Question

A buffer solution consists of $N{H_3}$ and $N{H_4}Cl$, total concentration of buffering agent$ = 0.6M$ . If the $pOH$ of buffer solution is $5.05$ then calculate $[N{H_3}]$ solution. ($p{K_b}$ of $N{H_3} = 4.75$)
A. $0.5$
B. $0.3$
C. $0.2$
D. $0.4$

Answer 1

Hint: A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount of strong acid or base is added to it. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications.

Complete answer:
In the above question, the buffer consists of ammonia and ammonium chloride and the concentration of this buffer solution is equal to $0.6M$ .
Given, the $pOH$ of the buffer solution = $5.05$
$[Base] + [salt] = 0.6M$
Let the concentration of the salt be $x$ .
Let the concentration of the base be $(0.6 - x)$.
As per the Henderson-Hasselbalch equation,
$pOH = p{K_b} + \log \left( {\dfrac{{[salt]}}{{[base]}}} \right)$
According to the question, the $p{K_b}$ of $N{H_3} = 4.75$
Thus, substituting the values in the above equation, we have:
$ \Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{x}{{0.6 - x}}} \right)$
Thus, on solving, we have:
$ \Rightarrow 0.3 = \log \dfrac{x}{{0.6 - x}}$
$ \Rightarrow anti\log (0.3) = \dfrac{x}{{0.6 - x}}$
Also, $anti\log (0.3) = 2$
Thus, substituting the value in the above equation, we have:
$ \Rightarrow \dfrac{x}{{0.6 - x}} = 2$
On solving, we get: $x = \dfrac{{1.2}}{3} = 0.4$
Thus, the concentration of the salt = $0.4M$
The concentration of the base = $0.6 - 0.4 = 0.2M$
Now, again applying the concentration of the salt in the Henderson-Hasselbalch equation, we can find the concentration of the base ($N{H_3}$ ) in the solution. Thus, the equation can be written as:
 \[ \Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{{0.6}}{{[base]}}} \right)\]
$ \Rightarrow 2 = \dfrac{{0.6}}{{[base]}}$
Thus, the concentration of $N{H_3}$ = $[N{H_3}] = 0.3M$

Thus option B is the correct answer.

Note:
The Henderson–Hasselbalch equation can be used to calculate the pH of a solution containing the acid and one of its salts, that is, of a buffer solution. With bases, if the value of equilibrium constant is known in the form of a base association constant, Kb the dissociation constant of the conjugate acid may be calculated from the given equation:
$p{K_a} + p{K_b} = p{K_w}$