H2S存在解离平衡：H2S ==可逆== H+ + HS- Ka1HS- ==可逆== H+ + S2- Ka2由于第2步电离程度远远小于第1步电离，所以第2步电离可以忽略，即 [HS-] = [H+]，Ka2 = [H+][S2-] / [HS-] = [S2-]即 [S2-] = Ka2
