Answer:
The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of HCl solution = 0.164 M
Volume of solution = 23.8 mL = 0.0238 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:

The chemical equation for the reaction of ammonia and HCl follows:

By Stoichiometry of the reaction:
1 mole of HCl reacts with 1 mole of ammonia
So, 0.0035 moles of HCl will react with =
of ammonia
- Calculating the initial concentration of ammonia by using equation 1:
Moles of ammonia = 0.0035 moles
Volume of solution = 25 mL = 0.025 L
Putting values in equation 1, we get:

By Stoichiometry of the reaction:
1 mole of ammonia produces 1 mole of ammonium ion
So, 0.0035 moles of ammonia will react with =
of ammonium ion
- Calculating the concentration of ammonium ion by using equation 1:
Moles of ammonium ion = 0.0035 moles
Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L
Putting values in equation 1, we get:

- To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

where,
= Ionic product of water = 
= Acid dissociation constant
= Base dissociation constant = 

The chemical equation for the dissociation of ammonium ion follows:

The expression of
for above equation follows:
![K_a=\frac{[NH_3][H^+]}{[NH_4^+]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BNH_3%5D%5BH%5E%2B%5D%7D%7B%5BNH_4%5E%2B%5D%7D)
We know that:
![[NH_3]=[H^+]=x](https://tex.z-dn.net/?f=%5BNH_3%5D%3D%5BH%5E%2B%5D%3Dx)
![[NH_4^+]=0.072M](https://tex.z-dn.net/?f=%5BNH_4%5E%2B%5D%3D0.072M)
Putting values in above expression, we get:

To calculate the pH concentration, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=6.32\times 10^{--6}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D6.32%5Ctimes%2010%5E%7B--6%7DM)

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20