Respuesta :
Answer : The value of [tex]K_{sp}[/tex] of the generic salt is, [tex]1.60\times 10^{-5}[/tex]
Explanation :
As we are given that, a solubility of salt is, 8.70 g/L that means 8.70 grams of salt present in 1 L of solution.
First we have to calculate the moles of salt [tex](AB_2)[/tex]
[tex]\text{Moles of }AB_2=\frac{\text{Mass of }AB_2}{\text{Molar mass of }AB_2}[/tex]
Molar mass of [tex]AB_2[/tex] = 345 g/mol
[tex]\text{Moles of }AB_2=\frac{8.70g}{345g/mol}=0.0252mol[/tex]
Now we have to calculate the concentration of [tex]A^{2+}\text{ and }B^-[/tex]
The equilibrium chemical reaction will be:
[tex]AB_2(s)\rightleftharpoons A^{2+}(aq)+2B^-(aq)[/tex]
Concentration of [tex]A^{2+}[/tex] = [tex]\frac{0.0252mol}{1L}=0.0252M[/tex]
Concentration of [tex]B^-[/tex] = [tex]\frac{0.0252mol}{1L}=0.0252M[/tex]
The solubility constant expression for this reaction is:
[tex]K_{sp}=[A^{2+}][B^-]^2[/tex]
Now put all the given values in this expression, we get:
[tex]K_{sp}=(0.0252M)\times (0.0252M)^2[/tex]
[tex]K_{sp}=1.60\times 10^{-5}[/tex]
Thus, the value of [tex]K_{sp}[/tex] of the generic salt is, [tex]1.60\times 10^{-5}[/tex]
