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Un-ionized ammonia (NH3) is toxic to fish at low concentrations. The dissociation of ammonia in water...

Un-ionized ammonia (NH3) is toxic to fish at low concentrations. The dissociation of ammonia in water has an equilibrium constant of pKa = 9.25 and is described by the reaction NH4+ NH3 + H+ Calculate and plot the concentrations of NH3 and NH4+ at pH values between 6 and 10 if the total ammonia concentration (NH3 + NH4+) is 1 mg/L as N.

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Answer #1

Using hessley henderbach equation

pH = pKa + log[NH3/NH4+]

at pH = 6

6 = 9.25 + log[NH3/NH4+]

-3.25 = log[NH3/NH4+]

[NH4+] = 10^(3.25) * [NH3] = 1778 [NH3]

[NH4+] = 1778/1779 [NH3] = 0.9994 mg/L

[NH3] = 0.000562 mg/L

now at pH = 10

10 = 9.25 + log[NH3/NH4+]

0.75 = log[NH3/NH4+]

[NH3] = 10^(0.75) * [NH4+] = 5.6253 [NH4+]

[NH3] = 5.625/6.625 * 1mg/L = 0.8490 mg/L

[NH4+] = 1 - 0.8490 = 0.1509 mg/L

Hence the value of [NH4+] will drop from 0.99994 mg/L to 0.1509 mg/L and [NH3] will increase from 0.000562 mg/L to 0.8490 mg/L

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