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.