Question

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.

Answer 1

Let's see

`\\ \rm\Rrightarrow [M_3X_3]\leftrightharpoons {[M^(3+)]\atop 3x}+{[X^(3-)]\atop 3x}`

So

As solubility and molar mass given

`\\ \rm\Rrightarrow [M_3X_3]=(Solubility)/(Molar\:mass)`

`\\ \rm\Rrightarrow x=(0.039)/(108.64)`

`\\ \rm\Rrightarrow x=0.00034M=3.4* 10^(-4)M`

So

`\\ \rm\Rrightarrow K_(sp)=(3x)(3x)=(3x)^2=9x²`

So

`\\ \rm\Rrightarrow K_(sp)=9(3.4* 10^(-4))^2=0.0000010404=104* 10^(-6)mol^2L^(-2)=104\mu mol^2L^(-2)`