If 0.075 g of MgF2 dissolves per liter of solution, calculate the solubility-product constant.
I got 2.0e-9 but it was not correct.
If you will post your work I will find the error. The correct answer is 6.9787E-9 which rounds to 7.0E-9 to two s.f.
To calculate the solubility-product constant (Ksp), we need to know the concentrations of the dissolved species in the solution. In this case, we are given the solubility of MgF2 as 0.075 g per liter of solution.
To begin, we need to convert the mass of MgF2 to moles. The molar mass of MgF2 is calculated as follows:
Molar mass of MgF2 = (molar mass of Mg) + 2 * (molar mass of F)
= (24.31 g/mol) + 2 * (19.00 g/mol)
= 62.31 g/mol
Next, we can convert the mass of MgF2 to moles:
Number of moles of MgF2 = mass of MgF2 / molar mass of MgF2
= 0.075 g / 62.31 g/mol
= 0.0012038 mol
Since one mole of MgF2 produces one mole of Mg2+ and two moles of F- ions in solution, the concentration of Mg2+ and F- in the solution is also given by 0.0012038 mol.
The solubility-product constant (Ksp) expression for MgF2 is given as:
MgF2 ⇌ Mg2+ + 2F-
Ksp = [Mg2+][F-]^2
Plugging in the concentrations, we get:
Ksp = (0.0012038)(0.0012038)^2
= 1.638 × 10^-9
Therefore, the solubility-product constant (Ksp) for MgF2 is 1.638 × 10^-9.