Given the two reactions H_2S rlhar + H^+, K_1 = 9.85 times 10^-8, and HS^- rlhar

Question

Given the two reactions H_2S rlhar + H^+, K_1 = 9.85 times 10^-8, and HS^- rlhar S^2- + H^+, K_2 = 1.27 times 10^-19, what is the equilibrium constant K_final for the following reaction? S^2- + 2H^+ rlhar H_2S Enter your answer numerically. K_final = __________ Given the two reactions PbCl_2 rlhar Pb^2+ +2Cl^-, K_3 = 1.76 times 10^-10, and AgCl rlhar Ag^+ + Cl^-, K_4 = 1.14 times 10^-4, what is the equilibrium constant K_final for the following reaction? PbCl_2 + 2Ag^+ rlhar 2AgCl + Pb^2+ Express your answer numerically.

Explanation / Answer

Part A -

H2S HS + H+,   K1 = 9.85x108,

HS S2 + H+,   K2 = 1.27×1019,

Adding the above two equations, we get -

H2S     S2   + 2H+,

We know that when the two equation are added, their equilibrium constants get multiplied.

So, K = K1 * K2

= 9.85x10^8 * 1.27x10^19

= 1.25x10^-26

Now, reversing the equation we get,

S2   + 2H+       H2S

K final = 1 / K

= 1 / (1.25x10^-26 )

K final = 8x10^25

Part B -

PbCl2 Pb2+ + 2Cl, K3 = 1.76x10^10, .......(1)

AgCl Ag+ + Cl,   K4 = 1.14x10^4,

Multiplying equation 2 we get,

2AgCl 2Ag+   + 2Cl,   

K' = (K4)2 = (1.14x10^4)2

K' = 1.29x10-8

Now reversing the resulted equation we get,

2Ag+   + 2Cl        2AgCl ...........(2)

K = 1/ K'

= 1 / 1.29x10-8

K = 7.75x107

Now adding equation (1) and (2),

PbCl2 +   2Ag+        2AgCl +     Pb2+  

K final = K3 * K

= 1.76x10^10 * 7.75x10^7

K final = 1.36x10^-2

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