To answer these questions, we need to use the principles of equilibrium and solubility. Let's go step by step:
1. K for the reaction:
The equilibrium constant, K, for the given reaction is obtained from the balanced equation. The equation shows a 1:1 stoichiometry between CaCO3 and H+ ions. The expression for K is given by:
K = [Ca2+][HCO3-] / [H+]
Without any additional information about the concentrations of the ions, the value of K cannot be determined.
2. Molar solubility of CaCO3 in pure water:
For a sparingly soluble salt like CaCO3, we can determine its solubility by setting up an equilibrium expression based on its dissociation. The solubility of CaCO3 in water is represented by the concentration of Ca2+ and CO3^2- ions, which is equal to x (assuming all CaCO3 dissolves).
The balanced equation, CaCO3(s) ↔ Ca2+(aq) + CO3^2-(aq), indicates that the concentration of Ca2+ ions is also x, and the concentration of CO3^2- ions is also x.
The solubility product constant, Ksp, is given by:
Ksp = [Ca2+][CO3^2-] = x * x = x^2
Since CaCO3 is considered almost pure, we can assume that the concentration of Ca2+ is equal to the solubility of CaCO3, x. Therefore, we have:
Ksp = x^2
3. Molar solubility of CaCO3 in acid rain water with pH of 4.00:
Since the pH of acid rain is given, we can use this information to determine the concentration of H+ ions in the acid rainwater.
The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions in the solution:
pH = -log[H+]
On rearranging the equation, we have:
[H+] = 10^(-pH)
[H+] = 10^(-4)
Now, using the balanced equation again:
CaCO3(s) + H+(aq) ↔ Ca2+(aq) + HCO3^-(aq)
Assuming all CaCO3 dissolves (concentration = x), the concentration of Ca2+ and HCO3^- ions will also be x. However, the concentration of H+ ions has changed due to the acid rain.
Now, we can use the equilibrium expression to solve for x:
K = [Ca2+][HCO3^-] / [H+]
K = (x)(x) / (10^(-4))
We still don't know the value of K, so we can't calculate the exact value of x. However, based on the magnitude of K and the dissociation of CaCO3, we can assume that the molar solubility of CaCO3 in acid rainwater with a pH of 4.00 will be lower than in pure water.
In summary:
1. The value of K for the reaction cannot be determined without additional information.
2. The molar solubility of CaCO3 in pure water is given by the square root of the solubility product constant, Ksp.
3. The molar solubility of CaCO3 in acid rainwater with a pH of 4.00 is expected to be lower than in pure water due to the increased concentration of H+ ions.