Question

Calculate the amount of ammonia (NH₃) that is in water at a pH of 7.4 if the total concentration of ammonia and ammonium (NH₄⁺) is 0.001M. Express the concentration of NH₃ as mg/L as N (or short-hand written as mg/L-N). For a pH range of 6 to 10 (in increments of 0.5 pH units), prepare a bar graph of total Nitrogen in mg/L-N versus pH (i.e., pH on the x-axis). Within each “bar” determine the concentration of NH₃ and NH₄⁺ for each pH value.

Answer 1

NH3 + H2O <----> NH4⁺ + OH^-

In above equation equal amount of OH^- and NH4⁺ are formed.

OH^- and then NH4⁺ can be had from pH

pH = 7.4

pOH + pH = 14

pOH = 14 - 7.4 = 6.6

pOH = - log[OH-]

[OH-] = -antilog pOH = -antilog 6.6 = 2.51 X 10^-7

Hence concentration of NH4⁺ = 2.51 X 10^-7 M

NH3 + NH4⁺ = 0.001

NH3 = 0.001 - 2.51 X 10^-7 = 0.00099M

It means 0.00099 moles of NH3 are present in one liter of water.

Mass of NH3 in grams = no. of moles X Molecular mass

Mass of NH3 in grams = 0.00099 X 17 = 0.01699g/L

Mass of NH3 in miligrams = 0.01699 X 10³ = 16.99mg/L

Same procedure can be applied to other pH values as well. For example

pH = 6

pOH = 8

[OH-] = antilog(-pOH) = antilog(-8) = 1.0 X 10^-8

Therefore NH4+ = 1.0 X 10^-8

NH3 = 0.001 - 1.0 X 10^-8 = 0.00099999 M

Concentration mg/L = 0.00099999 X 17 X 1000 = 16.99983mg/L

pH = 6.5

pOH = 7.5

[OH-] or [NH4+] = antilog [-7.5] = 3.1622 X 10^-8

[NH3] = 0.001 - 3.1622 X 10^-8 = 0.000999968

mass of NH3 in mg/l = 0.000999968 X 17 X 1000 = 16.9994628mg/L

pH = 7

pOH = 7

[OH-] or [NH4+] = antilog(-7) = 1.0 X 10^-7

[NH3] = 0.001 - 1.0 X 10^-7 = 0.0009999

concentration of NH3 in mg/L = 0.0009999 X 17 X 1000 = 16.9983mg/L

pH = 7.5

pOH = 6.5

[OH-] or [NH4+] = antilog (-6.5) = 3.16227 X 10^-7

[NH3] = 0.001 - 3.16227 X 10^-7 = 0.00099968M

concentration of NH3 in mg/L = 0.00099968 X 17 X 1000 = 16.99462mg/L

pH = 8

pOH = 6

[OH-] or [NH4+] = antilog (-6) = 1.0 X 10^-6

[NH3} = 0.001 - 1.0 X 10^-6 = 0.000999 M

mass of NH3 in mg/l = 0.00099 X 17 X 1000 = 16.83mg/L

mass of NH3 in mg/l = 0.000999 X 17 X 1000 = 16.983mg/L

pH 8.5

pOH = 5.5

[OH-] or [NH4+] = antilog (-5.5) = 3.16227 X 10^-6

[NH3] = 0.001 - 3.16227 X 10^-6 = 0.0009968 mg/L

pH = 9

pOH = 5

[NH4+] = antilog (-5) = 1.0 X 10^-5

[NH3] = 0.001 - 1.0 X 10^-5 = 0.00099M

pH = 9.5

pOH = 4.5

[NH4+] = 3.16227 X 10^-5

[NH3] = 0.001 -3.16227 X 10^-5 = 0.0009683M

Amount of NH3 in mh/L = 0.0009683 X 17 X 1000 = 16.4624mg/L

pH = 10

pOH = 4

[NH4+] = 1.0 X 10^-4

[NH3] = 0.001 - 1.0 X 10^-4 = 0.0009M

amount of NH3 in mg/L = 0.0009 X 17 X 1000 = 15.mg/L

NOTE; THE BAR DIAGRAM CAN NOT BE PLOTTED IN QA BOARD. PLEASE INSERT HE VALUES IN AN APPROPRIATE SOFTWARE TO GET THE PLOT.