If the Ksp value of MX,M2X and MX3 are same then find out the solubility order.?

Let s₁, s₂ and s₃ be the solubility of MX, M₂X and MX₃ respectively.

_______ MX(s) ⇌ Mⁿ⁺(aq) + Xⁿ⁻(aq) ___ Ksp
At eqm ________ s₁ M ____ s₁ M
Hence, Ksp = s₁²
Then, s₁ = √(Ksp)

_______ M₂X(s) ⇌ 2M⁺(aq) + X²⁻(aq) ___ Ksp
At eqm _________ 2s₂ M ____ s₂ M
Hence, Ksp = (2s₂)² * s₂ = 4s₂³
Then, s₂ = ∛(Ksp/4)

_______ MX₃(s) ⇌ M³⁺(aq) + 3X⁻(aq) ___ Ksp
At eqm _________ s₃ M ____ 3s₃ M
Hence, Ksp = s₃ * (3s₃)³ = 27s₃⁴
Then, s₃ = ∜(Ksp/27)

As Ksp is very small, i.e. Ksp ≪ 1,
then [s₃ = ∜(Ksp/27)] > [s₂ = ∛(Ksp/4)] > [s₁ = √(Ksp)]
Solubility : MX₃ > M₂X > MX

The answer : 2) MX₃ > M₂X > MX

...
let's take Ksp as equal to 1 for convenience
let solubility in mol/L be S

for MX
Ksp = [M] [X]
[M] and [X] are both S (1 to 1)
1 = S^2 so S = 1

for M2X
Ksp = [M]^2 [X]
[X] = S and [M] = 2S
1 = (2S)2 S = 4 S^3 so S = 0.630

for MX3
Ksp = [M] [X]^3
[M] = S and [X] = 3S
1 = S (3S)^3 = 27 S^4 so S = 0.439

so the correct order is choice (1)

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Ksp is very small. Therefore, taking Ksp = 1 would cause mistakes. For example, take small value of Ksp (e.g. 1 x 10^-6) would give an correct answer.