Question

At 25 °C, only 0.0410 0.0410 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the K sp Ksp of the salt at 25 °C? AB 3 ( s ) − ⇀ ↽ − A 3 + ( aq ) + 3 B − ( aq ) AB3(s)↽−−⇀A3+(aq)+3B−(aq)

Answer 1

Answer: The solubility product of the given salt is $7.63\times 10^{-5}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.0410 mol

Volume of solution = 1.00 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{0.0410mol}{1.00L}=0.0410M$

The given chemical equation follows:

$AB_3(s)\rightleftharpoons A^{3+}(aq.)+3B^-(aq.)$

1 mole of the $AB_3$ salt produces 1 mole of $A^{3+}$ ions and 3 moles of $B^-$ ions

So, concentration of $A^{3+}\text{ ions}=(1\times 0.0410)M=0.0410M$

Concentration of $B^{-}\text{ ions}=(3\times 0.0410)M=0.123M$

Expression for the solubility product of  will be:

$K_{sp}=[A^{3+}][B^-]^$

Putting values in above equation, we get:

$K_{sp}=(0.0410)\times (0.123)^3\\\\K_{sp}=7.63\times 10^{-5}$

Hence, the solubility product of the given salt is $7.63\times 10^{-5}$