Question

Given the two reactions

H2S⇌HS−+H+,   K1 = 9.24×10⁻⁸, and

HS−⇌S2−+H+,   K2 = 1.15×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−+2H+⇌H2S

Part B

Given the two reactions

PbCl2⇌Pb2++2Cl−,   K3 = 1.74×10⁻¹⁰, and

AgCl⇌Ag++Cl−,   K4 = 1.26×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

Answer 1

Answer – Part A) We are given the two reactions with equilibrium constant values and we need to calculate the equilibrium constant value for the third reaction. We know the when two reaction added their equilibrium constant value gets multiply to each other and when reaction gets reversed then equilibrium constant value gets inverse. When reaction multiply by 2 then equilibrium constant value become square.

We need the H₂S in the product side and S^2- in the reactant side ,so we need reverse the both reaction and added them

HS^- + H⁺ <----> H₂S , K’ = 1/ 9.24×10⁻⁸ = 1.08*10⁷

S^2- + H⁺ <----> HS^- , K” = 1/1.15×10⁻¹⁹ = 8.69*10¹⁸

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S^2- + 2H⁺ <----> H₂S, K = K’*K” = 9.41*10²⁵

Part B) In this one we need 2 AgCl in the product side, but it is in the reactant side in the second reaction and there is only AgCl, so we need to multiply by 2 and reverse the reaction second then added it in to the first reaction –

PbCl2 <-----> Pb2++2Cl−,   K3 = 1.74×10⁻¹⁰

2Ag⁺ + 2 Cl^- <-----> 2AgCl K4 = 6.30*10⁷

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PbCl₂+2Ag⁺ <-----> 2AgCl +Pb²⁺ , K = K3*K4 = 1.1*10^-2