I\'m kinda confused on how to do this Given the two reactions H_ 2S equivalence

Question

I'm kinda confused on how to do this

Given the two reactions H_ 2S equivalence HS^ - +H^ +, K_ 1 = 9.72 times 10^ -8, and HS^ - equivalence S^ 2- + H^ +, K_ 2 = 1.30 times 10^ -19 what is the equilibrium constant K_ final for the following reaction? Enter your answer numerically. Given the two reactions. Pb Cl_ 2 equivalence Pb^ 2+ + 2CI^ -, K_ 3 = 1.76 times 10^ -10, and AgCl equivalence Ag^+ + Cl^ -, K_ 4 = 1.29 times 10^ -4. What is the equilibrium constant K_ final for the following reaction? Express your answer numerically.

Explanation / Answer

For part A, we can first get the overall constant for the overall reaction:

H2S <-> 2H+ + S-2                  Kfinal = K1 * K2 = (9.72 x 10-8) * (1.3 x 10-19) = 1.2636 x 10-26

We flip the equation and constant to meet what's asked for:

2H+ + S-2 <-> H2S                 Kfinal = 1/(1.2636 x 10-26) = 7.9139 x 1025

For part B, we manipulate second equation, first, by multiplying it by 2:

AgCl <-> Ag+ + Cl-                         K4 = 1.29 x 10-4

2AgCl <-> 2Ag+ + 2Cl-                    K'4 = (1.29 x 10-4)2 = 1.6641 x 10-8

Then we turn it around:

2Ag+ + 2Cl- <-> 2AgCl                    K''4 = 1/(1.6641 x 10-8) = 60092542.52

Now we add both K3 and K''4 up, and get final constant:

PbCl2 + 2Ag+ <-> 2 AgCl + Pb+2      Kfinal = K3 * K''4 = 60092542.52 * (1.76 x 10-10) = 0.01058

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