Question

In: Chemistry

Part A. Given the two reactions: H2S⇌HS−+H+, K1 = 9.52×10−8, HS−⇌S2−+H+, K2 = 1.18×10−19, what is...

Part A. Given the two reactions:

H2S⇌HS−+H+, K1 = 9.52×10−8,

HS−⇌S2−+H+, K2 = 1.18×10−19, what is the equilibrium constant Kfinal for the following reaction? S2−+2H+⇌H2S

Enter your answer numerically.

Part B. Given the two reactions:

PbCl2⇌Pb2++2Cl−, K3 = 1.87×10⁻¹⁰

AgCl⇌Ag++Cl−, K4 = 1.22×10⁻⁴

what is the equilibrium constant Kfinal for the following reaction? PbCl2+2Ag+⇌2AgCl+Pb2+

Express your answer numerically.

Expert Solution

Part - A: The given equilibrium reactions are

H2S <-----> HS^- + H⁺ ; K1 = 9.52x10^-8 -------(1)

HS^- <-----> S^2- + H⁺ ; K2 = 1.18x10^-19 ------(2)

Adding equation(1) and (2) we get

H2S <-----> S^2- + 2H⁺ ; K' = K1xK2 = 9.52x10^-8x1.18x10^-19 = 1.12336x10^-26 ----- (3)

Now reversing the equation(3) we will get the desired equation.

S^2- + 2H⁺ ------ > H2S, K'' = 1 / K' = 1 / 1.12336x10^-26 = 8.90x10²⁵ (answer)

Part - B: The given equilibrium reactions are

PbCl2 <-----> Pb²⁺ + 2Cl^- ; K3 = 1.87x10^-10 -------(4)

AgCl <-----> Ag⁺ + Cl^- ; K4 = 1.22x10^-4 ------(5)

Adding equation(5) by 2 and then reversing the multiplied equation we get

2Ag⁺ + 2Cl^- <-----> 2AgCl ; K' = 1 / (1.22x10^-4 )² = 6.72x10⁷ ----- (6)

Now we will get the desired equation by adding equation (4) and equation (6)

PbCl2 + 2Ag⁺ + 2Cl^- <------> Pb²⁺ + 2Cl^- + 2AgCl

or PbCl2 + 2Ag⁺ <------> Pb²⁺ + 2AgCl; K" = K3 x K' = 1.87x10^-10 x 6.72x10⁷ = 1.26x10^-2 (answer)

queen_honey_blossom answered 2 years ago

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