what is solubility, in mol/L, of MgCO3 in a 0.65 mol/L solution of MgCL2 if the ksp of MgCO3 is 2.5*10^-5
MgCO3 ==> Mg^2+ + CO3^2-
....x......x........x
where x = solubility in mols/L.
........MgCl2 --> Mg^2+ + 2Cl^-
I........0.65.....0..........0
C.......-0.65....0.65.....2*0.65
E .......0.........0.65....1.30
Substitute from the above ICE charts into Ksp for MgCO3 and solve for x.
(Mg^2+) = x+0.65 (that's x from MgCO3 and 0.65 from MgCl2.)
(CO3^2-) = x
To determine the solubility of MgCO3 in a 0.65 mol/L solution of MgCl2 using the given Ksp value, we need to follow these steps:
Step 1: Write the balanced chemical equation for the dissolution of MgCO3.
MgCO3(s) ⇌ Mg2+(aq) + CO3^2-(aq)
Step 2: Write the Ksp expression for MgCO3.
Ksp = [Mg2+][CO3^2-]
Step 3: Determine the molar solubility of MgCO3 by assuming it completely dissociates in the solution.
Let's assume the molar solubility of MgCO3 is 'x'.
Hence, [Mg2+] = x mol/L and [CO3^2-] = x mol/L.
Step 4: Substitute the molar concentrations into the Ksp expression.
Ksp = x * x = x^2
Step 5: Substitute the given Ksp value into the equation and solve for x.
2.5 × 10^-5 = x^2
Taking the square root of both sides:
√(2.5 × 10^-5) = x
x ≈ 0.005
So, the solubility of MgCO3 in a 0.65 mol/L solution of MgCl2 is approximately 0.005 mol/L.
To determine the solubility of MgCO3 in a solution of MgCl2, you need to consider the solubility product constant (Ksp) and use the concept of common ion effect.
The common ion effect states that the solubility of a slightly soluble salt is reduced when a common ion is present in the solution. In this case, both MgCO3 and MgCl2 contribute Mg2+ ions. Therefore, the presence of Mg2+ ions from MgCl2 will reduce the solubility of MgCO3.
The equation for the solubility product constant (Ksp) of MgCO3 is:
MgCO3 ⇌ Mg2+ + CO3^2-
The Ksp expression can be written as:
Ksp = [Mg2+][CO3^2-]
Given that the value of Ksp for MgCO3 is 2.5*10^-5, we can set up the following equation:
2.5*10^-5 = [Mg2+][CO3^2-]
Since the concentration of Mg2+ ions in the 0.65 mol/L MgCl2 solution is already known, we can substitute it into the equation:
2.5*10^-5 = (0.65 mol/L)[CO3^2-]
Rearranging the equation to solve for [CO3^2-]:
[CO3^2-] = (2.5*10^-5) / (0.65 mol/L)
Evaluating the expression:
[CO3^2-] ≈ 3.85*10^-5 mol/L
Therefore, the solubility of MgCO3 in a 0.65 mol/L solution of MgCl2, considering the common ion effect, is approximately 3.85*10^-5 mol/L.