Calculate the molar solubility of MgCO3 in a solution that already contains 1.7×10−3 M sodium carbonate.
MgCO3 Ksp = 6.82 x 10-6
IMPORTANT NOTE: You must use the quadratic equation to solve this problem.
(DrBob222- thank you so much for all your help!)
.........MgCO3 ==>Mg^2+ + CO3^2-
I........solid....0.......0.0017
C.......for x.....x.........x
E.......solid.....x.........x+0.0017
6.82E-6 = (x)(0.0017+x)
Solve for x = solubility in mols/L.
To calculate the molar solubility of MgCO3 in a solution containing sodium carbonate, we need to use the given solubility product constant (Ksp) and the concentration of sodium carbonate.
The solubility product constant (Ksp) for MgCO3 is 6.82 x 10^-6.
The balanced equation for the dissolution of MgCO3 is:
MgCO3(s) ⇌ Mg2+(aq) + CO3 2-(aq)
Let's assume the molar solubility of MgCO3 is "s".
Then the equilibrium concentration of Mg2+ would be "s" and the carbonate ion (CO32-) concentration would be 2s (according to the stoichiometry of the balanced equation).
However, we're given that the solution already contains 1.7 x 10^-3 M sodium carbonate. This means that at equilibrium, the concentration of the carbonate ion would be 1.7 x 10^-3 M (since sodium carbonate dissociates completely to produce carbonate ions).
So, we have the following concentrations at equilibrium:
[Mg2+] = s
[CO32-] = 2s = 1.7 x 10^-3
Now we can write the expression for the solubility product constant (Ksp) and substitute the equilibrium concentrations:
Ksp = [Mg2+][CO32-] = s * (2s) = 2s^2 = 6.82 x 10^-6
Now we can solve the quadratic equation 2s^2 = 6.82 x 10^-6 for the molar solubility, "s".
To calculate the molar solubility of MgCO3 in a solution that already contains 1.7×10−3 M sodium carbonate, we need to use the concept of the solubility product constant (Ksp) and apply the following steps:
Step 1: Write the balanced chemical equation for the dissociation of magnesium carbonate (MgCO3):
MgCO3(s) ⇌ Mg2+(aq) + CO32-(aq)
Step 2: Write the expression for the equilibrium constant (Ksp) using the concentrations of the products raised to the power of their stoichiometric coefficients:
Ksp = [Mg2+][CO32-]
Step 3: Since the sodium carbonate will react to produce carbonate ions, [CO32-] in the expression is already known as 1.7×10−3 M.
Step 4: Let's assume the molar solubility of MgCO3 is "x" M. This will be the concentration of magnesium ions ([Mg2+]) as well.
Step 5: Substitute the known values into the expression for the equilibrium constant (Ksp):
Ksp = (x)(1.7×10−3)
Step 6: Rearrange the equation to solve for "x":
x = Ksp / (1.7×10−3)
Step 7: Calculate the value for "x" to find the molar solubility of MgCO3.
By substituting the given Ksp value of 6.82 x 10-6 and the sodium carbonate concentration of 1.7×10−3 M into the equation, we have:
x = (6.82 x 10-6) / (1.7×10−3)
Evaluating this expression will give us the molar solubility of MgCO3 in the given solution.