4.1 缓冲溶液与缓冲原理
Buffers are mixtures containing a common ion: either weak acids and their conjugate bases or weak bases and their conjugate acid.
Two common buffers: ammoniumammonia, carbonate-bicarbonate
V dpH
4.3 缓冲容量和缓冲范围
dna( b)
V dpH
where β is the buffer capacity and has units of moles per liter per pH (mol·L-1·pH-1); dna (or b) stands for moles of strong acid or strong base which are added to a buffer solution to cause the change in pH, dpH.
4.1 缓冲溶液与缓冲原理
4.1 缓冲溶液与缓冲原理
4.1.1 The Essential Feature of a Buffer
Buffer Solution:
— solutions that resist change in hydronium ion, H+, and the hydroxide ion,OH-, concentration (and consequently pH) upon addition of small amounts of acid or base, or upon dilution.
➢ NH4+(aq)
H (aq) +
+
NH3
(aq)
➢ CO32- (aq) &# -(aq)
+
OH -
(aq)
4.1 缓冲溶液与缓冲原理
缓冲溶液组成示意图
共轭酸
HAc NH4Cl H2PO4-
共轭碱
NaAc NH3·H2O HPO42-
抗碱成分
缓冲系
抗酸成分
4.1 缓冲溶液与缓冲原理
[B–]
pH = pKa + lg [HB]
= pKa + lg [conjugate base]
[conjugate acid]
pKa ：the –log of Ka of the conjugate acid
[B–]、[HB]：equilibrium concentration
[B–] / [HB]：buffer ratio
H2O + CH3COOH H3O+ + CH3COO-
CH3COOH + OH- H2O + CH3COO-
4.1 缓冲溶液与缓冲原理
anti-acid mechanism
H+
Shift left
+
HAc + H2O
H3O+ + Ac–
Shift right Anti-base
+ Anti-acid OH–
4.2 缓冲溶液pH值的计算
Sample Problem 4-1 Calculating the pH of a Buffer Solution-1
SOLUTION
Before the addition of 0.005mol of NaOH(s) to 500 mL of the buffer solution
nB– pH= pKa + lg nHB
Fight dilution
If the concentrations of conjugate acid and base used are equal, i.e. cB－ = cHB .
cB– • VB–
VB–
pH = pKa + lg
= pKa + lg
[ A ]
0.1
pH pKa lg [HA] 4.74 lg 0.2 4.44
4.2 缓冲溶液pH值的计算
Sample Problem 4-1 Calculating the pH of a Buffer Solution-1
SOLUTION
After the addition of 0.005mol of NaOH(s) to 500 mL of the buffer solution
Participatory Learning
缓 冲 溶 液 的 pH 计 算
4.2 缓冲溶液的pH值计算
4.2 缓冲溶液pH值的计算
4.2 缓冲溶液pH值的计算
For a HB-NaB buffer system,
HB + H2O NaB
[H+][B–] Ka = [HB]
H3O+ + B– Na+ + B–
pH
pK a
lg
n( A ) n(OH ) n(HA) n(OH )
0.2mol L1 250mL 5mmol
4.74 lg
4.50
0.4mol L1 250mL 5mmol
4.2 缓冲溶液pH值的计算
例4-2： 将0.10 mol·L-1 的 H3PO4溶液20 mL与 0.10 mol·L-1 的 NaOH溶液30 mL混合，混合溶液是缓冲溶液吗? 求混合 溶液的pH。已知：pKa1=2.16, pKa2=7.21, pKa3=12.32。
2.89 9.25 10.63 7.21
4.1 缓冲溶液与缓冲原理
Tris: Tris(Hydroxymethy)methanamin
三羟甲基氨基甲烷
NH2 HOH2C C CH2OH
CH2OH
4.1 缓冲溶液与缓冲原理
下列情况均需pH一定的缓冲溶液： ❖ 大多数为酶所控制的生化反应 ❖ 微生物的培养 ❖ 组织切片 ❖ 细胞染色 ❖ 药物调剂、研制等
Conjugate acid
HAc
Conjugate base
Ac-
pK ( at 25℃) a 4.76
H2CO3 H3PO4 Tris·H+
HCO3H2PO4-
Tris
6.35 2.16 7.85
H2C8H4O4 NH4+
CH3NH3+ H2PO4-
HC8H4O4NH3
CH3NH2 HPO42-
anti-base mechanism
H2O
4.1 缓冲溶液与缓冲原理
The amounts of weak acid and weak base in the buffer must be significantly larger than the amounts of H3O+ or OH- that will be added, otherwise the pH cannot remain approximately constant. Thus addition of limited amounts of a strong acid or base is counteracted by the species present in the buffer solution, and the pH changes very little. No solution can keep the pH approximately constant if you add larger amounts of either acid or base that are present in the original buffer.
Or, more specifically,
Buffer capacity is defined as the amount of strong acid or base needed to change the pH of one liter of buffer by 1 unit.
d na(b)
解: 反应1： H3PO4 + NaOH 反应前 20×0.10mmol 30×0.10mmol
NaH2PO4 + H2O
反应后
1.0mmol
2.0mmol
反应2： NaH2PO4 + NaOH
反应前 2.0mmol
1.0mmol
Na2HPO4 + H2O
反应后 1.0mmol
1.0mmol
pH
pK a2
cHB• VHB
VHB
pH pKa lg VBVHB
three different types of of Henderson-Hasselbalch Equation
4.2 缓冲溶液pH值的计算
如果用活度代替浓度，
pH pKa
lg
aB_ aHB
pKa
lg
[B_ ] B_ [HB] HB
[H+] = Ka
[HB] [B–]
Apply –log on both sides of above equation,
[B–] pH = pKa + lg [HB]
The HendersonHasselbalch Equation
4.2 缓冲溶液pH值的计算
The Henderson-Hasselbalch Equation
lg [[HH2PPOO442-]]
7.21
lg
1.0mmol 1.0mmol
7.21
Participatory Learning
缓 冲 容 量 和 缓 冲 范 围
4.3 缓冲容量和缓冲范围
4.3 缓冲容量和缓冲范围