show why ka(kb)=kw for this pair of conjugate acid-base: H2C6H6O6 and HC6H66 ^-1
It's too long and too confusing using long formulas. Let's cut that to ANY acid and its conjugate base, HA.
The conjugate base is A^- so if we add that to water we have
..........A^- + HOH ==> HA^- + OH^-
(HA)(OH^-)
----------- = Kb for the conjugate base.
(A^-)
Now let's multiply that by (H^+)/(H^+)
I must use ..... for spacing and it still may not look very good.
(HA)(OH^-)......(H^+)
----------- x ------ = Kb
(A^-)............(H^)
Now look at the equation.
In the numerator, (H^+)(OH^-) = Kw.
In the denominator, we have the reciprocal of Ka. Note that is (A^-)/(H^+)(A^-) which is just 1/Ka. Therefore, Kb = Kw*1/Kb or just Kw/Kb.
To understand why the product of the acid dissociation constant (Ka) of an acid and the base dissociation constant (Kb) of its conjugate base is equal to the water dissociation constant (Kw), let's consider the dissociation reactions of the given acid and its conjugate base.
1. Acid dissociation reaction (H2C6H6O6):
H2C6H6O6 ⇌ H+ + C6H6O6^2-
2. Base dissociation reaction (HC6H6O6^-1):
HC6H6O6^-1 + H2O ⇌ H3O+ + C6H6O6
The conjugate acid-base pair in this case includes H2C6H6O6 (acid) and HC6H6O6^-1 (conjugate base).
The equilibrium constant expression for the acid dissociation reaction is given by:
Ka = [H+][C6H6O6^2-] / [H2C6H6O6]
The equilibrium constant expression for the base dissociation reaction is given by:
Kb = [H3O+][C6H6O6] / [HC6H6O6^-1]
According to the principles of acid-base conjugate pairs, the concentration of the acid and its conjugate base are related by the equation:
[H+][C6H6O6^2-] = [H3O+][HC6H6O6^-1] = Kw
Since [H+][C6H6O6^2-] = Kw and [H3O+][HC6H6O6^-1] = Kw, it follows that:
Ka * Kb = Kw
Therefore, for any conjugate acid-base pair, the product of Ka and Kb is equal to Kw, as demonstrated above.
To understand why the product of the acid dissociation constant (Ka) of an acid and the base dissociation constant (Kb) of its conjugate base equals the ion product of water (Kw), let's break it down step by step.
1. Define the variables:
- Ka: Acid dissociation constant. It describes the strength of an acid.
- Kb: Base dissociation constant. It describes the strength of a base.
- Kw: The ion product of water. It represents the self-ionization of water.
2. Write the chemical equation for the acid-base pair:
- H2C6H6O6 (Acid) + H2O ⇌ HC6H6O6^- (Conjugate base) + H3O+ (Hydronium ion)
- Here, H2C6H6O6 is the acid, and HC6H6O6^- is its conjugate base.
3. Write the expression for the acid dissociation constant (Ka) and base dissociation constant (Kb):
- Ka = [HC6H6O6^-][H3O+] / [H2C6H6O6]
- Kb = [H2C6H6O6][OH^-] / [HC6H6O6^-]
4. Relate Kb to Ka using the ion product of water (Kw):
- Kw = [H3O+][OH^-]
- At 25°C, the concentration of H3O+ and OH^- in pure water are both 1 x 10^-7 M. Therefore, Kw = 1 x 10^-14.
5. Substitute the expressions for Ka and Kb into Kw:
- Ka * Kb = ([HC6H6O6^-][H3O+] / [H2C6H6O6]) * ([H2C6H6O6][OH^-] / [HC6H6O6^-]) = [H3O+][OH^-]
- The [HC6H6O6^-] and [H2C6H6O6^-] concentrations cancel out.
- This simplifies to: Ka * Kb = Kw.
Therefore, for the pair of conjugate acid-base H2C6H6O6 and HC6H6O6^-, the product of their acid dissociation constant (Ka) and base dissociation constant (Kb) is equal to the ion product of water (Kw).