Question

part a: Given the two reactions

1. H2S⇌HS−+H+, K1 = 9.22×10⁻⁸, and
2. HS−⇌S2−+H+, K2 = 1.01×10⁻¹⁹,

what is the equilibrium constant Kfinal for the following reaction?

S2−+2H+⇌H2S

part b: Given the two reactions

3. PbCl2⇌Pb2++2Cl−, K3 = 1.71×10⁻¹⁰, and
4. AgCl⇌Ag++Cl−, K4 = 1.13×10⁻⁴,

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+2Ag+⇌2AgCl+Pb2+

PbCl2+2Ag+⇌2AgCl+Pb2+

Answer 1

Answer – Part a) Given the reaction with equilibrium constant and we need to calculate the equilibrium constant for the third reaction. We need to use Hess’s law and we know when two reaction added then equilibrium constant gets multiplied. When reaction get reversed then, equilibrium constant gets inversed. When we multiply by 2 to reaction then equilibrium constant gets square and when we divvied by 2 then there is equilibrium constant gets square root.

H₂S <-----> HS⁻+H⁺,   K1 = 9.22×10⁻⁸ ……1

HS^- <-----> S²⁻+H⁺,   K2 = 1.01×10⁻¹⁹ …….2

Now we need to reverse the both reaction and then added as follow –

HS⁻+H⁺ <-----> H₂S    K^’ = 1.1*10⁷

S²⁻+H⁺ <-----> HS^-     K^’’_­ = 9.9*10^-18

S²⁻+2H⁺ <-----> H₂S   K_final = 1.07*10²⁶

Part b)

PbCl2 <----> Pb²⁺+2Cl^−,   K₃ = 1.71×10⁻¹⁰ ……1

AgCl<----> Ag⁺+Cl⁻,   K₄ = 1.13×10⁻⁴_­………..2

Multiply the reaction number 3 and then reverse it

PbCl2 <----> Pb²⁺+2Cl^−,   K₃ = 1.71×10⁻¹⁰

2 Ag⁺+ 2 Cl⁻ <----> 2 AgCl , K^’ = 7.8*10⁷

PbCl₂+2Ag⁺ <---->2AgCl + Pb²⁺ K_final = 1.34*10^-2