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

Therefore the equilibrium constant
`K_(final)` is 9.35× 10²⁵

Step-by-step 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=([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 * 10^(-8)`

`K_1=([HS^-][H^+])/([H_2S])`

2.
`HS^-(aq)\rightleftharpoons S^(2-)(aq)+H^+(aq)`
`K_2= 1.18 * 10^(-19)`

`K_2=([S^(2-)][H^+])/([HS^-])`

3.

`S^(2-)(aq)+2H^+(aq)\rightleftharpoons H_2S(aq)`

`K_3=([H_2S])/([S^(2-)][H^+]^2)`

Therefore,

`k_1k_2=([HS^-][H^+])/([H_2S])([S^(2-)][H^+])/([HS^-])`

`\Rightarrow k_1k_2=([S^(2-)][H^+]^2)/([H_2S])`

`\Rightarrow k_1k_2=(1)/(K_3)`

`\Rightarrow K_3=(1)/(K_1K_2)`

`\Rightarrow K_3=(1)/(9.06* 10^(-8)* 1.18* 10^(-19))`

⇒K₃=9.35× 10²⁵

`K_3=K_(final)`

Therefore the equilibrium constant
`K_(final)` is 9.35× 10²⁵