#26
The Ksp of lead(II) carbonate, PbCO3, is 7.40 × 10-14. Calculate the solubility of this compound in g/L.
See your Ni(OH)2 post.
To calculate the solubility of lead(II) carbonate (PbCO3) in grams per liter (g/L), we will use the Ksp expression and the molar mass of PbCO3.
The balanced equation for the dissociation of PbCO3 is:
PbCO3(s) ⟶ Pb2+(aq) + CO3^2-(aq)
The solubility of PbCO3 is equal to the concentration of Pb2+ and CO3^2-. Since PbCO3 is a 1:1 salt, the solubility of Pb2+ and CO3^2- will be the same.
Let's assume the solubility of PbCO3 is "s". Therefore, the concentration of Pb2+ and CO3^2- will also be "s".
The Ksp expression for PbCO3 is given as follows:
Ksp = [Pb2+][CO3^2-]
Given that the Ksp of PbCO3 is 7.40 × 10^-14, we can substitute the concentrations (s) into the Ksp expression:
7.40 × 10^-14 = s * s
Next, we solve the equation for s:
s^2 = 7.40 × 10^-14
s = √(7.40 × 10^-14)
Calculating the square root gives:
s ≈ 8.60 × 10^-8 M
The solubility of PbCO3 is approximately 8.60 × 10^-8 M.
To convert it to grams per liter (g/L), we need to consider the molar mass of PbCO3. The molar mass of PbCO3 is given by adding the atomic masses of lead (Pb), carbon (C), and oxygen (O).
Pb: 207.2g/mol
C: 12.0g/mol
O: 16.0g/mol (since there are three oxygen atoms, we multiply this by 3)
Molar mass of PbCO3 = 207.2 + 12.0 + (16.0 * 3)
= 323.2g/mol
Now, we can convert the solubility from molar concentration to grams per liter:
8.60 × 10^-8 M * 323.2 g/mol = 2.78 × 10^-5 g/L
Therefore, the solubility of lead(II) carbonate (PbCO3) in grams per liter (g/L) is approximately 2.78 × 10^-5 g/L.
To calculate the solubility of lead(II) carbonate (PbCO3) in grams per liter (g/L), we need to use the solubility product constant (Ksp) value.
The solubility product constant (Ksp) is an equilibrium constant that describes the extent to which a compound dissolves in water. For PbCO3, the Ksp value is given as 7.40 × 10-14.
The formula for calculating the molar solubility (s) of a compound is given by:
Ksp = [Pb2+][CO32-]
Since the solubility of the compound is in terms of grams per liter (g/L), we need to convert the molar solubility (s) to grams per liter.
To do this, we need to know the molar mass of PbCO3, which is the sum of the atomic masses of lead (Pb), carbon (C), and three oxygen (O) atoms.
The atomic mass of Pb is 207.2 g/mol, the atomic mass of C is 12.0 g/mol, and the atomic mass of O is 16.0 g/mol.
So, the molar mass of PbCO3 is:
(207.2 g/mol Pb) + (12.0 g/mol C) + (3 * 16.0 g/mol O) = 267.2 g/mol PbCO3
Now, we can calculate the solubility in g/L using the molar solubility (s) and the molar mass of PbCO3 as:
Solubility (g/L) = (s * molar mass of PbCO3) / 1 L
Therefore, the solubility of PbCO3 in g/L can be calculated by substituting the values into the equation. However, to calculate the molar solubility (s), we need to know the stoichiometry of the compound. If you have that information, please provide it, and I can assist you further in obtaining the solubility value.