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Question: The value of \({{\text{K}}_{\text{b}}}\) for \({\text{N}}{{\text{H}}_{\text{3}}}\) is \(1.8 \times {...

The value of Kb{{\text{K}}_{\text{b}}} for NH3{\text{N}}{{\text{H}}_{\text{3}}} is 1.8×10−51.8 \times {10^{ - 5}}. Calculate the value of pKa{\text{p}}{{\text{K}}_{\text{a}}} for its conjugate acid. [log⁡1.8=0.255]\left[ {\log 1.8 = 0.255} \right]

Explanation

Solution

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 Ka{{\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 A - {{\text{A}}^{\text{ - }}} while every weak base B has a strong conjugate base BH + {\text{B}}{{\text{H}}^{\text{ + }}}.
Consider the following reaction:
HA + H2O→H3O +  + A - {\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
A− + H2O→HA + OH−{{\text{A}}^ - }{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{HA + O}}{{\text{H}}^ - } here A - {{\text{A}}^{\text{ - }}} acts as a base
Applying the equilibrium to both these equations we get,
Ka = [H3O] + [A - ][H2O]{{\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 Kb = [HA][OH - ][A - ]{{\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}}}iscalledtheionicproductofwater=is called the ionic product of water = 1.0 \times {10^{ - 14}}{\text{mo}}{{\text{l}}^{\text{2}}}{{\text{L}}^{{\text{ - 2}}}}Therefore, Therefore,{{\text{K}}{\text{a}}} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}}.Takingnegativelogarithmonbothsidesweget,. 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 + 0.255So,thevalueof So, the value of{\text{p}}{{\text{K}}{\text{a}}}is:is: {\text{p}}{{\text{K}}_{\text{a}}} = 9.255$

Note:
The conjugate acid of ammonia is NH4 + {\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.