + 2\ \mathrm{H}_3\mathrm{O}^{+},</math>
-předpokládáme tedy:
+we assume, then:
-*úplnou disociaci, tedy <math>c_{\mathrm{H}_2\mathrm{A}} = [\mathrm{A}^{2-}];</math>
+*complete dissociation, i.e. <math>c_{\mathrm{H}_2\mathrm{A}} = [\mathrm{A}^{2-}];</math>
-*ovšem látkové množství oxoniových kationtů a látkové množství vzniklého <math>\mathrm{A}^{2-}</math> je – na rozdíl od jednosytných kyselin – v poměru 1:2, tedy <math>[\mathrm{H}_3\mathrm{O}^{+}] = 2 \cdot c_{\mathrm{H}_2\mathrm{A}}.</math>
+*however, the amount of oxonium cations and the amount of formed <math>\mathrm{A}^{2-}</math> is - in contrast to monosaturated acids - in a ratio of 1:2, i.e. <math>[\mathrm{H}_3\mathrm{O}^{+}] = 2 \cdot c_{\mathrm{H}_2\mathrm{A}}.</math>
-Z toho odvodíme <math>\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}}</math>
+From this we derive <math>\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}}</math>
-a pH spočítáme podle vzorce
+and the pH is calculated according to the formula
   <math> \mathrm{pH} = - \log\ c_{\mathrm{H}_2\mathrm{A}} - \log\ 2.</math>
-===Silné dvousytné zásady===
+===Strong dibasic bases===
-U '''silných dvousytných zásad''' probíhá disociace podle rovnice
+'''Strong dibasic bases''' dissociate according to the equation
 <math>\mathrm{B(OH)}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{B}^{2+} + 2\ \mathrm{OH}^{-} + \mathrm{H}_2\mathrm{O}</math>
-jako u jednosytných zásad a dvousytných kyselin předpokládáme:
+as we assume for monosaturated bases and dibasic acids:
-*úplnou disociaci, tedy <math>c_{\mathrm{B(OH)}_2} = [\mathrm{B}^{2+}];</math>
+*complete dissociation, i.e. <math>c_{\mathrm{B(OH)}_2} = [\mathrm{B}^{2+}];</math>
-*koncentrace vzniklého <math>\mathrm{B}^{2+}</math> a koncentrace hydroxidových aniontů je v poměru 1:2, tedy <math>[\mathrm{OH}^{-}] = 2 \cdot [\mathrm{B}^{2+}]</math>, podle předchozího předpokladu navíc <math>[\mathrm{OH}^{-}] = 2 \cdot c_{\mathrm{B(OH)}_2}</math>
+*concentration of the formed <math>\mathrm{B}^{2+}</math> and the concentration of hydroxide anions is in the ratio 1:2, i.e.<math>[\mathrm{OH}^{-}] = 2 \cdot [\mathrm{B}^{2+}]</math>, in addition, according to the previous assumption <math>[\mathrm{OH}^{-}] = 2 \cdot c_{\mathrm{B(OH)}_2}</math>
-*hydroxidové anionty odčerpávají z prostředí oxoniové kationty, <math>[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}</math>.
+*hydroxide anions drain oxonium cations from the environment, <math>[\mathrm{H}_3\mathrm{O}^{+}] = \frac{K_w}{[\mathrm{OH}^{-}]}</math>.
-Poté odvodíme
+Then we derive
 <math>- \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</math>
-a pH při 25&nbsp;°C se spočítá podle vzorce
+and the pH at 25 °C is calculated according to the formula
   <math>\mathrm{pH} = 14 + \log 2 + \log\ c_{\mathrm{B(OH)}_2}.</math>
 <noinclude>
-==Odkazy==
+==References==
-===Související články===
+===Related articles===
-*[[pH slabých kyselin a zásad]]
+*[[pH of weak acids and bases]]
-*[[pH-metrie]]
+*[[pH-metry]]
-*[[Měření pH]]
+*[[pH measurment]]
-*[[Pufry, pufrační kapacita, oxidoredukce, elektrodové děje|pH pufrů]]
+*[[Buffers, buffering capacity, oxidation and reduction, electrode processes]]
-*[[pH moči]]
+*[[pH of urine]]
-*[[pH solí]]
+*[[pH of salts]]
-</noinclude>[[Kategorie:Vložené články]] [[Kategorie:Biochemie]] [[Kategorie:Chemie]]
+</noinclude>[[Category:Inserted articles]] [[Category:Biochemistry]] [[Category:Chemistry]]