Question

) 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)

Answer 1

`NH_4^+ = 2.5 mg/lt`

Step-by-step explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

`NH_4 ^(+) ⇄ H^(+) + NH_3`

Rate constant is given as

`K_a = ([H^(+)] [NH_3])/(NH_4^(+))` ...........1

`K_a = 5.6 * 10^(-10)`

Total ammonia as NItrogen is given as 30 mg/l

`\%NH_4^(+) = ( [NH_4] * 100)/([NH_4^(+)] + [NH_3])`

=
`(100)/((NH_4^+)/(NH_4^+) +(NH_3^+)/(NH_4^+))`

`= (100)/(1+ (NH_3^+)/(NH_4^+))` .....2

from equation 1 we have

`(NH_3^+)/(NH_4^+) =(K_a)/([H^+]) = \frac{5.6* 10^(-10)}`{10^{8.5}}

plug this value in equation 2 we get

`\%NH_4^(+) = 84.96 \%`

Total ammonia as N = 30 mg/lt

`NH_4^+ = (84.96)/(100) * 30 = 25.5 mg/lt`