Question

What are \[p{K_a}\] and \[p{K_b}\] in acids and bases?

Answer 1

Hint: 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 [A], [B], [C] and [D] refer to the molar concentration of species A, B, C, 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 \[p{K_b}\] 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.
The relationship between \[{K_a}\] and \[{K_b}\] through ion constant for water, \[{K_w}\] is:
$ {K_a} \times {K_b} = {K_w} $
Where, \[{K_a}\] is acid dissociation constant and \[p{K_a} = - \log {K_a}\] . Similarly, \[{K_b}\] is base dissociation constant, and \[p{K_b} = - \log {K_b}\] . The above given relation is valid for conjugate acid-base pairs. When an acid gets dissolved in water:
 \[HA \rightleftharpoons {H^ + } + {A^ - }\]
 $ {K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{HA}} $
We can say the greater the value of \[{K_a}\] , stronger is the acid.
For most of the weak acids, \[{K_a}\] ranges from \[{10^{ - 2}}\;\] to \[{10^{ - 14}}\] .
We can convert the exponential numbers into the normal range if we take their negative logarithm.
As we know \[p{K_a} = - \log {K_a}\]
For most of the weak acids, \[p{K_a}\] ranges from \[2{\text{ }}to{\text{ }}14\] .
Thus, we can say the smaller the value of \[p{K_a}\] , stronger is the acid.
Similarly, when base gets dissolved in water:
$
  B + {H_2}O \rightleftharpoons B{H^ + } + O{H^ - }{\text{ }} \\
  {K_b} = \dfrac{{\left[ {B{H^ + }} \right]\left[ {O{H^ - }} \right]}}{B} \\
$
We can say greater the value of \[{K_b}\] , stronger is the base.
For most of the weak acids, \[{K_b}\] ranges from \[{10^{ - 2}}\;\] to \[{10^{ - 13}}\] .
We can convert the exponential numbers into the normal range if we take their negative logarithm.
As we know \[p{K_b} = - \log {K_b}\]
For most of the weak acids, \[p{K_b}\] ranges from \[2{\text{ }}to{\text{ }}13\] .
Thus, we can say smaller the value of \[p{K_b}\] , stronger is the base.

Note: The relation $ {K_a} \times {K_b} = {K_w} $ is valid for conjugate acid-base pairs. 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.