Question

One of the most common types of fertilizers used by farmers is a form of the highly soluble salt ammonium chloride (NH4Cl). At a certain farm, water runoff from fields carries dissolved ammonium chloride to a nearby pond, resulting in a dissolved concentration of NH4Cl in the pond of 5 mg/L. At the same time, ammonia gas (NH3) is being produced by cows that feed around the farmer’s pond, resulting in an atmospheric level of 500 ppb above the water surface. In water, the following reactions take place:

NH4Cl → NH4+ + Cl-
NH4+ ↔ NH3 + H+

The first equation represents complete dissociation (dissolution of a salt), and the second equation is reversible with equilibrium constant [NH3] [H+ ] / [NH4 + ] = 5.89 x 10-10 M. Assume that the ambient temperature is 250 C and the air pressure is 1 atm. Neglect the effects of other impurities such as CO2. Calculate the pH of the pond water.

Given: MW of NH4Cl=53.5 g/mol; KH (Henry’s constant) = 62 M/atm; [H+ ][OH- ]= 10^-14 Molecular weights: H=1, N=14, Cl=35.5, O=16 pH=-log10[H+ ]; H2O = H+ + OH- ; [NH3]=K_HP_NH3; P_HH3= mole fraction x Ptotal; ppb= Vi/Vtotal x 109 = molesi/molestotal x 10^9

Answer 1

pH of the pond water

The chemistry of the pond and surrounding air consists of the following elements: The fertilizer is a salt, which dissociates completely in the water. Thus,

NH₄Cl --------> NH₄⁺ + Cl^-

While the ammonium ion NH4+ undergoes additional chemical changes, the Cl ion remains inactive. It's concentration is then entirely due to the NH4Cl input. To get that concentration, we first need to convert the given mass amount of NH4Cl into moles.

For this, we begin with the molecular weight of this molecule:

MW of NH4Cl = 14 + (4x1) + 35.50 = 53.50 g/mol.

The number of moles of NH4Cl in a 5 mg/L solution is:

[NH₄Cl] = 5x10^-3 g/L / 53.50 g/mol = 9.35x10^-5 mol/L = 9.35x10^-5 M.

Since one molecule of NH4Cl generates one ion of Cl, one mole of NH4Cl generates one mole of Cl, and

[Cl^-] = 9.35 x 10^-5 M.

Another piece of information that we have is that the NH₃ in the pond is in equilibrium with an atmospheric concentration of 550 ppb = 550x10^-9 atm.

By virtue of Henry's Law, with constant KH: [NH₃] = KH PNH3 = (62 M/atm)(550 x 10^-9 atm) = 3.41x10^-5 M .

Now, we turn to the chemical reactions that occur in the water. There are two chemical equilibria, each with its constant of equilibrium:

K = [NH₃] [H⁺] / [NH₄⁺]

[H⁺] = K[NH₄⁺] / [NH₃]

[NH₄⁺] = [NH₃] [H⁺] / K = 3.41x10^-5 / 5.89x10^-10 = 5.79x10⁴[H⁺] = [NH₄⁺] (1)

1x10^-14 = [H⁺] [OH^-] ---> [OH^-] = 1x10^-14 / [H⁺] (2)

To close the set of equations, we add the requirement of electroneutrality, which takes the form:

[NH4⁺] + [H⁺] = [Cl^-] + [OH^-]. (3)

Since we already know [NH₃] and [Cl^-], we have three equations for the following three unknowns: [NH4⁺], [H⁺] and [OH^-]. We are after the pH of the water which demands that we know [H⁺]. For this, we need to eliminate all other unknowns. Using (1) and (2) in (3)we have:

5.7895x10⁴[H⁺] + [H⁺] = 9.35x10^-5 + 1x10^-14/[H⁺]

5.7896x10⁴[H⁺] - 9.35x10^-5 = 1x10^-14 / [H⁺] [H⁺] = x

5.7896x² - 9.35x10^-5x - 1x10^-14 = 0

From here, solving this cuadratic equation, we'll get to the value of 1.71x10^-9 M

The pH is:

pH = -log[H⁺] = -log(1.71x10^-9)

pH = 8.77.

Hope this helps Tell me in a comment if there is something you don't understand or something to be fixed.