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

`Ksp=2.59x10^(-3)`

Step-by-step 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 (molAB_3)/(L)*(1molA)/(1molAB_3) =0.0099M`

`[B]=0.099 (molAB_3)/(L)*(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.