Question

H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.36×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.06×10−19, what is the equilibrium constant Kfinal for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)

Answer 1

Solution

₂S ⇌ HS⁻ + H⁺,   K1 = 9.36x10⁻⁸,

HS⁻ ⇌ S²⁻ + H⁺,   K2 = 1.06×10⁻¹⁹,

Adding the above two equations, we get -

H₂S ⇌    S²⁻   + 2H⁺,

We know that when the two equation are added, their equilibrium constants get multiplied.

So, K = K1 * K2

= 9.36x10⁻⁸ * 1.06x10⁻¹⁹

= 9.922x10^-27

Now, reversing the equation we get,

S²⁻   + 2H⁺    ⇌   H₂S

K _final = 1 / K

= 1 / (9.922x10^-27)

K _final = 1.078x10²⁶

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