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