Calculate the molar solubility of calcium carbonate in 0.10 M Ca(NO3)2. (Ksp= 2.8 x 10-8)
CaCO3 ==> Ca^+2 + CO3^-2
Ksp = (Ca^+2)(CO3^-2)
Let S = solubility CaCO3, then
Ca^+2 comes from two places. It is S from CaCO3 and it is 0.1 M from Ca(NO3)2 so total Ca^+2 concn is S+0.01. Finally, S = concn CO3^-2
Solve for S which is solubility CaCO3.
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To calculate the molar solubility of calcium carbonate, we need to find the equilibrium concentration of calcium and carbonate ions in the solution.
The balanced equation for the dissolution of calcium carbonate is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
Let's assume the molar solubility of calcium carbonate is represented as "x" in mol/L.
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
Initial Conc: 0 0 0
Change in Conc: x x x
Equilibrium Conc: x x x
Since calcium nitrate (Ca(NO3)2) dissociates to provide Ca2+ ions, the concentration of Ca2+ ions in the equilibrium is 0.10 M.
Using the solubility product constant (Ksp) expression, Ksp = [Ca2+][CO3^2-]
We can substitute the equilibrium concentrations for [Ca2+] and [CO3^2-] into the Ksp expression:
Ksp = (x)(x) = x^2 = 2.8 x 10^-8
Now we can solve for "x" to find the molar solubility of calcium carbonate.
Taking the square root of both sides:
√(x^2) = √(2.8 x 10^-8)
x = 1.67 x 10^-4 M
Therefore, the molar solubility of calcium carbonate in 0.10 M Ca(NO3)2 is 1.67 x 10^-4 M.
To calculate the molar solubility of calcium carbonate in 0.10 M Ca(NO3)2, we need to use the solubility product constant (Ksp) of calcium carbonate.
The chemical equation for the dissolution of calcium carbonate in water is:
CaCO3(s) ⇌ Ca2+(aq) + CO3^2-(aq)
The equilibrium expression for this dissolution reaction is as follows:
Ksp = [Ca2+][CO3^2-]
We know the value of Ksp, which is 2.8 x 10^(-8).
Since we are given the concentration of Ca(NO3)2, which is 0.10 M, we can assume that the concentration of Ca2+ is also 0.10 M, since one mole of Ca(NO3)2 dissociates into one mole of Ca2+ and two moles of NO3^-.
Let's assume that the molar solubility of calcium carbonate is represented by S.
From the balanced equation, we know that every 1 mole of CaCO3 produces 1 mole of Ca2+ and 1 mole of CO3^2-. Thus, at equilibrium, the concentration of Ca2+ is S M, and the concentration of CO3^2- is also S M.
Now, substitute these values into the Ksp expression:
Ksp = (S)(S) = S^2
Since Ksp = 2.8 x 10^(-8), we can solve for S:
S^2 = 2.8 x 10^(-8)
Taking the square root of both sides:
S ≈ 1.67 x 10^(-4) M
Therefore, the molar solubility of calcium carbonate in 0.10 M Ca(NO3)2 is approximately 1.67 x 10^(-4) M.