Question

At 25 °c only 0.0260 mol of the generic salt ab3 is soluble in 1.00 l of water. what is the ksp of the salt at 25 °c?

Answer 1

Answer: The $K_{sp}$ of the given salt at 25°C is $1.23\times 10^{-5}$

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 $AB_3$ = 0.0260 mol

Volume of solution = 1.00 L

Putting values in above equation, we get:

$\text{Molarity of }AB_3=\frac{0.0260mol}{1L}=0.0260M$

The chemical equation for the ionization of $AB_3$ salt follows:

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

The expression of $K_{sp}$ for above reaction is written as:

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

We are given:

$[A^{3+}]=0.0260M$

$[B^-]=(3\times 0.0260)=0.078M$

Putting values in above expression, we get:

$K_{sp}=(0.0260)\times (0.078)^3\\\\K_{sp}=1.23\times 10^{-5}$

Hence, the $K_{sp}$ of the given salt at 25°C is $1.23\times 10^{-5}$

Answer 2

Given:

At 25 degrees Celcius:

amount of generic salt AB3 = 0.0260 moles
Volume of solvent = 1.0 L water

Set up a balanced chemical equation:

AB3 =========> A3+ + 3B-
0.0260M             0.0260   0.078

Ksp = [A][B]^3
Ksp = 1.23 x 10^-5<span />