Answer:
Approximately 
.
Explanation:
Initial volume of the solution: 
.
Initial quantity of 
:
.
Ammonia reacts with hydrochloric 
acid at a one-to-one ratio:
.
Hence, approximately 
of 
molecules would be required to exactly react with the 
in the original solution and hence reach the equivalence point of this titration.
Calculate the volume of that 
solution required for reaching the equivalence point of this titration:
.
Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately 
.
If no hydrolysis took place, 
of 
would be produced. Because 
is a soluble salt, the solution would contain 
of 
ions. The concentration of 
would be approximately:
.
However, because 
is a weak base, its conjugate would be a weak base.
.
Hence, the following reversible reaction would be take place in the solution at the equivalence point:
.
Let 
be the increase in the concentration of 
in this solution because of this reversible reaction. (Notice that 
.) Construct the following 
table:
.
Thus, at equilibrium:
- Concentration of the weak acid:
![[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20%7BNH_4%7D%5E%7B%2B%7D%7D%5D%20%5Capprox%20%280.255782%20-%20x%29%20%5C%3B%20%5Crm%20M)
. - Concentration of the conjugate of the weak acid:
![[{\rm NH_3}] = x\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20NH_3%7D%5D%20%3D%20x%5C%3B%20%5Crm%20M)
. - Concentration of :
![[{\rm {H}^{+}}] \approx x\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20%7BH%7D%5E%7B%2B%7D%7D%5D%20%5Capprox%20x%5C%3B%20%5Crm%20M)
.
![\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5B%7B%5Crm%20NH_3%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20H%5E%7B%2B%7D%7D%5D%7D%7B%5B%7B%20%5Crm%20%7BNH_4%7D%5E%7B%2B%7D%7D%5D%7D%20%3D%2010%5E%7BpK_%5Ctext%7Ba%7D%28%7B%5Crm%20%7BNH_4%7D%5E%7B%2B%7D%7D%29%7D)
.
Solve for 
. (Notice that the value of 
is likely to be much smaller than 
. Hence, the denominator on the left-hand side 
.)
.
Hence, the concentration of at the equivalence point of this titration would be approximately 
.
Hence, the 
at the equivalence point of this titration would be:
![\begin{aligned}pH &= -\log_{10}[{\rm {H}^{+}}] \\ &\approx -\log_{10} \left(1.19929 \times 10^{-5}\right) \approx 4.92\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DpH%20%26%3D%20-%5Clog_%7B10%7D%5B%7B%5Crm%20%7BH%7D%5E%7B%2B%7D%7D%5D%20%5C%5C%20%26%5Capprox%20-%5Clog_%7B10%7D%20%5Cleft%281.19929%20%5Ctimes%2010%5E%7B-5%7D%5Cright%29%20%5Capprox%204.92%5Cend%7Baligned%7D)
.
