Answer:
Therefore the equilibrium constant
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}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC%5D%5Ez%7D%7B%5BA%5D%5Ex%5BB%5D%5Ey%7D)
Here [A] is equilibrium concentration of A
[B] is equilibrium concentration of B
[C] is equilibrium concentration of C.
1.

![K_1=\frac{[HS^-][H^+]}{[H_2S]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BHS%5E-%5D%5BH%5E%2B%5D%7D%7B%5BH_2S%5D%7D)
2.

![K_2=\frac{[S^{2-}][H^+]}{[HS^-]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BS%5E%7B2-%7D%5D%5BH%5E%2B%5D%7D%7B%5BHS%5E-%5D%7D)
3.

![K_3=\frac{[H_2S]}{[S^{2-}][H^+]^2}](https://tex.z-dn.net/?f=K_3%3D%5Cfrac%7B%5BH_2S%5D%7D%7B%5BS%5E%7B2-%7D%5D%5BH%5E%2B%5D%5E2%7D)
Therefore,
![k_1k_2=\frac{[HS^-][H^+]}{[H_2S]}\frac{[S^{2-}][H^+]}{[HS^-]}](https://tex.z-dn.net/?f=k_1k_2%3D%5Cfrac%7B%5BHS%5E-%5D%5BH%5E%2B%5D%7D%7B%5BH_2S%5D%7D%5Cfrac%7B%5BS%5E%7B2-%7D%5D%5BH%5E%2B%5D%7D%7B%5BHS%5E-%5D%7D)
![\Rightarrow k_1k_2=\frac{[S^{2-}][H^+]^2}{[H_2S]}](https://tex.z-dn.net/?f=%5CRightarrow%20k_1k_2%3D%5Cfrac%7B%5BS%5E%7B2-%7D%5D%5BH%5E%2B%5D%5E2%7D%7B%5BH_2S%5D%7D)



⇒K₃=9.35× 10²⁵
Therefore the equilibrium constant
is 9.35× 10²⁵
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