What is the value for the equilibrium constant for the following reaction?

Question

Reactions of cadmium with thiosulfate anion:

Cd^2+(aq) + S2O3^2-(aq) <==> Cd(S2O3)(aq); K1 = 8.3 x 10^3

Cd(s2O3)(aq) + S2O3^2-(aq) <==> Cd(S2O3)2^2-(aq); K2 = 2.5 x 10^2

What is the equilibrium constant for the following reaction?

Cd^2+(aq) + 2S2O3^2-(aq) <==> Cd(S2O3)2^2-(aq)

A.) 0.030

B.) 33

C.) 8.1 x 10^3

D.) 8.6 x 10^3

E.) 2.1 x 10^6

Answer 1

(1) Cd^2+ &amp; S2O3^2- &lt;==&gt; Cd(S2O3)     K1 =8.3e3

(2) Cd(s2O3) &amp;  S2O3^2-&lt;==&gt; Cd(S2O3)2^2-   K2 = 2.5 e2

(3) Cd^2+   &amp;     2 S2O3^2- &lt;==&gt; Cd(S2O3)2^2-  K3 = ?

======================================...

K1 times K2   gives   your K = 2.1e6

your answer is :E.) 2.1 x 10^6

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K1 = [Cd(S2O3)] / [Cd^2+] [S2O3^2]

K2 = [ Cd(S2O3)2^2-] / [Cd(s2O3)] [S2O3^2-]

when you multiply K1&#039;s ratio by K2&#039;s ratio:

[Cd(S2O3)] / [Cd^2+] [S2O3^2] times [ Cd(S2O3)2^2-] / [Cd(s2O3)] [S2O3^2-]

everything cancels out except for what you wanted :

K3 = [Cd(S2O3)2^2-] / [Cd^2+] [S2O3^2-]^2

so  K3 = (K1) (K2)