xmol?L?1 pH?3.90?3.74?lg?1?10.400mol?L?xmol?Lc(HCOO-) =x mol·L-1=0.236mol·L-1
c(HCOOH)=0.400mol·L-1-0.236mol·L-1=0.164mol·L-1
0.20g/40g?mol?11000mL6. n(NaOH) == 0.050mol·L-1 ?100mL1L因加入NaOH后，
[B?]?0.050mol?L-1 pH?5.60?5.30?lg-1-10.25mol?L-0.050mol?L[B-] = 0.35mol·L-1
0.35mol?L?1固原溶液 pH?5.30?lg?5.45 ?10.25mol?L??n(Asp)n(Asp)7. pH?pKa?lg?3.50?lg?2.95 n(HAsp)n(HAsp)n(Asp?)?0.28
n(HAsp)又 n(Asp?)?n(HAsp)?0.65g ?0.0036mol180.2g?mol-1n(HAsp)?0.0028 mol
可吸收阿司匹林的质量=0.0028mol×180.2g·mol-1=0.50g 8. 求近似pH
?-1n(CHCOO)1.8g/40g?mol25pH = pKa?lg?4.87?lg?4.78 n(C2H5COOH)0.500L?0.20mol?L-1?1.8g/40g?mol-1求精确pH，丙酸钠(C2H5COONa)是强电解质
I =
110.045mol20.045mol2∑ciZi2 = (×1+×1) = 0.09 mol·L-1≈0.1mol·L-1 220.500L0.500L??(B)当Z = 0，I = 0.10，校正因数 lg??0.11 ?(HB)??pH = pKa?lgn(C2H5COO)?lg?(C2H5COO)?4.78?(?0.11)?4.67
n(C2H5COOH)?(C2H5COOH)9. HAc溶液和NaOH溶液的体积分别为3V和V，
c(HAc) =(0.10 mol·L-1×3V - 0.10 mol·L-1×V)/(3V+V) = 0.050 mol·L-1
c(Ac-) =0.10 mol·L-1×V/(3V+V) = 0.025 mol·L-1
0.025mol?L-1pH?4.75?lg?4.45 -10.050mol?L
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0.050mol?L-1?0.025mol?L-1??2.303??0.038mol?L?1 -1-10.050mol?L?0.025mol?L0.10mol?L?110. (1) pH?3.85?lg?3.9 ?10.10mol?L0.10mol?L-1?0.10mol?L-1?1 ??2.303??0.12mol?L-1-10.10mol?L?0.10mol?L(2) pH?3.85?lg0.010mol?L?1?3.9 ?10.010mol?L0.010mol?L-1?0.010mol?L-1??2.303??0.12mol?L?1 -1-10.010mol?L?0.010mol?L0.10mol?L?1(3) pH?3.85?lg?4.9 ?10.010mol?L0.10mol?L-1?0.010mol?L-1?1 ??2.303??0.21mol?L0.10mol?L-1?0.010mol?L-111.
