Question

A generic salt, AB 2 , has a molar mass of 345 g/mol and a solubility of 8.70 g/L at 25 °C. AB 2 (s) − ⇀ ↽ − A 2 + (aq) + 2B − (aq) What is the K sp of this salt at 25 °C

Answer 1

Answer : The value of $K_{sp}$ of the generic salt is, $1.60\times 10^{-5}$

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 $(AB_2)$

$\text{Moles of }AB_2=\frac{\text{Mass of }AB_2}{\text{Molar mass of }AB_2}$

Molar mass of $AB_2$ = 345 g/mol

$\text{Moles of }AB_2=\frac{8.70g}{345g/mol}=0.0252mol$

Now we have to calculate the concentration of $A^{2+}\text{ and }B^-$

The equilibrium chemical reaction will be:

$AB_2(s)\rightleftharpoons A^{2+}(aq)+2B^-(aq)$

Concentration of $A^{2+}$ = $\frac{0.0252mol}{1L}=0.0252M$

Concentration of $B^-$ = $\frac{0.0252mol}{1L}=0.0252M$

The solubility constant expression for this reaction is:

$K_{sp}=[A^{2+}][B^-]^2$

Now put all the given values in this expression, we get:

$K_{sp}=(0.0252M)\times (0.0252M)^2$

$K_{sp}=1.60\times 10^{-5}$

Thus, the value of $K_{sp}$ of the generic salt is, $1.60\times 10^{-5}$