Question

The value of ${{\text{K}}_{\text{b}}}$ for ${\text{N}}{{\text{H}}_{\text{3}}}$ is $1.8 \times {10^{ - 5}}$. Calculate the value of ${\text{p}}{{\text{K}}_{\text{a}}}$ for its conjugate acid. \[\left[ {\log 1.8 = 0.255} \right]\]

Answer 1

Hint:This question can be answered from the concept of Bronsted-Lowry’s Conjugate acid-base pair. An acid is called a proton donor and a base, proton acceptor. Keeping this in mind, the Bronsted-Lowry’s Conjugate acid-base pair theory was put forward for weak acids and weak bases. We shall apply the law of equilibrium to find the value of ${{\text{K}}_{\text{a}}}$.

Complete step by step answer:
According to the conjugate acid-base theory, every weak acid, HA has a strong conjugate base ${{\text{A}}^{\text{ - }}}$ while every weak base B has a strong conjugate base ${\text{B}}{{\text{H}}^{\text{ + }}}$.
Consider the following reaction:
${\text{HA + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ + }}{{\text{A}}^{\text{ - }}}$ here HA acts as an acid
${{\text{A}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{HA + O}}{{\text{H}}^ - }$ here ${{\text{A}}^{\text{ - }}}$ acts as a base
Applying the equilibrium to both these equations we get,
${{\text{K}}_{\text{a}}}{\text{ = }}\dfrac{{{{\left[ {{{\text{H}}_{\text{3}}}{\text{O}}} \right]}^{\text{ + }}}\left[ {{{\text{A}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}$ and ${{\text{K}}_{\text{b}}}{\text{ = }}\dfrac{{\left[ {{\text{HA}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{A}}^{\text{ - }}}} \right]}}$
Multiplying both these equations we get, \[{{\text{K}}_{\text{a}}}{{\text{K}}_{\text{b}}}{\text{ = }}\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]${\text{ = }}{{\text{K}}_{\text{w}}}$
${{\text{K}}_{\text{w}}}$is called the ionic product of water = $1.0 \times {10^{ - 14}}{\text{mo}}{{\text{l}}^{\text{2}}}{{\text{L}}^{{\text{ - 2}}}}$
Therefore, ${{\text{K}}_{\text{a}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}$.
Taking negative logarithm on both sides we get,
${\text{pKa}} = - \log \left[ 1 \right] + \log \left[ {1.8} \right] + 9$
$ \Rightarrow {\text{p}}{{\text{K}}_{\text{a}}} = 9 + 0.255$
So, the value of ${\text{p}}{{\text{K}}_{\text{a}}}$ is:
${\text{p}}{{\text{K}}_{\text{a}}} = 9.255$

Note:
The conjugate acid of ammonia is ${\text{NH}}_{\text{4}}^{\text{ + }}$ is a very weak acid and will thus not produce a very acidic solution.
The conjugate acid-base pairs find a lot of usage as buffer solutions. A buffer solution consists of either a weak base or its conjugate acid or a weak acid and its conjugate base that act to maintain the pH of a chemical change or a titrimetric process.
Blood is a good example of a buffer solution in which the carbonic acid-bicarbonate system acts as the buffer and prevents drastic changes when carbon dioxide is introduced in the body.
Acetic acid and sodium acetate pair also act as another example of a buffer that is commonly used in the laboratories.