The Ksp of nickel(II) carbonate, NiCO3, is 1.42×10−7. Calculate the molar solubility, S, of this compound.
To calculate the molar solubility (S) of nickel(II) carbonate (NiCO3) using its solubility product constant (Ksp), we need to set up an equilibrium expression and then solve for S.
The solubility product constant (Ksp) expression for a slightly soluble salt like nickel(II) carbonate is as follows:
Ksp = [Ni2+][CO3 2-]
In this equation, [Ni2+] represents the molar concentration of nickel(II) ions (in moles/L) in the solution at equilibrium, and [CO3 2-] represents the molar concentration of carbonate ions (in moles/L) in the solution at equilibrium.
Since nickel(II) carbonate dissociates into one Ni2+ ion and one CO3 2- ion, the equilibrium concentration of both ions is the same, and we can represent it as "S."
Therefore, the equilibrium expression for nickel(II) carbonate becomes:
Ksp = S * S
Now we can solve for S.
Rearranging the equation:
S^2 = Ksp
Taking the square root of both sides:
S = √(Ksp)
Substituting the given value of Ksp = 1.42×10^−7 into the equation, we can calculate the molar solubility:
S = √(1.42×10^−7)
S ≈ 1.19×10^−3 mol/L
Therefore, the molar solubility (S) of nickel(II) carbonate is approximately 1.19×10^−3 mol/L.
To calculate the molar solubility (S) of nickel(II) carbonate (NiCO3), we need to use the information given in the Ksp expression.
The Ksp expression for nickel(II) carbonate is:
NiCO3(s) ⇌ Ni2+(aq) + CO32-(aq)
The Ksp expression can be written as:
Ksp = [Ni2+][CO32-]
From the balanced equation, we can see that every 1 mole of nickel(II) carbonate dissociates into 1 mole of Ni2+ and 1 mole of CO32- ions:
NiCO3(s) ⇌ 1 Ni2+(aq) + 1 CO32-(aq)
Let's assume the molar solubility of nickel(II) carbonate is x mol/L. Since every 1 mole of NiCO3 dissociates into 1 mole of Ni2+ ions, the concentration of Ni2+ ions is also x mol/L. Similarly, the concentration of CO32- ions is also x mol/L.
Substituting these values into the Ksp expression, we get:
Ksp = x * x
1.42×10^(-7) = x^2
To solve for x, we take the square root of both sides of the equation:
√(1.42×10^(-7)) = x
Calculating this value, we find:
x ≈ 3.77×10^(-4) mol/L
So, the molar solubility (S) of nickel(II) carbonate is approximately 3.77×10^(-4) mol/L.
...........NiCO3 ==> Ni^2+ + CO3^2-
I..........solid......0.......0
C..........solid......S.......S
E..........solid......S.......S
Ksp = (Ni^2+) + (CO3^-)
Plug in the values from the ICE chart and calculate S in mols/L. .