At 25 Â°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 Â°c? ab(s)â½âââa+(aq)+bâ(aq)

Answers

Answer 1

Answer:

Answer: $4.09* 10^(-3)$

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 = $(moles)/(Volume)=(0.0640)/(1.00L)=0.0640M$

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.

$K_(sp)=[A^(+)][B^(-)]$

$K_(sp)=[0.0640][0.0640]=4.09* 10^(-3)$

Thus $K_(sp)$ of the salt at $25^0C$ is $4.09* 10^(-3)$

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Answers

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Explanation:

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