Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
=14-56.17
=8.823
The volume of the 
= 40.00 ml
convert into the liter= 0.040L
The value of the concentrated =0.10 M
The volume of the 
= 50.00 ml
convert into the liter= 0.050L
The value of concentrated = 0.10 M
The volume of the 
= 30 ml
convert into the liter= 0.030L
The value of concentrated 
=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of 
= 
calculating the new concentrated value of 
= 
calculating the new concentrated value of 
when 1 mol produced 2 mols 
so, 0.0125 in produced:
create the ICE table:
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
![P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769](https://tex.z-dn.net/?f=P%5E%7BH%7D%3D%20p%5E%7Bkb%7D%7C%2B%20%5Clog%5Cfrac%7B%5BNH_4%5E%7B%2B%7D%5D%7D%7B%5BNH_3%5D%7D%5C%5C%5C%5C8.83%3Dp%5E%7Bkb%7D%2B%5Clog%5Cfrac%7B0.0667%7D%7B8.3%20%5Ctimes%2010%5E%7B-3%7D%7D%5C%5C%5C%5Cp%5E%7Bkb%7D%3D8.83-0.9069%5C%5C%5C%5C%20%5C%20%5C%20%5C%20%3D7.7231%20%5C%5C%5C%5C%5C%20The%20P%5E%7Bkb%7D%20%5C%20for%20%5C%20NH_3%20%5C%20is%20%3D7.7231%5C%5C%5C%5C%5C%20The%20P%5E%7Bkb%7D%20%5C%20for%20N%5E%7B%2B%7DH_4%3D14-7.7231%5C%5C%5C%5C%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%3D6.2769)
