Answer:
[tex]NH_4^+ = 2.5 mg/lt[/tex]
Explanation:
Given data:
Ammonia Nitrogen 30 mg/L
pH = 8.5
-log[H +] = 8.5
[H +] = 10^{-8.5}
[tex]NH_4 ^{+} ⇄ H^{+} + NH_3[/tex]
Rate constant is given as
[tex]K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}}[/tex] ...........1
[tex]K_a = 5.6 \times 10^{-10}[/tex]
Total ammonia as NItrogen is given as 30 mg/l
[tex]\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}[/tex]
= [tex]\frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}[/tex]
[tex]= \frac{100}{1+ \frac{NH_3^+}{NH_4^+}}[/tex] .....2
from equation 1 we have
[tex]\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}[/tex]{10^{8.5}}
plug this value in equation 2 we get
[tex]\%NH_4^{+} = 84.96 \%[/tex]
Total ammonia as N = 30 mg/lt
[tex]NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt[/tex]