If you place 5 mg NiCO3 in 1 L of pure water, does all the salt dissolve? If not, how much salt remains undissolved? The Ksp for NiCO3 is 6.6x10^-9 M.
........NiCO3 ==> Ni^2+ + CO3^2-
I.......solid.....0........0
C.......solid.....x........x
E.......solid.....x........x
Substitute the E line into the Ksp expression and solve for x = (NiCO3) in mols/L.
Then grams = mols x molar mass = ?
Compare the solubility with the 5 mg you have.
To determine if all the NiCO3 salt will dissolve in water, we need to compare the value of the solubility product constant (Ksp) for NiCO3 with the initial concentration of the salt in the solution.
The expression for the solubility product constant is: Ksp = [Ni2+][CO32-]
Given the Ksp value for NiCO3 as 6.6x10^-9 M, we can assume that at equilibrium, [Ni2+] and [CO32-] are both equal to x M (since 1 mole of NiCO3 forms 1 mole of Ni2+ and 1 mole of CO32-).
Therefore, at equilibrium, the concentration of Ni2+ and CO32- ions will be x M, and the expression for Ksp becomes: Ksp = x * x = x^2
Solving for x, we have: x^2 = Ksp
x = sqrt(Ksp)
Now, substitute the given Ksp value into the equation: x = sqrt(6.6x10^-9 M)
x ≈ 8.124x10^-5 M
Since the solubility of NiCO3 is approximately 8.124x10^-5 M, if we place 5 mg (0.005 g) of NiCO3 in 1 L of pure water, the maximum amount of NiCO3 that can dissolve is determined by its solubility.
Using the molar mass of NiCO3 (58.7 g/mol), we can calculate the number of moles of NiCO3 in 0.005 g:
moles of NiCO3 = 0.005 g / 58.7 g/mol ≈ 8.516x10^-5 mol
Therefore, the amount of NiCO3 that will dissolve is limited by its solubility, which is approximately 8.124x10^-5 M or 8.124x10^-5 mol/L.
So, all 0.005 g of NiCO3 will dissolve since the solubility is greater than the initial concentration of the salt.