Marble is almost pure CaCO3. Acid rain has a devastating effect on marble statuary left outdoors. Assume that the reaction which occurs as:
CaCO3(s)+ H+(aq)->Ca2+(aq) + HCO3(aq)
Neglecting all other competing, equilibria, Calculate:
1. K for the reaction
2. molar solubility of CaCO3 in pure water
3. Molar solubility of CaCO3 in acid rain water with pH of 4.00
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.
To calculate the values you need, we will first calculate the equilibrium constant (K) for the reaction and then use it to determine the molar solubility of CaCO3 in pure water and in acid rain water with a pH of 4.00.
1. K for the reaction:
The equilibrium constant (K) of the reaction can be determined using the following formula:
K = [Ca2+][HCO3-] / [H+]
Since the reaction involves a 1:1 molar ratio between Ca2+ and H+, and between HCO3- and H+, we can simplify the equation to:
K = [Ca2+][HCO3-]
2. Molar solubility of CaCO3 in pure water:
In pure water, the concentration of H+ ions is very low, so we can assume it to be negligible. Therefore, the equation becomes:
K = [Ca2+][HCO3-] = [Ca2+][CO3^2-]
The solubility product constant (Ksp) for CaCO3 is the value at which the reaction is at equilibrium:
K = [Ca2+][CO3^2-] = Ksp
The solubility product constant for CaCO3 is 4.8 x 10^-9 at 25°C. Therefore, the molar solubility of CaCO3 in pure water is given by the square root of Ksp:
√Ksp = √(4.8 x 10^-9) = 6.93 x 10^-5 M
3. Molar solubility of CaCO3 in acid rain water with pH of 4.00:
In acid rain water, the pH is 4.00, which means the concentration of H+ ions is 10^-4 M.
Using the reaction equation:
CaCO3(s) + H+(aq) -> Ca2+(aq) + HCO3-(aq)
We know that [H+] = 10^-4 M and we need to find [Ca2+].
To calculate the molar solubility, we assume that the concentration of HCO3- is negligible, as it is a product of the reaction. Therefore, the equation becomes:
K = [Ca2+][HCO3-] = [Ca2+][CO3^2-]
Also, we know that [CO3^2-] = [HCO3-] since CaCO3 is a weak base, and the carbonate and bicarbonate will exist in equilibrium.
Thus, we have:
K = [Ca2+][CO3^2-] = [Ca2+][HCO3-] = [Ca2+](H+)
Substituting the values:
4.8 x 10^-9 = [Ca2+](10^-4)
Rearranging the equation, we can find [Ca2+]:
[Ca2+] = (4.8 x 10^-9) / (10^-4) = 4.8 x 10^-5 M
Therefore, the molar solubility of CaCO3 in acid rain water with a pH of 4.00 is 4.8 x 10^-5 M.
To summarize:
1. The equilibrium constant (K) for the reaction is given by [Ca2+][HCO3-].
2. The molar solubility of CaCO3 in pure water is 6.93 x 10^-5 M.
3. The molar solubility of CaCO3 in acid rain water with a pH of 4.00 is 4.8 x 10^-5 M.