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The compound M3X3 has a solubility of 0.039 g/L and a molar mass, ℳ = 108.64 g/mol. Calculate the Ksp for this compound.
Report your answer to TWO significant figures. You may (but are not required to) enter your answer in scientific notation; for example, if your answer is 2.4 × 10−2, you would enter 2.4E-2. The E must be uppercase.

Respuesta :

Let's see

[tex]\\ \rm\Rrightarrow [M_3X_3]\leftrightharpoons {[M^{3+}]\atop 3x}+{[X^{3-}]\atop 3x}[/tex]

So

As solubility and molar mass given

[tex]\\ \rm\Rrightarrow [M_3X_3]=\dfrac{Solubility}{Molar\:mass}[/tex]

[tex]\\ \rm\Rrightarrow x=\dfrac{0.039}{108.64}[/tex]

[tex]\\ \rm\Rrightarrow x=0.00034M=3.4\times 10^{-4}M[/tex]

So

[tex]\\ \rm\Rrightarrow K_{sp}=(3x)(3x)=(3x)^2=9x²[/tex]

So

[tex]\\ \rm\Rrightarrow K_{sp}=9(3.4\times 10^{-4})^2=0.0000010404=104\times 10^{-6}mol^2L^{-2}=104\mu mol^2L^{-2}[/tex]