Question

At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−⇀A3+(aq)+3B−(aq)

Answer 1

Answer:

$Ksp=2.59x10^{-3}$

Explanation:

Hello,

In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:

$AB_3\rightleftharpoons A^{3+}+3B^-$

The concentrations of the A and B ions in the solution are:

$[A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M$

$[B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M$

Then, as the solubility product is defined as:

$Ksp=[A][B]^3$

Due to the given dissociation, it turns out:

$Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}$

Regards.