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Question: What is the concentration of Ammonia in the solution? A solution is prepared by dissolving \[0.20m...

What is the concentration of Ammonia in the solution?
A solution is prepared by dissolving 0.20mol0.20mol mol of acetic acid and 0.20mol0.20mol of ammonium chloride in enough water to make 1.0L1.0L of solution. Find the concentration of Ammonia in the solution.

Explanation

Solution

Hint : Ammonia is a nitrogen-hydrogen compound with the formula NH3N{H_3}. While abundant in nature, both on Earth and the Solar System's outer planets, and widely used, Ammonia is caustic and hazardous in condensed form.

Complete Step By Step Answer:
While both acetic and ammonium chloride are weak acids, ammonium chloride is nearly a million times weaker. As a result, Ammonium Chloride has a "common-ion effect" equilibrium problem.
The KaKa value of acetic acid determines how it dissociates. The H+{H^ + } ions in the solution then enter into the Ammonium Chloride equilibrium, and you get a reduced dissociation into Ammonia.
For Acetic Acid:
HOAc=H++OAc−HOAc = {H^ + } + OA{c^ - }
Initial concentration,
0.2M$$$$0M$$$$0M
Let xx mole per litre of each of the product be formed.
At equilibrium:
\left( {C - x} \right)M$$$$xM$$$$xM
\left( {E0.2} \right)M$$$$xM$$$$xM
Where xx is the amount of H+{H^ + } and 0Ac−0A{c^ - } at equilibrium.
Hence, equilibrium constant can be written as,
Ka=x20.2M=1.8×10−5Ka = \dfrac{{{x^2}}}{{0.2M}} = 1.8 \times {10^{ - 5}}
After solving this, we got x=0.0019Mx = 0.0019M
Therefore, H+=0.0019M{H^ + } = 0.0019M
Now, consider Ammonium Chloride with Ka=5.6×10−10Ka = 5.6 \times {10^{ - 10}}
NH4+=NH3+H+NH_4^ + = N{H_3} + {H^ + }
Initial concentration,
0.2M$$$$0M$$$$0.0019M
Let xx mole per litre of each of the product be formed.
At equilibrium:
\left( {C - x} \right)M$$$$xM$$$$xM
\left( {E0.2 - x} \right)$$$$x0.0019$$$$x
Where xx is the amount of NH3N{H_3} and H+{H^ + } at equilibrium.
Hence, equilibrium constant can be written as,
Ka=x(0.0019+x)0.2=5.6×10−10Ka = \dfrac{{x\left( {0.0019 + x} \right)}}{{0.2}} = 5.6 \times {10^{ - 10}}
That is, x2+0.0019x−1.12×10−12=0{x^2} + 0.0019x - 1.12 \times {10^{ - 12}} = 0
After solving this, we got x=5.89×10−10M=[NH3]x = 5.89 \times {10^{ - 10}}M = \left[ {N{H_3}} \right]
So, the concentration of Ammonia is 5.89×10−10M5.89 \times {10^{ - 10}}M .

Additional Information:
The acid dissociation constant, abbreviated as KaKa, is the equilibrium constant of an acid's dissociation reaction. This equilibrium constant is a numerical representation of an acid's power in a solution.

Note :
Ammonia is a precursor for amino acid and nucleotide synthesis and is needed for many biological processes. Ammonia is the component of the nitrogen cycle in the ecosystem, and it is formed in soil by bacterial processes. Ammonia is also formed naturally when organic matter, such as plants, livestock, and animal waste, decomposes.