One of the issues caused by acid rain is the lowering of pH of small lakes, which leads to the death of numerous aquatic fish and plant species. Calcium oxide, commonly known as lime, is frequently used to temporarily remediate lakes that have been acidified by acid rain. Suppose you are tasked with restoring an acid lake with a volume of 3.00 × 1010𝐿 and a pH of 5.0. Using the equilibria below, what mass of calcium oxide will be needed to restore the lake to an ideal pH of 6.5?

Question

One of the issues caused by acid rain is the lowering of pH of small lakes, which leads to the death of numerous aquatic fish and plant species. Calcium oxide, commonly known as lime, is frequently used to temporarily remediate lakes that have been acidified by acid rain. Suppose you are tasked with restoring an acid lake with a volume of 3.00 × 1010𝐿 and a pH of 5.0. Using the equilibria below, what mass of calcium oxide will be needed to restore the lake to an ideal pH of 6.5?

𝐶𝑎𝑂(𝑠) + 𝐻2𝑂(𝑙) ↔ 𝐶𝑎(𝑂𝐻)2(𝑠)

𝐾𝑒𝑞 = 1.33 × 1010

𝐶𝑎(𝑂𝐻)2(𝑠) ↔ 𝐶𝑎2+(𝑎𝑞) + 2𝑂𝐻−(𝑎𝑞)

𝐾𝑠𝑝 = 5.61 × 10−5

Answer 1

How much OH in pH=5 and How much OH in pH=6.5. Determine difference and that's how much CaO is needed.

Answer 2

To find the mass of calcium oxide (CaO) needed to restore the lake's pH, you can use the equilibrium constants and the stoichiometry of the reaction.

First, let's write down the balanced chemical equation for the reaction between calcium oxide and water:

CaO(s) + H2O(l) ↔ Ca(OH)2(s)

Now, we can set up an ICE (initial, change, equilibrium) table to find the concentration of Ca2+ and OH- ions at equilibrium.

Let x be the change in the concentration (in mol/L) of Ca2+ and OH- ions in the equilibrium. Since CaO is a solid, its concentration remains constant.

Initial:
Ca2+: 0 mol/L
OH-: 0 mol/L

Change:
Ca2+: +x mol/L
OH-: +2x mol/L

Equilibrium:
Ca2+: x mol/L
OH-: 2x mol/L

The equilibrium constant expression for the first equilibrium is given by:

Keq = [Ca(OH)2] / [H2O]

Since the concentration of water remains essentially constant, we can simplify the expression to:

Keq = [Ca(OH)2]

Now, let's calculate the equilibrium constant (Keq1) using the given value:

Keq1 = 1.33 × 10^10

Next, the equilibrium constant expression for the second equilibrium is given by:

Ksp = [Ca2+][OH-]^2

Substituting the equilibrium concentrations, we get:

Ksp = (x)(2x)^2 = 4x^3

The value of the equilibrium constant (Ksp) is given:

Ksp = 5.61 × 10^-5

To relate Keq1 and Ksp, we can use the fact that the Ca(OH)2 concentration (Keq1) is equal to twice the OH- concentration (Ksp).

Ca(OH)2 concentration = 2x = Keq1 / 2

Plugging in the values:

2x = 1.33 × 10^10 / 2

Simplifying:

x = 6.65 × 10^9

Now, we need to convert the concentration of Ca2+ (x) to mass of calcium oxide (CaO) using the given volume of the lake:

Mass of CaO = Concentration of Ca2+ × volume of lake

To convert x from mol/L to mass, we need to multiply by the molar mass of CaO.

The molar mass of CaO is approximately 56 g/mol.

Mass of CaO = 6.65 × 10^9 mol/L × 56 g/mol × 3.00 × 10^10 L

Calculating this expression will give you the mass of calcium oxide needed to restore the lake to an ideal pH of 6.5.