Question

H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.06×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.18×10−19, what is the equilibrium constant Kfinal for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)

Answer 1

Answer:

Therefore the equilibrium constant  $K_{final}$ is 9.35× 10²⁵

Explanation:

Equilibrium constant:

Equilibrium constant is used to find out the ratio of the concentration of the product to that of reactant.

xA+yB→zC

The equilibrium constant K,

$k=\frac{[C]^z}{[A]^x[B]^y}$

Here [A] is equilibrium concentration of A

[B] is equilibrium concentration of B

[C] is equilibrium concentration of C.

1.  $H_2S(aq)\rightleftharpoons HS^-(aq)+H^+(aq)$          $K_1= 9.06 \times 10^{-8}$

$K_1=\frac{[HS^-][H^+]}{[H_2S]}$

2.   $HS^-(aq)\rightleftharpoons S^{2-}(aq)+H^+(aq)$      $K_2= 1.18 \times 10^{-19}$

$K_2=\frac{[S^{2-}][H^+]}{[HS^-]}$

3.

$S^{2-}(aq)+2H^+(aq)\rightleftharpoons H_2S(aq)$

$K_3=\frac{[H_2S]}{[S^{2-}][H^+]^2}$

Therefore,

$k_1k_2=\frac{[HS^-][H^+]}{[H_2S]}\frac{[S^{2-}][H^+]}{[HS^-]}$

$\Rightarrow k_1k_2=\frac{[S^{2-}][H^+]^2}{[H_2S]}$

$\Rightarrow k_1k_2=\frac{1}{K_3}$

$\Rightarrow K_3=\frac{1}{K_1K_2}$

$\Rightarrow K_3=\frac{1}{9.06\times 10^{-8}\times 1.18\times 10^{-19}}$

⇒K₃=9.35× 10²⁵

$K_3=K_{final}$          

Therefore the equilibrium constant  $K_{final}$ is 9.35× 10²⁵