Equilibrium Constant

Based on the reactions and equilibrium constants given below, determine the value of the equilibrium constant (Kf) for the formation of the soluble PbI42- complex:

𝑃𝑏𝐼2(𝑠)𝑃𝑏2+(𝑎𝑞)+2𝐼(𝑎𝑞)𝐾𝑠𝑝=7.9×109𝑃𝑏𝐼2(𝑠)+2𝐼(𝑎𝑞)𝑃𝑏𝐼42(𝑎𝑞)𝐾=2.37×104𝑃𝑏2+(𝑎𝑞)+4𝐼(𝑎𝑞)𝑃𝑏𝐼42(𝑎𝑞)𝐾𝑓=?\begin{alignedat}{} &𝑃𝑏𝐼_2 (𝑠) ⇌𝑃𝑏^{2+}(𝑎𝑞)+2 𝐼^− (𝑎𝑞) &&𝐾_{𝑠𝑝}=7.9×10^{−9}\\ &𝑃𝑏𝐼_2(𝑠)+2 𝐼^− (𝑎𝑞) ⇌𝑃𝑏𝐼_4^{2−} (𝑎𝑞) &&𝐾=2.37×10^{−4}\\ &𝑃𝑏^{2+} (𝑎𝑞)+4 𝐼^− (𝑎𝑞)⇌𝑃𝑏𝐼_4^{2−} (𝑎𝑞) &&𝐾_𝑓= ? \end{alignedat}
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