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Given the following equilibria:

PbCl2(aq) ⇌ Pb²⁺(aq) + 2Cl⁻(aq), K3 = 1.89×10⁻¹⁰
AgCl(aq) ⇌ Ag⁺(aq) + Cl⁻(aq), K4 = 1.19×10⁻⁴
What is the equilibrium constant Kfinal for the following reaction?
PbCl2(aq) + 2Ag⁺(aq) ⇌ 2AgCl(aq) + Pb²⁺(aq)

User Zotta
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1 Answer

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Final answer:

The equilibrium constant Kfinal for the reaction PbCl2(aq) + 2Ag⁺(aq) ⇌ 2AgCl(aq) + Pb²⁺(aq) is calculated using the formula Kfinal = K3 / K4², resulting in a value of 1.334×10⁻².

Step-by-step explanation:

To calculate the equilibrium constant (Kfinal) for the reaction PbCl2(aq) + 2Ag⁺(aq) ⇌ 2AgCl(aq) + Pb²⁺(aq), we need to use the individual equilibrium constants given for separate reactions and apply the principle of combining equilibrium constants.

The equilibrium constant for the dissociation of PbCl2 is K3 = 1.89×10⁻¹⁰, and for the dissociation of AgCl, it is K4 = 1.19×10⁻⁴. When inverse processes are combined, as seen in the desired reaction, the overall equilibrium constant is found by dividing the individual constants for the forward reactions. The reaction involving Ag⁺ ions is the reverse of the given AgCl equilibrium, so we invert that equilibrium constant (1/K4).

The combined equilibrium constant is then calculated by multiplying the constants for forward processes and dividing by the constants for reverse processes. The formula for this would be Kfinal = K3 / K4². Squaring K4 is because the stoichiometry for Ag⁺ in the final equation involves two moles of AgCl for one mole of PbCl2.

Calculation

Kfinal = (1.89×10⁻¹⁰) / (1.19×10⁻⁴)² = (1.89×10⁻¹⁰) / (1.4161×10⁻⁸) = 1.334×10⁻²

Therefore, the equilibrium constant for the reaction PbCl2(aq) + 2Ag⁺(aq) ⇌ 2AgCl(aq) + Pb²⁺(aq) is 1.334×10⁻².

User Bartonjs
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