The solubility of nickel (II) carbonate at 25 degrees C is 0.047 g/L. Calculate Ksp for nickel (II) carbonate.
To calculate the solubility product constant (Ksp) for nickel (II) carbonate (NiCO3), we need to first write the balanced chemical equation for the dissociation of the compound in water.
The balanced equation for the dissociation of nickel (II) carbonate is as follows:
NiCO3(s) ⇌ Ni2+(aq) + CO3^2-(aq)
The solubility of nickel (II) carbonate, given in the question, is 0.047 g/L. This means that for every liter of water, 0.047 grams of nickel (II) carbonate will dissolve.
To determine the concentration of nickel ions (Ni2+) in solution, we need to convert the solubility from grams per liter to moles per liter. We can do this by using the molar mass of nickel (II) carbonate, which is approximately 118.71 g/mol.
First, calculate the number of moles of NiCO3 in 0.047 g:
Number of moles = mass / molar mass
Number of moles of NiCO3 = 0.047 g / 118.71 g/mol
Next, since NiCO3 dissociates to form one mole of Ni2+ ions, the concentration of Ni2+ in solution is equal to the number of moles of NiCO3.
Therefore, the concentration of Ni2+ ions is 0.047 mol/L.
Similarly, since NiCO3 dissociates to form one mole of CO3^2- ions, the concentration of CO3^2- in solution is also 0.047 mol/L.
Finally, to calculate the solubility product constant (Ksp), we need to multiply the concentrations of the nickel and carbonate ions.
Ksp = [Ni2+][CO3^2-]
= (0.047 mol/L)(0.047 mol/L)
Therefore, the Ksp for nickel (II) carbonate at 25 degrees Celsius is the square of 0.047 mol/L, which equals 0.002209 mol^2/L^2.
NiCO3 ==> Ni^+2 + CO3^-2
Ksp = (Ni^+2)(CO3^-2)
Convert solubility opf 0.047 g/L to moles/L, insert M into the Ksp expression and solve for Ksp.