Calculate the solubility product constant for calcium carbonate, given that it has a solubility of 5.3×10^7 g/L of water.
To calculate the solubility product constant (Ksp) for calcium carbonate (CaCO3), we need to know the balanced chemical equation for the dissolution of CaCO3 in water.
The balanced equation is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
The Ksp expression for this equilibrium is:
Ksp = [Ca2+][CO3^2-]
Since calcium carbonate dissociates into one Ca2+ ion and one CO3^2- ion, the solubility of CaCO3 can be used to determine the concentrations of each ion.
Given that the solubility of CaCO3 is 5.3×10^7 g/L, we can convert this into moles per liter (M) by dividing by the molar mass of CaCO3:
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3*16.00 g/mol) = 100.09 g/mol
Solubility in moles per liter (M) = (5.3×10^7 g/L) / (100.09 g/mol) = 5.296×10^5 mol/L
Since CaCO3 dissociates into one Ca2+ ion and one CO3^2- ion, their concentrations (in mol/L) will be the same as the solubility:
[Ca2+] = 5.296×10^5 mol/L
[CO3^2-] = 5.296×10^5 mol/L
Plugging these values into the Ksp expression:
Ksp = [Ca2+][CO3^2-] = (5.296×10^5 mol/L) × (5.296×10^5 mol/L)
Calculating this product,
Ksp = (5.296×10^5)^2 = 2.803×10^11
Therefore, the solubility product constant (Ksp) for calcium carbonate (CaCO3) is 2.803×10^11.
To calculate the solubility product constant (Ksp) for calcium carbonate, we need to know the balanced chemical equation for its dissolution. In this case, calcium carbonate dissolves in water to form calcium ions (Ca2+) and carbonate ions (CO32-). The balanced equation is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
The Ksp expression for calcium carbonate can be written as:
Ksp = [Ca2+][CO32-]
Since the stoichiometric coefficient of calcium carbonate is 1, the molar solubility (x) of calcium carbonate will be equal to the concentrations of Ca2+ and CO32- ions.
The solubility of calcium carbonate is given as 5.3 × 10^7 g/L. To convert this to molarity, we need to know the molar mass of calcium carbonate, which can be found by adding up the atomic masses of calcium (Ca), carbon (C), and oxygen (O).
CaCO3:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 O atoms)
Therefore, the molar mass of CaCO3 is:
40.08 + 12.01 + (16.00 × 3) = 100.09 g/mol
Now, we can convert the solubility from grams per liter (g/L) to molarity (mol/L).
Given that:
Solubility = 5.3 × 10^7 g/L
Molar mass of CaCO3 = 100.09 g/mol
The molar solubility (x) of calcium carbonate = 5.3 × 10^7 g/L ÷ 100.09 g/mol
Simplifying:
x = 5.297 × 10^5 mol/L
Since x represents the moles per liter of both Ca2+ and CO32- ions, we can substitute it into the Ksp expression:
Ksp = [Ca2+][CO32-] = (5.297 × 10^5 mol/L) × (5.297 × 10^5 mol/L) = 2.801 × 10^11 (mol/L)^2
Therefore, the solubility product constant (Ksp) for calcium carbonate is 2.801 × 10^11 (mol/L)^2.