θθ解：已知298K时，Ka1(H2S) = 9.1×10-8 Ka2(H2S) = 1.1×10-12
θθ>>Ka2，计算H+ 浓度时只考虑一级离解 Ka1
H2S ?H+ + HS-
θθ又c/Ka1= 0.10/(9.1×10-8) >>500，c(H+)?c?Ka1?0.10?9.1?10-8?9.5?10-5mol?L-1
c(HS-) = c(H+) = 9.5×10-5 mol·L-1 因S2-是二级产物，设c(S2-) = x mol·L-1 HS- ＝ H+ + S2-
平衡时 9.5×10-5-x 9.5×10-5+x x
c2(H?)?c(S2?)K?K?c(H2S)θa1θa2θc(S2?)?Kaθ1?Ka2
c(H2S)0.10?9.1?10?8?1.1?10?12??1.0?10?18mol?L?12?2c(H)(0.10)θ由于Ka2极小，9.5×10-5±x ≈ 9.5×10-5，则有
θ Ka2= c(H+)·c(S2-) / c(HS-) = 9.5×10-5·c(S2-)/ 9.5×10-5 = 1.1×10-12 θ故c(S2-) =Ka2=1.1×10-12 (mol·L-1)
习题6-20 Calculate the equilibrium concentration of sulfide ion in a saturated solution of hydrogen sulfide to which enough hydrochloric acid has been added to make the hydronium ion concentration of the solution 0.10 mol·L-1 at equilibrium. (A saturated H2S solution is 0.10 mol·L-1 in hydrogen sulfide)
解：
c2(H?)?c(S2?)K?K?c(H2S)
0.10θθc(H2S)?8?12c(S2?)?Ka1?Ka2?9.1?10?1.1?10??1.0?10?18mol?L?12?2c(H)(0.10)θa1θa2