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Question: A buffer solution consists of \(N{H_3}\) and \(N{H_4}Cl\), total concentration of buffering agent\( ...

A buffer solution consists of NH3N{H_3} and NH4ClN{H_4}Cl, total concentration of buffering agent=0.6M = 0.6M . If the pOHpOH of buffer solution is 5.055.05 then calculate [NH3][N{H_3}] solution. (pKbp{K_b} of NH3=4.75N{H_3} = 4.75)
A. 0.50.5
B. 0.30.3
C. 0.20.2
D. 0.40.4

Explanation

Solution

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 0.6M0.6M .
Given, the pOHpOH of the buffer solution = 5.055.05
[Base]+[salt]=0.6M[Base] + [salt] = 0.6M
Let the concentration of the salt be xx .
Let the concentration of the base be (0.6−x)(0.6 - x).
As per the Henderson-Hasselbalch equation,
pOH=pKb+log⁡([salt][base])pOH = p{K_b} + \log \left( {\dfrac{{[salt]}}{{[base]}}} \right)
According to the question, the pKbp{K_b} of NH3=4.75N{H_3} = 4.75
Thus, substituting the values in the above equation, we have:
⇒5.05=4.75+log⁡(x0.6−x)\Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{x}{{0.6 - x}}} \right)
Thus, on solving, we have:
⇒0.3=log⁡x0.6−x\Rightarrow 0.3 = \log \dfrac{x}{{0.6 - x}}
⇒antilog⁡(0.3)=x0.6−x\Rightarrow anti\log (0.3) = \dfrac{x}{{0.6 - x}}
Also, antilog⁡(0.3)=2anti\log (0.3) = 2
Thus, substituting the value in the above equation, we have:
⇒x0.6−x=2\Rightarrow \dfrac{x}{{0.6 - x}} = 2
On solving, we get: x=1.23=0.4x = \dfrac{{1.2}}{3} = 0.4
Thus, the concentration of the salt = 0.4M0.4M
The concentration of the base = 0.6−0.4=0.2M0.6 - 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 (NH3N{H_3} ) in the solution. Thus, the equation can be written as:
⇒5.05=4.75+log⁡(0.6[base])\Rightarrow 5.05 = 4.75 + \log \left( {\dfrac{{0.6}}{{[base]}}} \right)
⇒2=0.6[base]\Rightarrow 2 = \dfrac{{0.6}}{{[base]}}
Thus, the concentration of NH3N{H_3} = [NH3]=0.3M[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:
pKa+pKb=pKwp{K_a} + p{K_b} = p{K_w}