Question

Which of the following is valid for $ {{K}_{a}}\times {{K}_{b}}={{K}_{w}} $ :
(A) $ HCl $ and $ NaOH $
(B) $ N{{H}_{4}} $ and $ HCOOH $
(C) $ HCOOH $ and $ HCO{{O}^{-}} $
(D) All of these.

Answer 1

Hint :We know that equilibrium refers to a condition when the rate of forward reaction is equal to the rate of reverse reaction. The equilibrium constant, denoted by $ K, $ expresses the relationship between reactants and products of a reaction at an equilibrium condition with respect to a specific unit.

Complete Step By Step Answer:
For a generalised chemical reaction taking place in a solution:
 $ aA+bB\rightleftharpoons cC+dD~ $
The equilibrium constant can be expressed as follows:
 $ K=\dfrac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}} $
where $ \left[ A \right],\text{ }\left[ B \right],\text{ }\left[ C \right] $ and $ \left[ D \right] $ refer to the molar concentration of species $ A,\text{ }B,\text{ }C,\text{ }D $ respectively at equilibrium. The coefficients like $ a,~b,~c, $ and $ d $ in the generalised chemical equation become exponents as seen in the above expression.
 $ {{K}_{a}},\text{ }p{{K}_{a}},\text{ }{{K}_{b}} $ and $ p{{K}_{b}} $ are mainly helpful while predicting whether any species will either donate or accept the protons at a specified $ pH $ value. They actually describe the degree of ionization of acid or a base. They are the true indicators of acidic or basic strength as adding water to any solution won’t alter the equilibrium constant. $ p{{K}_{a}} $ and $ {{K}_{a}} $ are related to acids, whereas $ p{{K}_{b}} $ and $ {{K}_{b}} $ are related to bases. Similar to $ pH $ and $ pOH,\text{ }{{K}_{a}} $ and $ p{{K}_{a}} $ also account for the hydrogen ion concentration or $ p{{K}_{b}} $ and $ {{K}_{b}} $ account for hydroxide ion concentration.
As mentioned in the question also, the relationship between $ {{K}_{a}} $ and $ {{K}_{b}} $ through ion constant for water $ {{K}_{w}} $ :
 $ {{K}_{a}}\times {{K}_{b}}={{K}_{w}} $
For acid and base i.e. $ HCl $ and $ NaOH, $ where, $ {{K}_{a}} $ is acid dissociation constant and $ p{{K}_{a}} $ is $ -\log $ of $ {{K}_{a}}. $ Similarly, $ {{K}_{b}} $ is base dissociation constant, and $ p{{K}_{b}} $ is the $ -\log {{K}_{b}} $ . The above given relation is valid for conjugate acid-base pairs. When an acid gets dissolved in water:
 $ HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}......\text{(}{{\text{K}}_{a}}) $
 $ {{\text{A}}^{-}}+\text{ }{{H}_{2}}O\rightleftharpoons HA~+\text{ }O{{H}^{-}}......\text{(}{{\text{K}}_{b}}) $
Similarly for $ N{{H}_{4}} $ and $ HCOOH $ , the dissociation of water can be represented as follows: $ {{H}_{2}}O\rightleftharpoons {{H}^{+}}+O{{H}^{-}} $
 $ {{K}_{w}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right] $
 $ {{K}_{w}}={{K}_{a}}\times {{K}_{b}} $
Likewise, $ HCOOH $ and $ HCO{{O}^{-}} $ for when base gets dissolved in water:
 $ MA\rightleftharpoons {{A}^{-}}+{{M}^{+}} $
 $ {{A}^{-}}+{{H}_{2}}O\rightleftharpoons AH+O{{H}^{-}}......\text{(}{{\text{K}}_{b}}^{'}) $
 $ AH\rightleftharpoons {{A}^{-}}+{{H}^{+}}.......\text{(}{{\text{K}}_{a}}^{'}) $
The dissociation of water can be represented as follows:
 $ {{H}_{2}}O\rightleftharpoons {{H}^{+}}+O{{H}^{-}} $
 $ {{K}_{w}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right] $
 $ {{K}_{w}}={{K}_{a}}^{'}\times {{K}_{b}}^{'} $
Hence, the correct answer is Option D.

Note :
Note that the conjugate acid-base pairs differ only by a proton. The conjugate base of any weak acid is generally a strong base. And, the conjugate base of an acid is usually the anion which results when an acid molecule loses its hydrogen to a base.