无机及分析化学第七章酸碱平衡与酸碱滴定法习题答案.doc

第七章酸碱平衡与酸碱滴定法
7-3 (1) Kb = =×10-5; (2) Kb = =×10-11
(3) Kb = =×10-5; (4) Kb = =×10-5
(5) Kb1 = =×10-2; (6) Kb2 = =×10-13
7-4 (1) ∵= % < 5% ∴= ×10-5 c(H+) == × = ×10-4 -1
(2) c(H+)==×10--1 = %
7-5 (1) c(H+) = c(OH-) + c(H2PO) + 2c(HPO) + 3c(PO); (2) c(H+) + c(H3PO4) = c(OH-) + c(HPO) + 2c(PO)
(3) c(H+) + c(HS-) + 2c(H2S) = c(OH-) + c(NH3); (4) c(H+) + c(H2CO3) = c(OH-) + c(CO) + c(NH3)
(5) c(H+) + c(H2PO) + 2c(HPO) + 3c(PO) = c(OH-)
(6) c(Na+) + c(NH) + c(H+) = c(OH-); (7) c(H+) = c(OH-) + c(Ac-) + c(H2BO)
(8) c(H+) = c(OH-) + c(Ac-) + c(Cl-); (9) c(H+) + c(HF) = c(OH-) + c(NH3)
7-6 c(NH3) = c×= ×10-8 -1
= c - c(NH3) = - ×10-8 ≈ -1
7-7 (1) ∵Kac = ×10-5×>>20Kw = 5556 > 500
∴c(H+) = =×10-3 -1 pH= -lgc(H+) =
(2) ∵Kac = ×10-2×>>20Kw = 3 < 500
∴c(H+) = = -1 pH = -lgc(H+) =
(3) Ka = = ×10-10 ∵Kac = ×10-10× > 20Kw = ×108 > 500
∴c(H+) = =×10-5 -1 pH= -lgc(H+) =
(4) ∵Ka1 >> Ka2 >> Ka3 Ka1c = ×10-3×>>20Kw = 53 < 500
∴c(H+) = = -1 pH = -lgc(H+) =
(5) Kb = = ×10-5 ∵Kbc = ×10-5× >> 20Kw = 6250 > 500
∴c(OH-) = = ×10-3 -1 pOH= -lgc(OH-) = pH =
(6) ∵ Kb1 = Kb2 = Kb3 =
Kb1 >> Kb2 >> Kb3 Kb1c = ×10-2×>>20Kw = < 500
∴c(OH-) = = 0.