Question

Define the relation between ${K_a}$ and ${K_b}$.

Answer 1

Hint: ${K_a}$ is dissociation constant for acid and ${K_b}$ is dissociation constant for base. Both can be found out by finding the equilibrium constant for acid and base respectively. Both constants are the measure of the strengths of acid and base respectively. The relation between these two constants gives the equation for autoionization of water or self-dissociation for water.
Formulas used:${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{\text{Conjugate base}}} \right]}}{{\left[ {Acid} \right]}}$, ${K_b} = \dfrac{{\left[ {Acid} \right]\left[ {{{O}}{{{H}}^ - }} \right]}}{{\left[ {{\text{Conjugate base}}} \right]}}$
Where $[{H_3}{O^ + }],[O{H^ - }],[{\text{Conjugate base}}]$ and $[Acid]$ represent the concentrations of the hydronium ions, hydroxyl ions, conjugate base and the acid respectively.

Complete answer:
Let us consider a weak acid $HA$. This will dissociate in water to give hydronium ions and the corresponding conjugate base. A conjugate base is that compound (ion) which is left behind after an acid (its conjugate acid) releases hydrogen ion into the solution. Thus,
$HA + {H_2}O \rightleftharpoons {H_3}{O^ + } + {A^ - }$
Now let us write the acid dissociation constant (${K_a}$) for this acid. This is nothing but the equilibrium constant of the given dissociation, that is, it is the ratio between product of concentrations of products to the product of concentrations of reactants. In this case, we take the concentration of water to be unity as it is a pure liquid. Hence, generally:
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{\text{Conjugate base}}} \right]}}{{\left[ {Acid} \right]}}$
Where $[{H_3}{O^ + }],[{\text{Conjugate base}}]$ and $[Acid]$ represent the concentrations of the hydronium ions, hydroxyl ions and the acid respectively.
Hence, applying this for our acid, we get:
${K_a} = \dfrac{{{{[{H_3}O]}^ + }[{A^ - }]}}{{[HA]}}$ ----------------($1$)
Now let us consider its conjugate base which gives the reaction as follows when present in the aqueous solution:
${A^ - } + {H_2}O \rightleftharpoons HA + O{H^ - }$
Similar to the case where we defined ${K_a}$, we can define ${K_b}$ as:
${K_b} = \dfrac{{\left[ {Acid} \right]\left[ {{{O}}{{{H}}^ - }} \right]}}{{\left[ {{\text{Conjugate base}}} \right]}}$
Where $[O{H^ - }],[{\text{Conjugate base}}]$ and $[Acid]$ represent the concentrations of the hydroxyl ions, conjugate base and the acid respectively.
Applying this to our equation for conjugate base, we get:
${K_b} = \dfrac{{[HA][O{H^ - }]}}{{[{A^ - }]}}$ -------------($2$)
Multiplying equation $1$ by equation $2$ we get
${K_a} \times {K_b} = \dfrac{{{{[{H_3}O]}^ + }[{A^ - }]}}{{[HA]}} \times \dfrac{{[HA][O{H^ - }]}}{{[{A^ - }]}}$
Cancelling the common terms, we get:
${K_a} \times {K_b} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]$
As we know the product of concentrations of hydronium and hydroxyl ions at a particular temperature is known as the ionic product of water, which is a constant at the given temperature. Hence, the relation between ${K_a}$ and ${K_b}$ can be stated as:
${K_a} \times {K_b} = {K_w} = \left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]$
Here ${K_w}$ is defined as the ionic product of water.

Note:Larger the value of ${K_a}$, stronger is the acid meaning that the acid can easily give protons and larger the value of ${K_b}$, stronger is the base meaning that the base can easily give hydroxyl ions. Thus, we can say that the strength of acid or base is related to the value of dissociation constant. Note that at room temperature the value of the ionic product of water, ${K_w} = {10^{ - 14}}$.