PH of Weak Acids and Bases

Weak single-celled acids
For the weak single-celled acids ' is in progress dissociation according to balanced reaction :

$$\ mathrm {HA} + \ mathrm {H}_2\ mathrm {O} \ rightleftharpoons \ mathrm {A}^{-} + \ mathrm {H}_3\ mathrm {O}^{+}$$,

s éväläžnou konstantou K, which is defined like :

$$K = \ frac {[\ mathrm {A}^{-}] \ cdot [\ mathrm {H}_3\ mathrm {O}^{+}] }{ [\ mathrm {HA}] \ cdot [ \ mathrm {H}_2\ mathrm {O}] }$$.

necessary to calculate the pH use other method than at pH strong acids and bases, because in weak acids can't to consider dissociation for complete.

On the contrary we assume that :
 * acid dissociates very little difference _ between $$[\ mathrm {HA}]$$ and $$c_{\ mathrm {HA}}$$ ( i.e. $$[\ mathrm {A}^{-}]$$) we neglect, and therefore $$[\ mathrm {HA}] \ approx c_{\ mathrm {HA}}$$;

Because you can concentration water to consider for constant, the so - called dissociative constant  $$K_A = K \ cdot [\ mathrm {H}_2\ mathrm {O}]$$ whose value is for each acid tabulated. We are getting thus :
 * dissociation acid is the only one source oxonium of cations in the system, that is according to ratios fabric amount chemical equation $$[\ mathrm {H}_3\ mathrm {O}^{+}] = [\ mathrm {A}^{-}]$$.

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

According to assumptions we substitute $$[\ mathrm {H}_3\ mathrm {O}^{+}]$$ after $$[\ mathrm {A}^{-}]$$ and $$c_{ \ mathrm {HA}}$$ after $$[\ mathrm {HA}]$$ and expression concentration oxonium cations we get :

$$[\ mathrm {H}_3\ mathrm {O}^{+}]^{2} = K_A \ cdot [\ mathrm {HA}] = K_A \ cdot c_{\ mathrm {HA}}$$.

root ( concentration is always positive number ), we logarithmize both parties equation and multiply by $$-1$$:

$$-\log\ [\ mathrm {H}_3\ mathrm {O}^{+}] = -\log \ sqrt {K_A \ cdot c_{\ mathrm {HA}}} = -\log\ (K_A \ cdot c_{\ mathrm {HA}})^{\ frac {1}{2}} = -\ frac {1}{2} \log\ (K_A \ cdot c_{\ mathrm {HA}}) = - \ frac {1}{2} (\log K_A + \log c_{\ mathrm {HA}}) = -\ frac {1}{2} \log K_A -\ frac {1}{2} \log c_{ \ mathrm {HA}}$$.

For "$$-\log\ K_A$$" (by analogy with pH) the symbol $$\ mathrm {p}K_A$$ was used. If we know dissociative constant or her negative decadent logarithm, we calculate the pH according to formulas :

$$\ mathrm {pH} = \ frac {1}{2} \ mathrm {p}K_A -\ frac {1}{2} \log\ c_{\ mathrm {HA}}$$.

Weak single-celled policy
For the weak single-celled principles  is ongoing dissociation according to balanced equation :

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

s tabelovanou Dissociation konstantou $$K_B = K,$$ which is defined like :

$$K_B = \ frac {[\ mathrm {B}^{+}] \ cdot [\ mathrm {OH}^{-}] }{ [\ mathrm {BOH}] }$$.

We assume that :
 * principle dissociates very little difference _ between $$[\ mathrm {BOH}]$$ and $$c_{\ mathrm {BOH}}$$ ( ie $$[\ mathrm {B}^{+}]$$) we neglect, and therefore $$[\ mathrm {BOH}] \ approx c_{\ mathrm {BOH}}$$;


 * dissociation acid is the only one source hydroxides of anions in the system, that is according to ratios fabric amount chemical equation $$[\ mathrm {OH}^{-}] = [\ mathrm {B}^{+}]$$;

Substitution according to assumptions into the definition dissociative constants we get :
 * emergence hydroxides of anions acts decrease oxonium cations according to equation for the ionic product of water : $$[\ mathrm {H}_3\ mathrm {O}^{+}] = \ frac { K_w }{ [\ mathrm {OH}^{-}]}$$.

$$ [\ mathrm {OH}^{-}]^{2} = K_B \ cdot [\ mathrm {BOH}] = K_B \ cdot c_{\ mathrm {BOH}}$$.

We square root ( concentration they are always positive numbers ):

$$ [\ mathrm {OH}^{-}] = \ sqrt {K_B \ cdot c_{\ mathrm {BOH}} } = (K_B \ cdot c_{\ mathrm {BOH}})^{\ frac {1 }{2}}$$.

We substitute in the equation for ionic product water :

$$ [\ mathrm {H}_3\ mathrm {O}^{+}] = \ frac { K_w }{ (K_B \ cdot c_{\ mathrm {BOH}})^{\ frac {1}{2} } }$$.

Zlogaritmujeme a vynaňíme $$-1$$:

$$ -\log [\ mathrm {H}_3\ mathrm {O}^{+}] = \ frac {1}{2} \log K_B + \ frac {1}{2} \log c_{\ mathrm {BOH}} - \log K_w $$

At 25 °C, we calculate the pH according to formulas

$$ \ mathrm {pH} = 14 + \ frac {1}{2} \log c_{\ mathrm {BOH}} - \ frac {1}{2} \ mathrm {p}K_B$$

Related articles

 * pH


 * pH of strong acids and bases


 * pH -metry


 * pH measurement


 * pH buffers


 * Urine pH


 * pH of salts

Used literature

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