第六章 酸碱平衡和酸碱滴定法习题答案

?(Na2HPO4)??[c(HCl)?V(HCl)?c(NaOH)?V(NaOH)]?M(Na2HPO4)ms

(0.1000?30.00?0.1000?20.00)?141.96?0.14201.000?1000T(N/NaOH) = 0.1000×14.01×10-3=1.401×10-3 (g·mL-1)

习题6-16 蛋白质试样0.2320 g经克氏法处理后，加浓碱蒸馏，用过量硼酸吸收蒸出的氨，然后用0.1200 mol·L-1 HCl 21.00 mL滴至终点，计算试样中氮的质量分数。

-解： NH3 + H3BO3 ? NH++H2BO3 4-

H+ + H2BO3= H3BO3

- n(N) = n(NH3) = n(H2BO3) = n(HCl)

ω(N)= c(HCl)·V(HCl)·M(N)/ms= 0.1200×21.00×14.01×10-3/ 0.2320 = 0.1522

习题6-17 称取土样1.000 g溶解后，将其中的磷沉淀为磷钼酸铵，用20.00mL0.1000mol·L-1 NaOH溶解沉淀，过量的NaOH用0.2000 mol·L-1 HNO37.50 mL滴至酚酞终点，计算土样中ω(P)、ω(P2O5)。已知：

H3PO4+12MoO2-+ 2NH4+ + 22H+?(NH4)2HPO4·12MOO3·H2O +11H2O 4(NH4)2HPO4·12MOO3·H2O +24OH- ?12MoO2-+HPO2-+ 2NH++13H2O 444解： H3PO4+12MoO2-+ 22H+ ? (NH4)2HPO4·12MoO3·H2O +11H2O 42-(NH4)2HPO4·12MoO3H2O +24OH- ?12MoO3+HPO2-+NH++13H2O 44n(P) = n(H3PO4) = n((NH4)2HPO4·12MoO3·H2O) = n(OH-)/24

1/24??0.1000?20.00?0.2000?7.50??30.97?6.45?10?41.000?1000

M?P2O5?141.95??P2O5?????P???6.45?10?4?1.48?10?32M?P?2?30.97??P??

习题6-18 Calculate the concentration of sodium acetate needed to produce a pH of 5.0 in a solution of

θacetic acid (0.10mol·L-1) at 25℃pKa for aceticacid is 4.756 at 25℃.

解：

c(HAc)c(NaAc)0.10 5.0?4.756?lgc(NaAc)pH?pKaθ?lgc(NaAc)?0.057(mol?L?1)

习题6-19 The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.10 mol·L-1.Calculate c(H+), c(HS-) and c(S2-) in the solution.

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