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 the calculation we assume:
- the substance quantity of
according to the above equation will be the same as 
, which, given an identical volume, is also true for the concentration, i.e. ![{\displaystyle [\mathrm {H} _{3}\mathrm {O} ^{+}]=[\mathrm {A} ^{-}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/884e76b2536ce7da3ed060d0cf71599568ce60e8)
;
- all acid - because it is a strong acid - is converted into a , therefore we will mark
![{\displaystyle [\mathrm {A} ^{-}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4081d3ad257c85039ca6a3864c7da1d76e5f59f2)
its concentration, i.e. ![{\displaystyle [\mathrm {A} ^{-}]=c_{\mathrm {HA} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbe129c33d984d5d97e1a4d051422e9971b03d59)
Let's deduce:
![{\displaystyle \mathrm {pH} =-\log \ [\mathrm {H} _{3}\mathrm {O} ^{+}]=-\log \ [\mathrm {A} ^{-}]=-\log \ c_{\mathrm {HA} },}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cb373abb61ba8564af3e7605fffe24a3c687f7b)
and to calculate the pH we get the formula
Strong monosaturated bases
For strong monosaturated bases the dissociation follows the equation
We assume, as in the case of strong monosaturates, that:
- the amount, or concentration, of hydroxide ions and the resulting
is the same according to the above chemical equation, i.e. ![{\displaystyle [\mathrm {OH} ^{-}]=[\mathrm {B} ^{+}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c39fe60b34a30c8b36a20fa693f8ec176a4b168c)
;
- dissociation occurs completely, i.e.
![{\displaystyle [\mathrm {B} ^{+}]=c_{\mathrm {BOH} }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42fd6f37d6fe4284f4047fd44a84d94d4d3c1da7)
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:
![{\displaystyle [\mathrm {H} _{3}\mathrm {O} ^{+}]={\frac {K_{w}}{[\mathrm {OH} ^{-}]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6d90de41502d92f8d6f48fd5ee9a825644cf47f)
and from these assumptions, we deduce
![{\displaystyle \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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23b05e7971fa6d480a740b21c4ed021d83272821)
Calculate the pH at 25 °C using the formula
Strong dibasic acids
Strong dibasic acids dissociate according to the equation
we assume, then:
- complete dissociation, i.e.
![{\displaystyle c_{\mathrm {H} _{2}\mathrm {A} }=[\mathrm {A} ^{2-}];}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca307ee2da428c7570828c88ad56dc3376e0add6)
- however, the amount of oxonium cations and the amount of formed
is - in contrast to monosaturated acids - in a ratio of 1:2, i.e. ![{\displaystyle [\mathrm {H} _{3}\mathrm {O} ^{+}]=2\cdot c_{\mathrm {H} _{2}\mathrm {A} }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e950708b61264ddb766badfdc6b805850a61e4d6)
From this we derive ![{\displaystyle \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} }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09618c3da3652498c06d482a300b3434f1ba7252)
and the pH is calculated according to the formula
Strong dibasic bases
Strong dibasic bases dissociate according to the equation
as we assume for monosaturated bases and dibasic acids:
- complete dissociation, i.e.
![{\displaystyle c_{\mathrm {B(OH)} _{2}}=[\mathrm {B} ^{2+}];}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be88aabbe2ccea758989f9b9888a89cd2fd279e9)
- concentration of the formed
and the concentration of hydroxide anions is in the ratio 1:2, i.e.![{\displaystyle [\mathrm {OH} ^{-}]=2\cdot [\mathrm {B} ^{2+}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7086384b32cd0574e14400d68f52ca51aa69eb3)
, in addition, according to the previous assumption ![{\displaystyle [\mathrm {OH} ^{-}]=2\cdot c_{\mathrm {B(OH)} _{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bfe4db4440e5eb3dd30535fbf486e61cac372d4)
- hydroxide anions drain oxonium cations from the environment, .
Then we derive
![{\displaystyle -\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffc7bb1b843d4d3c6dda7b269d6594fbb38f48e6)
and the pH at 25 °C is calculated according to the formula
References
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