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At 25°C, an aqueous solution has an equilibrium concentration of 0.00331 M for a generic cation, [tex]\( A^+(aq) \)[/tex], and 0.00331 M for a generic anion, [tex]\( B^-(aq) \)[/tex].

What is the equilibrium constant, [tex]\( K_{sp} \)[/tex], of the generic salt [tex]\( AB(s) \)[/tex]?

[tex]\[ K_{sp} = \square \times 10^y \][/tex]



Answer :

GinnyAnswer
To solve for the equilibrium constant, [tex]\( K_{sp} \)[/tex], of the generic salt [tex]\( AB(s) \)[/tex], we need to understand and apply the relationship between the ion concentrations at equilibrium and the solubility product constant. The solubility product constant, [tex]\( K_{sp} \)[/tex], is given by the product of the molar concentrations of the ions in a saturated solution.

Given:
- The concentration of the cation [tex]\( A^+ \)[/tex] in the solution is [tex]\( 0.00331 \, M \)[/tex].
- The concentration of the anion [tex]\( B^- \)[/tex] in the solution is [tex]\( 0.00331 \, M \)[/tex].

The equilibrium expression for the solubility product of the salt [tex]\( AB \)[/tex] dissociating in water is:
[tex]\[ AB(s) \rightleftharpoons A^+(aq) + B^-(aq) \][/tex]

The [tex]\( K_{sp} \)[/tex] expression for this equilibrium is:
[tex]\[ K_{sp} = [A^+][B^-] \][/tex]

Substituting the given ion concentrations into this expression:
[tex]\[ K_{sp} = (0.00331 \, M) \times (0.00331 \, M) \][/tex]

By performing the multiplication:
[tex]\[ K_{sp} = 0.00331 \times 0.00331 \][/tex]

This results in:
[tex]\[ K_{sp} = 1.09561 \times 10^{-5} \][/tex]

Therefore, the equilibrium constant [tex]\( K_{sp} \)[/tex] of the generic salt [tex]\( AB(s) \)[/tex] is:
[tex]\[ K_{sp} = 1.09561 \times 10^{-5} \][/tex]

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