Question

At 25 °c, only 0.0640 mol of the generic salt ab is soluble in 1.00 l of water. what is the ksp of the salt at 25 °c? ab(s)â½âââa+(aq)+bâ(aq)

Answer 1

Answer: $4.09\times 10^{-3}$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

The equation for the ionization of the  is given as:

$AB\rightarrow A^++B^-$

Molar concentration = $\frac{moles}{Volume}=\frac{0.0640}{1.00L}=0.0640M$

By stoichiometry of the reaction:

1 mole of  $AB$ gives 1 mole of $A^+$ and 1 mole of $B^-$

When the solubility of  $AB$ is S moles/liter, then the solubility of $A^+$  will be S moles\liter and solubility of $B^-$ will be S moles/liter.

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

$K_{sp}=[0.0640][0.0640]=4.09\times 10^{-3}$

Thus $K_{sp}$ of the salt at $25^0C$ is $4.09\times 10^{-3}$