Question

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), H2S(aq)⇌HS−(aq)+H+(aq), K1K1K_1 = 9.39×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), HS−(aq)⇌S2−(aq)+H+(aq), K2K2K_2 = 1.45×10−19, what is the equilibrium constant KfinalKfinalK_final for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)

Answer 1

Answer: The value of $K_{final}$ for the net reaction is $7.34\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1:  $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^+(aq.);K_1$

Equation 2:  $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^+(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_{final}$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_{final}=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:  

$K_1=9.39\times 10^{-8}$

$K_2=1.45\times 10^{-19}$

Putting values in above equation, we get:

$K_{final}=\frac{1}{(9.39\times 10^{-8})}\times \frac{1}{(1.45\times 10^{-19})}=7.34\times 10^{25}$

Hence, the value of $K_{final}$ for the net reaction is $7.34\times 10^{25}$