Question

Ammonium nitrate, NH4NO3, is used as a N fertilizer and has the potential to degrade water quality due to its solubility. The US EPA mean contaminant level (MCL) for nitrate concentrations in drinking water is 10 mg NO3 L-1.

First, calculate the Ksp for NH4NO3. Express your answer as a number rather than in a power of 10 (e.g., 100 instead of 10^2). The Ksp does not need units.

Next, calculate the molarity of the equilbirum concentration of NO3. Units required.

Answer 1

Balanced equation:

NH₄NO₃ ----> NH₄⁺   + NO₃^-

Ksp = ]NH₄⁺] * [NO₃^-]

= (10 ) (10 )

= 100

Since Ammonium nitrate gives as many moles of NO₃^- as many as moles of NH₄⁺

equilibrium concentration of NO₃^- would be same as that of NH₄⁺

So initial concentration of NH₄⁺ as many as moles of NO₃^-

10 mg = 0.01 g per litre

Molarity of NO₃^-  = (0.01 g) / (62.0 g/mole)

= 1.613 * 10^-4 moles/litre

NH₄⁺   + H2O ------> NH₃ + H₃O⁺

1.613 * 10^-4   - - I

-x +x +x C

(1.613 * 10^-4 - x) +x +x E

This represents base dissociation of ammoniua Kb ( = 1.8 * 10^-5 )

Kb = 1.8 * 10^-5 = x * x

(1.613 * 10^-4 - x)

Ignoring x in the denominator since its too small indictaing dissociation of weak acid - ammonia and solving for x, we get

x = 4.91 * 10^-5 M   

Equilibrium concentration of NH₄⁺  = ( 1.613 * 10^-4  - 4.91 * 10^-5)

   = (0.0001613 - 0.0000491)

= 0.0001122

   = 1.22 * 10^-4 M

This would be the equilibrium concentration of NO₃^-  = 1.22 * 10^-4 M