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$={\displaystyle \frac{mol}{V}}$…….$\left(1\right)$
$mol=$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 $\left(1\right)$
$\Rightarrow 0.3M={\displaystyle \frac{mol}{0.3}}$
$\Rightarrow mol=0.09mol$

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.
$pOH=p{K}_b+\log {\displaystyle \frac{salt}{base}}$
$\Rightarrow pOH=4.75+\log {\displaystyle \frac{0.25}{0.09}}$
$\Rightarrow pOH=4.75-0.493$
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.
$1L=1000cc$