Equilibrium constant questions

Question

What is the numerical equilibrium constant of the following reaction at 25˚C? HCNO(aq) + RbOH(aq) ⇄ RbCNO(aq) + H₂O(l) Assume that RbCNO(aq) is a strong electrolyte.

For the following two I was hoping you could tell me whether I'm right or wrong.

What is the numerical equilibrium constant of the following reaction at 25˚C?<br /> AlPO₄(s) + 3NaCl(aq) ⇄ AlCl₃(aq) + Na₃PO₄(aq)

9.84x10^-21 because that is the ksp of AlPO4.

What is the numerical equilibrium constant of the following reaction at 25˚C?<br /> AlCl₃(aq) + K₃PO₄(aq) ⇄ 3KCl(aq) + AlPO₄(s)

1.02x10^-20 because you take 1/ ksp of AlPO4 which is 9.84x10^-21

Why is it that you take 1/ksp instead of just the ksp since the precipitate is in the products?

Answer 1

HCNO(aq) + RbOH(aq) <==> RbCNO(aq) + H₂O(l) and since RbOH is a strong base and dissociates 100%...

HCNO(aq) + OH^-(aq) <==> CNO^-(aq) + H₂O(l)

K' = [CNO^-]/[HCNO][OH^-]

multiplying numerator and denominator by [H⁺], we obtain...

K' = [H⁺][CNO^-]/[HCNO][OH^-][H⁺] = Ka/Kw so the Keq for the weak acid is Ka/Kw

K = 2x10^-4/1x10^-14 = 2x10¹⁰

To my way of thinking, you would be correct in your answers to the next two questions.

The ionic equation for the reaction is...

AlPO₄(s) + 3Na⁺(aq) + 3Cl^-(aq) <=> Al³⁺(aq) +3Cl^-(aq) + 3Na⁺(aq) + PO₄^3-(aq) and net ionic is...

AlPO₄(s) <==> Al³⁺(aq) + PO₄^3-(aq)

K' = [Al³⁺][PO₄^3-]/[AlPO₄] = [Al³⁺][PO₄^3-] = Ksp

You take 1/Ksp because the reaction as written, is the reverse of the original.