Question

What are $p{K_a}$ and $p{K_b}$ in acids and bases?

Answer 1

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.