溶 液 Na2HCit+HCl Na2HCit+HCl Na2HCit+NaOH
缓 冲 系
抗 酸成 分 H2Cit- Cit3-
抗 碱成 分 H2Cit- H3Cit HCit2-
有效缓冲范围 3.77~5.77 2.14~4.14 5.39~7.39
β最大时 配制体积比
2:1 2:3 2:1
H2Cit--HCit2- HCit2- H3Cit-H2Cit- HCit2--Cit3-
12. 设需加入HCl溶液体积 V，
n(Tris)0.0500mol?L-1?100mL-0.0500mol?L-1?V
7.40?7.85?lg?7.85?lgn(Tris ?HCl)0.0500mol?L-1?100mL?0.0500mol?L-1?VV(HCl) = 47.6mL
13. 设需NH4Cl 质量m、1.00mol·L-1的NaOH溶液体积V，
pKa(NH4+) = 14-4.75 = 9.25
1.00mol?L?1?V 9.00?9.25?lgm/53.5g?mol-1?1.00mol?L?1?V得 1.00mol·L-1×V = 0.562[m/53.5 g·mol-1-1.00mol·L-1×V] 又 [1.00mol·L-1×V + 0.562(m/53.5 g·mol-1-1.00mol·L-1×V)] /1L=0.125mol·L-1 则 m = 6.69g；V = 0.045L 14. (1) HCl与NaOH完全反应需HCl 50mL。 (2) HCl + NH3·H2O = NH4Cl + H2O NH4+的pKa = 14-4.75= 9.25，设需HCl体积V，
0.10mol?L?1?50mL?0.10mol?L?1?V 7.00?9.25?lg?10.10mol?L?VV = 0.0497 L = 49.7 mL
(3) HCl + Na2HPO4 = NaH2PO4 + NaCl
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H3PO4的pKa2=7.21，设需HCl体积V，
0.10mol?L?1?50mL?0.10mol?L?1?V 7.00?7.21?lg?10.10mol?L?VV = 0.0309 L = 30.9 mL
第一种混合溶液无缓冲作用；第二种pH H3PO4 + NaOH = NaH2PO4 + H2O H3PO4完全反应，生成0.020mol·L-1×V1的NaH2PO4，剩余NaOH 0.020mol·L-1×V2 - 0.020mol·L-1×V1 = 0.020mol·L-1×(V2 - V1) NaH2PO4 + NaOH = Na2HPO4 + H2O 生成Na2HPO4的物质的量等于NaOH的物质的量0.020mol·L-1×(V2 - V1)，剩余NaH2PO4 0.020mol·L-1×V1 - 0.020mol·L-1×(V2 - V1) = 0.020mol·L-1×(2V1 - V2) 0.020mol?L?1?(V2?V1) 7.40?7.21?lg0.020mol?L?1?(2V1?V2)解得 V1 = 38mL，V2 = 62mL。 ??L?1?0.9016. pH= pKa1′?lg[HCO3]?6.10?lg24mmol?7.36 ?1[CO2]溶解1.2mmol?LpH虽接近7.35，但由于血液中还有其他缓冲系的协同作用，不会引起酸中毒。 Exercises 1. A buffer solution consists of a conjugate acid-base pair. The conjugate base can consume the added strong acid, and the conjugate acid can consume the added strong base, to maintain pH。The conjugate acid-base pair of weak electrolytes presents in the same solution at equilibrium. 2. The pH of a buffer depends on the buffer component ratio. When [B-]/[HB]=1, pH = pKa, the buffer is most effective. The further the buffer-component ratio is from 1，the less effective the buffering action is. Practically, if the [B-]/[HB] ratio is greater than 10 or less than 0.1, the buffer is poor. The buffer has a effective range of pH = pKa±1. 3. Choose (d) and (e). Buffer capacity depends on both the concentration of the reservoirs and the buffer-component ratio. The more concentrated the components of a buffer, the greater the buffer capacity. When the component ratio is close to one, a buffer is most effective. 4.（a）The pH increase to a small extent;（b）The pH decrease to a small extent;（c）The pH increase to a small extent;（d）The pH increase to a larger extent. 13 5. (b) and (d) 6. The pKa of CH3NH2·HCl is 10.65. The pKa of NH4+ is 9.25. The pKa of the former is more close to 10.5. A buffer is more effective when the pH is close to pKa. She should choose CH3NH2. The other is not a good choice. 7. c(CH3COO-) +c(CH3COOH)=0.150mol·L-1 n(CH3COO-)+n(CH3COOH)=0.150mol·L-1×500mL×10-3L·mL-1=0.0750mol n(CH3COO?) 5.00?4.75?lgn(CH3COOH)n(CH3COO?)?1.78 n(CH3COOH)1.78n(CH3COOH)=n(CH3COO-) n(CH3COOH)=0.0270mol, n(CH3COO-)=0.0480mol Mass of sodium acetate = 0.0480mol×136.1g·mol -1=6.53g 0.0270mol?103ml?L?1V(CH3COOH)??1.54ml 17.45mol?L?12?[HPO4] 8. 7.40?6.80?lg[H2PO?4]?[HPO24]?3.98?4 ?[H2PO4] 14