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charle [14.2K]
3 years ago
14

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)â½â

ââa+(aq)+bâ(aq)
Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0

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}

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