Question

How are the pK's of the conjugate acid and base are related to each other?​

Answer 1

Answer:

Increase in pH

This means that whenever the pH increases more than 2.0 pH units, then the major chemical in the solution would be the conjugate base. Remember that when the pH is equal to the pKa value, the proportion of the conjugate base andconjugate acid are equal to each other.

Answer 2

Answer:

The pK's of the conjugate acid and base are related to each other by the equation :

$\sf pK_a+pK_b=pK_w=14$

Derivation of above equation :

Let us take an example of a weak acid which ionises in aqueous :

$\sf HA+H_2O \rightleftharpoons H_3O^+(aq)+A^-(aq)$

$\sf K_a =\dfrac{\left[H_3O^+ \right]\left[ A^- \right]}{\left[HA \right]}...(i)$

Now dissociation of weak base written as :

$\sf A^- +H_2O \rightleftharpoons HA +OH^-$

$\sf K_B =\dfrac{\left[HA \right]\left[ OH^- \right]}{\left[A^- \right]}...(ii)$

Multiply ( i )  and  ( ii ) we get :

$\sf K_a\times K_b=\dfrac{\left[H_3O^+ \right]\left[ A^- \right]}{\left[HA \right]}\times \dfrac{\left[HA \right]\left[ OH^- \right]}{\left[A^- \right]}.$

$\sf K_a\times K_b=\left[H^3O^+ \right]\left[OH^- \right]=K_w$

Taking negative log both side :

$\sf -\log \left(K_a\times K_b\right)=-\log (K_a)$

$\sf -\log \left(K_a) - \log (K_b\right)=-\log (K_a)$

$\sf \rightarrow pK_a+pK_b=PK_w=14$

Hence derived.