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* 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=(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+)` =
`(0.0252mol)/(1L)=0.0252M`

Concentration of
`B^-` =
`(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)* (0.0252M)^2`

`K_(sp)=1.60* 10^(-5)`

Thus, the value of
`K_(sp)` of the generic salt is,
`1.60* 10^(-5)`