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* 10^(-3)`

Step-by-step 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)`