Calculate the molar solubility of MX2, if the Ksp=3.3x10^-8
how can i solve this problem?
To solve this problem, you can follow these steps:
Step 1: Write the balanced equation for the dissociation of the MX2 compound into its ions. If the compound MX2 dissociates into M2+ and 2X- ions, the balanced equation would be:
MX2 ↔ M2+ + 2X-
Step 2: Write the expression for the solubility product constant (Ksp) using the concentrations (in molar) of the ions. In this case, since the molar solubility of MX2 is defined as the number of moles of MX2 that dissolve in one liter of solution, let's assume that 's' is the molar solubility of MX2. Therefore, the concentrations of M2+ and X- ions will be 2s and 2s, respectively. Finally, the expression for Ksp will be:
Ksp = [M2+] * [X-]^2 = (2s) * (2s)^2 = 4s^3
Step 3: Substitute the given Ksp value into the equation and solve for 's':
3.3x10^-8 = 4s^3
Step 4: Solve the equation for 's' by taking the cube root of both sides:
s = (3.3x10^-8 / 4)^(1/3)
Step 5: Calculate the numerical value of 's' using a calculator:
s ≈ 0.00628 or 6.28x10^-3 M
Therefore, the molar solubility (s) of MX2 is approximately 0.00628 M or 6.28x10^-3 M.