) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, determine the ammonium nitrogen concentration (mg/L) in the sample assuming a temperature of 25ºC. (Note: The lab analysis represents the sum of the NH4-N and the NH3-N.) Answer: 25.9 mg/L as N (NH4-N)

Respuesta :

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]

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