Question

In a basic buffer, 0.0025 mole of ${\text{NH}}{ _4}{\text{Cl}}$ and 0.15 mole of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ are present. The pH of the solution will be:$\left( {{\text{p}}{{\text{K}}_{\text{a}}}{\text{ = }}4.74} \right)$
A. ${\text{11}}{\text{.04}}$
B. ${\text{10}}{\text{.24}}$
C. $6.62$
D. $5.48$

Answer 1

We will use Henderson equation to determine the ${\text{pOH}}$ of a basic buffer solution. Then the ${\text{pH}}$ and ${\text{pOH}}$relation to determine the ${\text{pH}}$.

Formula used: ${\text{pOH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{b}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{salt}}}}{{{\text{base}}}}} \right]$
${\text{pH}}\,\, + {\text{pOH = }}\,14$

Complete step by step answer:
Buffer is a mixture of a weak base and its conjugate acid or a weak acid and its conjugate base.
The Henderson equation is used to determine the ${\text{pH}}$ and ${\text{pOH}}$ of buffer which is as follows:
${\text{pOH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{b}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{salt}}}}{{{\text{base}}}}} \right]$
Where,
${\text{p}}{{\text{K}}_{\text{b}}}$ is the negative logarithm of dissociation constant of the base.
The volume of the solution is not given so, assume volume of solution is one liter so, the concentration of salt is,
$= \dfrac{{0.0025\,{\text{mol}}}}{{1\,{\text{L}}}}$
$= 0.0025\,{\text{M}}$
The concentration of base is,
$= \dfrac{{0.15\,{\text{mol}}}}{{1\,{\text{L}}}}$
$= 0.15\,{\text{M}}$
Substitute 4.74 for ${\text{p}}{{\text{K}}_{\text{b}}}$, $0.0025\,{\text{M}}$ for concentration of salt and $0.15\,{\text{M}}$ for concentration of base.
${\text{pOH}}\,\,{\text{ = }}\,{\text{4}}{\text{.74}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{0.0025}}{{{\text{0}}{\text{.15}}}}} \right]$
${\text{pOH}}\,\,{\text{ = }}\,{\text{4}}{\text{.74}}\, - 1.78$
${\text{pOH}}\,\,{\text{ = }}\,2.96$
Now we will use the ${\text{pH}}$and ${\text{pOH}}$ relation to determine the ${\text{pH}}$ as follows:
${\text{pH}}\,\,{\text{ = }}\, - \,{\text{log }}\left[ {{{\text{H}}^{\text{ + }}}} \right]$
Substitute 2.96 for ${\text{pOH}}$.
${\text{pH}}\,\, + {\text{2}}{\text{.96 = }}\,14$
${\text{pH}}\,\,{\text{ = }}\,14 - 2.96$
${\text{pH}}\,\,{\text{ = }}\,11.04$
So, the ${\text{pH}}$ of the solution is 11.04.

Therefore, option (A) 11.04 is correct.

Additional Information: Henderson equation to determine the ${\text{pH}}$ of an acidic buffer solution is as follows:
${\text{pH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{a}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{salt}}}}{{{\text{acid}}}}} \right]$ Where, ${\text{p}}{{\text{K}}_{\text{a}}}$ is the negative logarithm of acid dissociation constant.

Note: Concentration of salt and base is determined as molarity. Molarity is defined as the number of moles of solute dissolved in a given volume of the solution. The formula to determine the molarity is, ${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Mole}}\,{\text{of}}\,{\text{solute(mol)}}}}{{{\text{Volume of solution(L)}}}}$ .