Question

$\text{p}{{\text{K}}_{\text{a}}}$ of $\text{NH}_{\text{4}}^{\text{+}}$ is $9.26$, Hence, effective range of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\,\text{-}\,\,\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$ buffer is about $\text{pH}$:
(A) $\text{8}\text{.26}\,\,\text{to}\,\,\text{10}\text{.26}$
(B) $\text{4}\text{.74}\,\,\text{to}\,\,\text{5}\text{.74}$
(C) $\text{3}\text{.74}\,\,\text{to}\,\,\text{5}\text{.74}$
(D) $\text{8}\text{.26}\,\,\text{to}\,\,\text{9}\text{.26}$

Answer 1

A solution which resists the change in$\text{pH}$ and $\text{pH}$ does not change significantly in addition to a small amount of strong acid and strong base is called a buffer solution. Mixed buffer solution is two types; acidic buffer solution and basic buffer solution.
- Change in $\text{pH}$ of a solution due to the change in concentration of present free ${{\text{H}}^{\text{+}}}$ and $\text{O}{{\text{H}}^{\text{-}}}$ ions.

Complete Solution :
$\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\,\text{-}\,\,\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$ Solution is basic buffer solution (the solution in which weak base and its conjugate acid are present).
An aqueous solution of mixture of weak base and salt of same weak base with any strong acid solution is called basic buffer solution.

& \text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\text{+}\,\text{N}{{\text{H}}_{\text{4}}}\text{Cl} \\\ & \text{N}{{\text{H}}_{\text{4}}}\text{OH}\,\,\rightleftharpoons \,\,\text{NH}_{\text{4}}^{\text{+}}\,\,\text{+}\,\,\text{O}{{\text{H}}^{\text{-}}} \\\ & \text{N}{{\text{H}}_{\text{4}}}\text{Cl}\,\,\rightleftharpoons \,\,\text{NH}_{\text{4}}^{\text{+}}\,\,\text{+}\,\,\text{C}{{\text{l}}^{\text{-}}} \\\ \end{aligned}$$ $\text{pH}$ of a buffer solution is calculated by Henderson’s equation – $\text{pH}\,\,\text{=}\,\,\text{p}{{\text{K}}_{\text{a}}}\,\text{+}\,\text{log}\dfrac{\text{ }\\!\\![\\!\\!\text{ salt }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ acid }\\!\\!]\\!\\!\text{ }}\,\,\text{or}\,\,\dfrac{\text{ }\\!\\![\\!\\!\text{ conjugate }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ acid }\\!\\!]\\!\\!\text{ }}$ $\text{pH}$ Range of basic buffer solution: it depends on the $\text{p}{{\text{K}}_{\text{a}}}$ value of base and ratio of its salt to base concentrations. $$\begin{aligned} & \,\,\text{ }\\!\\![\\!\\!\text{ N}{{\text{H}}_{\text{4}}}\text{OH }\\!\\!]\\!\\!\text{ }\,\,\,\text{:}\,\,\,\,\text{ }\\!\\![\\!\\!\text{ N}{{\text{H}}_{\text{4}}}\text{Cl }\\!\\!]\\!\\!\text{ } \\\ & \,\,\,\,\begin{matrix} \text{(i) if} & 1 & 10 \\\ \end{matrix} \\\ & \,\,\,\text{(ii})\,\,\,\,\begin{matrix} \text{if} & 10 & 1 \\\ \end{matrix} \\\ & \text{so,}\,\,\text{pH}\,\,\text{range}\,\,\text{is}\,\, \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\text{pH}\,\,\text{=}\,\,\text{p}{{\text{K}}_{\text{a}}}\,\text{ }\\!\\!\pm\\!\\!\text{ }\,\,\text{1} \\\ \end{aligned}$$ So, $\text{pH}$ range of the buffer solution $\begin{aligned} & \text{pH}\,\,\text{=}\,\,\,\text{p}{{\text{K}}_{\text{a}}}\,\text{ }\\!\\!\pm\\!\\!\text{ }\,\text{1} \\\ & \text{pH=}\,\,\,\,\text{9}\text{.26}\,\text{ }\\!\\!\pm\\!\\!\text{ }\,\text{1} \\\ & \text{pH}\,\,\text{range}\,\,\text{will}\,\text{be}\,\,\text{8}\text{.26}\,\,\text{to}\,\,\text{10}\text{.26} \\\ \end{aligned}$ **So, the correct answer is “Option A”.** **Additional information** - Buffer action in basic solution – If a small amount of acid is added to the buffer solution, the $${{\text{H}}^{\text{+}}}$$ ions are consumed by $$\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,$$ so there is no appreciable change in$\text{pH}$of the buffer solution and whenever a small amount of base is added in the buffer solution, the $\text{O}{{\text{H}}^{\text{-}}}$ ion react with the $\text{NH}_{\text{4}}^{\text{+}}$ ion to produce $$\text{N}{{\text{H}}_{\text{4}}}\text{OH}\,$$molecule. $\text{O}{{\text{H}}^{\text{-}}}$ **Note:** To calculate the $\text{pOH}$ rang of basic buffer solution we will use the same Henderson’s equation, but instate of $\,\,\text{p}{{\text{K}}_{a}}\,$ value we will use $\,\,\text{p}{{\text{K}}_{b}}\, $ value of the solution $\text{pOH}\,\,\text{=}\,\,\text{p}{{\text{K}}_{\text{b}}}\,\text{+}\,\text{log}\dfrac{\text{ }\\!\\![\\!\\!\text{ salt }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ acid }\\!\\!]\\!\\!\text{ }}\,\,\text{or}\,\,\dfrac{\text{ }\\!\\![\\!\\!\text{ conjugate }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ acid }\\!\\!]\\!\\!\text{ }}$ So, the $\text{pOH}$ range will be $\text{pOH}\,\,\text{=}\,\,\text{p}{{\text{K}}_{\text{b}}}\,\text{ }\\!\\!\pm\\!\\!\text{ }\,\text{1}$