Calculate the molar solubility of MX2, if the Ksp=3.3x10^-8
how can i solve this problem?
Hi roseanne,
First, we write out the equation when MX2 dissolves in water (or a suitable solvent):
MX2 --> (reversible arrow) M2+ + 2X-
So Ksp = [M2+][X-]^2
Solubility would refer to the concentration of an ion produced in 1:1 ratio with the original solid. In this case, we can take [M2+] to be the solubility since M2+ is in a 1:1 ratio with MX2. Note that we CANNOT use [X-] as the solubility since X- is in a 2:1 ratio with MX2.
So we let [M2+] = x
Now [X-] = 2x since ratio of M2+:X- = 1:2
So Ksp = (x)(2x)^2 = 4x^3 = 3.3X10^-8
Then x = solubility = 0.00202M.
Hope I helped! (:
-J
that helped a lot, thanks!
To calculate the molar solubility of MX2, we can use the stoichiometry of the compound in the equilibrium expression. The general form of the equilibrium expression for the dissolution of a sparingly soluble salt MX2 is:
MX2 ⇌ M^2+ + 2X^-
The solubility product constant (Ksp) expression for MX2 can be written as:
Ksp = [M^2+][X^-]^2
Given the value of Ksp for MX2 as 3.3x10^-8, we can assume that the compound fully dissociates in water, resulting in the following relationship:
[M^2+] = [X^-] = x
Substituting these assumptions into the Ksp expression, we get:
Ksp = x * (x)^2 = x^3
To solve for x, we can take the cube root of both sides:
x = (Ksp)^(1/3)
Now, we can substitute the given value of Ksp into the equation to find the molar solubility of MX2:
x = (3.3x10^-8)^(1/3)
Calculating this expression will give us the molar solubility of MX2.
To solve the problem and calculate the molar solubility of MX2, you will need to make use of the given Ksp value. The Ksp is the solubility product constant, and it represents the product of the ions' concentrations when the compound is in a saturated solution.
Let's assume that the formula for the compound MX2 represents one mole of M cations (M2+) and two moles of X anions (X-). Therefore, the dissociation equation for the compound can be written as:
MX2 ⇌ M^2+ + 2X^-
The expression for the solubility product constant (Ksp) can be written as:
Ksp = [M^2+][X^-]^2
Here, [M^2+] represents the concentration of M2+ ions, and [X^-] represents the concentration of X- ions in the saturated solution. Since the stoichiometry of the compound is 1:2, the concentration of X- ions is twice that of M2+ ions.
Now, let's solve the problem:
1. Write the expression for Ksp:
Ksp = [M^2+][X^-]^2 = 3.3x10^-8
2. Assuming that the molar solubility of MX2 is represented by s, the molar concentrations of M^2+ and X^- in terms of s would be:
[M^2+] = s
[X^-] = 2s
3. Substitute the concentrations into the expression for Ksp:
Ksp = [M^2+][X^-]^2 = (s)(2s)^2 = 4s^3 = 3.3x10^-8
4. Solve for s by rearranging the equation:
4s^3 = 3.3x10^-8
s^3 = (3.3x10^-8)/4
s^3 = 8.25x10^-9
s = ∛(8.25x10^-9)
5. Calculate the molar solubility, s, using a calculator:
s ≈ 0.00203
Therefore, the molar solubility of MX2 is approximately 0.00203 M.