Question

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00343M for a generic cation, A+(aq), and 0.00343M for a generic anion, B−(aq). What is the equilibrium constant, sp, of the generic salt AB(s)?

Answer 1

`\:\:\:\:\:\:\star`For the general solubility equilibrium
`\sf \underline{AB \longrightarrow A^+ + B^-}`the solubility product has the following expression-

`\:\:\:\:\:\:\star\longrightarrow \sf\underline{K_((sp)) = [A^+] * [B^-]}\\`

As per question, we are given that-

• Equilibrium concentration for generic cation,
  `\sf [A^+]`= 0.00343M

• Equilibrium concentration for generic anion,
  `\sf [B^-]`= 0.00343M

`\:\:\:\:\:\:\star` Now that we have all the required values, so we can substitute these values into the Ksp expression and solve for Ksp-

`\:\:\:\:\:\:\star\longrightarrow \sf\underline{K_((sp)) = [A^+] * [B^-]}\\`

`\:\:\:\:\:\:\longrightarrow \sf K_((sp)) = 0.00343 \:M* 0.00343\:M\\`

`\:\:\:\:\:\:\longrightarrow \sf K_((sp)) = 0.00343 \:molL^(-1)* 0.00343\:molL^(-1)\\`

`\:\:\:\:\:\:\longrightarrow \sf \underline{K_((sp)) = 1.17649* 10^(-5) \: mol^2L^(-2)}\\`

• Hence, the equilibrium constant(Ksp) of the generic salt AB(s) is
  `\sf\underline{\boxed{\sf1.17649* 10^(-5) \: mol^2L^(-2)}}.\\`