Question

What is the $pH$ of the buffer prepared by mixing $300cc$ of $0.3M$$N{H_3}$ with $500cc$ of $0.5M$ $N{H_4}Cl$ ? ${K_b}$ for $N{H_3}$ is $1.8 \times {10^{ - 5}}$
A. $8.1187$
B. $9.8117$
C. $8.8117$
D. None of the above

Answer 1

Hint: A buffer solution is a mixture of a weak acid or base with its salt. Here we are given an alkaline buffer where $N{H_3}$ is a weak base and its salt is $N{H_4}Cl$. ${K_b}$ is a base dissociation constant, from here we can calculate $p{K_b}$ which will help us calculate the value of $pOH$. Sum of $pH$ and $pOH$ is $14$.

Complete Step by step answer:
A buffer solution is a mixture of a weak acid or base with its salt. A buffer solution resists change in its $pH$ when a strong base or an acid is added to it. There are two types of buffer solutions that are acidic and alkaline. An acidic buffer is a mixture of a weak acid with its salt. Here we are given an alkaline buffer where $N{H_3}$ is a weak base and its salt is $N{H_4}Cl$. ${K_b}$ is a base dissociation constant, it tells us how much base dissociates into its consequent ions. Here ${K_b}$ for $N{H_3}$ is $1.8 \times {10^{ - 5}}$
Molarity is defined as the number of moles of solute in a given volume of solution.
Molarity$ = \dfrac{{mol}}{V}$…….$\left( 1 \right)$
$mol = $Number of moles
$V = $ Volume of solution $\left( L \right)$
Here molarity of $N{H_3}$ is $0.3M$ and volume is $300cc$. Putting these values in $\left( 1 \right)$
$\Rightarrow 0.3M = \dfrac{mol}{0.3}$
$\Rightarrow mol = 0.09mol$

Similarly, we will calculate the number of moles of $N{H_4}Cl$
$
 \Rightarrow mol = 0.5M \times 0.5L
 \Rightarrow mol = 0.25mol
$
$p{K_b}$ is defined as the negative of $\log {K_b}$. It tells about the strength of the base.
$p{K_b} = - \log {K_b}$. Putting $${K_b} = 1.8 \times {10^{ - 5}}$$
$ \Rightarrow p{k_b} = 4.75$
Since we are given an alkaline buffer, we will calculate $pOH$ of the solution. $pOH$ tells the amount of $O{H^ - }$ in the solution.
$pOH = p{K_b} + \log \dfrac{{salt}}{{base}}$
$\Rightarrow pOH = 4.75 + {\log} \dfrac{{0.25}}{{0.09}}$
$\Rightarrow pOH = 4.75 - 0.493$
From here we get $pOH = 4.307$. We know that sum of $pH$ and $pOH$ is $14$
$\Rightarrow pH = 14 - 4.307$
$ \Rightarrow pH = 8.8117$

Therefore the correct option is A.

Note: It is important to note that correct $SI$ units should be used. For example, the volume of solution is given in $cc$. It should be converted into liters. The following conversion can be used.
$1L = 1000cc$