Calculate the pH of a 3.0 M H2S solution.

`H_2S + H_2O <=> H_3O^+ + HS^-`

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H2S is a weak acid. It's Ka or acid dissociation constant is 9.6x10^-8. It is placed in water, H2S forms H3O+ and its conjugate base HS-. Since we have the concentration of H2S, we can do the ICE table.

`K_a = ([H_3O^+][HS^-])/([H_2S])`

`H_2S + H_2O -> H_3O^+ + HS^-`

I 3.0 0 0

C -x +x +x

E 3.0-x x x

Substitute the expression derived from the ICE table to the Ka expression.

`x =[H_3O^+] = [HS^-]`

`K_a = ([H_3O^+][HS^-])/([H_2S])`

`K_a = (x*x)/(3.0-x)`

`9.6x10^-8 = (x^2)/(3.0-x)`

Assumption: Since the Ka value is extremely small, we can let x = 0 in the denominator. So we have:

`9.6x10^-8 = (x^2)/(3.0)`

`9.6x10^-8 * 3.0= (x^2)`

`sqrt(9.6x10^-8 * 3.0= (x^2))`

`x = 5.36656x10^-4 =[H_3O^+]`

`pH = -log[H_3O^+]`

`pH = -log[5.36656x10^-4]`

**pH = 3.27**

**Sources:**

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