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Chapter Three Buffer Solution

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1 Chapter Three Buffer Solution
Chapter Three Buffer Solution 3-1 Concept of Buffer Solution 3-2 The pH of Buffer Solution 3-3 Capacity of Buffer Solution 3-4 Designing and Preparation of Buffers 3-5 Buffer Systems in the Body 2018/11/27

2 3-1 Concept of Buffer Solution
3-1.1 Definition of buffer solution A solution that can resist changes in pH when limited amounts of strong acid or base are added to it is called buffer solution (buffer). How do buffers resist changes in [H+] or [OH- ] ? 3-1.2 Theory of buffer solution How buffer solutions work ? To understand this process, we can analyze the buffer solution of HAc and NaAc. We can write such an equilibrium as: 2018/11/27

3 ● Adding a small amount of acid.
little HAc H+ + Ac- large NaAc = Na+ + Ac- large ● Adding a small amount of acid. If an amount of H+ ions are added, the above reaction will shift to the left. ● Adding a small amount of base. If OH- ions are added they will remove H+ ions to form water. Therefore, no matter adding small amount of acid or base can not change the pH of buffers significantly. 2018/11/27

4 3-1.3 Composition and Type of buffer solution
anti-base part ● Composition anti-acid part ● Type of buffer solution Weak acid and its conjugate base HAc-NaAc, H2CO3-NaHCO3 Weak base and its conjugate acid NH3·H2O-NH4Cl Polyprotic acid salt and its conjugate base: NaHCO3- Na2CO3 ; 2018/11/27

5 3-2 The pH of Buffer Solution
3-1.1 Henderson-Hasselbalch equation HA (aq) H+ (aq) + A- (aq) [H+][A-] Ka = [HA] [HA] [H+] = Ka [A-] We first take the negative log of both sides: 2018/11/27

6 [HA] -log[H+] = -logKa - log------- [A-]
[HA] [A-] pH = pKa - log = pKa + log [A-] [HA] In general: [base] pH = pKa + log [acid] This is known as the Henderson-Hasselbalch equation. It relates pH to pK a and the concentrations of the acid and conjugate base, and can be used to design buffers to maintain any desired pH. 2018/11/27

7 [base] pH = pKa + log --------- [acid]
Buffer ratio From above equation, we can know: ● The pH of a buffer solution depends on the Ka of conjugate acid and the ratio of conjugate acid and conjugate base. The ratio is called buffer ratio. ● When buffer ratio =1 , [A-] = [HA], pH = pKa . ● For certain buffer solution, at a given temperature, Ka is a constant. pH depends on the buffer ratio. We can design different buffer solutions by changing the buffer ratio. 2018/11/27

8 NH4+ - NH3, ● When buffer solution is diluted, the buffer ratio is not
changed, so pH of buffer solution is not changed. ● pKa = -logKa H2PO4- - HPO42-, NH4+ - NH3, 2018/11/27

9 3-1.2 Calculating the pH of buffer solution
Assuming that the volume of buffer is V L, amount of substance of conjugate acid and base is nA, nB respectively. [acid]=nA / V, [base]=nB / V So, nB pH = pKa + log ---- nA variable constant 2018/11/27

10 (a) According to the buffer equation [A-] 1.0
Example 3-1 (a) Calculate the pH of buffer system containing mol/L HAc and 1.0mol/L NaAc. (b) What is the pH of the buffer system after the addition of mole of gaseous HCl to one liter of the solution? Assume that the volume of the solution does not change when the HCl is added and pKa HAc= 4.75 Solution (a) According to the buffer equation [A-] pH = pKa + log = log----- [HA] = 4.75 2018/11/27

11 (b) After the addition of 0. 10mol HCl to one liter of the solution, 0
(b) After the addition of 0.10mol HCl to one liter of the solution, 0.1mol HCl can consume 0.1mol Ac so, the amount-of-substance of acetic acid and acetate ions present are nHAc = =1.1 mol nAc- = 1.0 – 0.10 = 0.90 mol The pH of the solution becomes nB pH = pKa + log ----- nA = log(0.90 / 1.1) = 4.66 2018/11/27

12 3-3 Capacity of Buffer Solution
3-3.1 Concept of buffer capacity The buffer capacity (β) is the amount-of-substance of strong acid or base per liter needed to produce a unit change in pH. Δb β= |ΔpH| where Δb: ΔpH: 2018/11/27

13 3-3.2 Factors of Influencing Buffer Capacity (β)
The buffer capacity of a buffer solution depends on the total concentration (ctotal = [B] +[HA]) and buffer ratio ( [B]/[HA]). When the buffer ratio is fixed, the greater total concentration, the greater the buffer capacity is (See figure 3-1 curve b and c). This can been seen also from following table (3-1). 2018/11/27

14 ---------------------------------------------------------------
Table 3-1 the relationship between capacity and concentration Buffer [Ac-](mol/L) [HAc] ratio ctotal β(molL-1pH-1) : : 2018/11/27

15 When the total concentration is fixed, if the ratio is equal to 1 , the capacity is the greatest; the pH = pKa at that time. The more ratio deviate 1, the more pH deviate pKa, the smaller the capacity is. This can been seen also from following table (3-2). 2018/11/27

16 ------------------------------------------------------------
Table 3-2 The relationship between capacity and buffer ratio Buffer [Ac-](mol/L) [HAc] ratio ctotal β(mol L-1pH-1) : : : : : 2018/11/27

17 3-4 Designing and Preparation of Buffers
3-4.1 Preparing Principle of Buffer Solution 1. Choose the suitable buffer system pH=4.2 buffer, HAc-NaAc, pKa=4.75, buffer range: 2. Choose the best buffer system To select a buffer system, we should first look for an acid with a pKa that is as closes possible to the desired pH. 2018/11/27

18 3. Choose the suitable total concentration ctotal = 0.05-0.2mol/L
pH= 5.2 buffer, HAc-NaAc, pKa=4.75, KOOCC6H4COOH- KOOCC6H4COONa, pKa=5.4 3. Choose the suitable total concentration ctotal = mol/L The concentrations of acid and conjugate base can then be adjusted slightly from the most desirable 1:1 ratio to give exactly the desired pH. 2018/11/27

19 3-4.2 Preparing Methods of Buffer Solution
Choose the suitable buffer system Choose the suitable total c Calculate the amount of acid and base Mix the amount of acid and base Example 3-2 Design a buffer system with pH 4.60. Solution: Step 1: Choose a suitable system, HAc- Ac-. Step 2: Calculation. 2018/11/27

20 [CH3COO-] pH = 4.60 = pKa + log ---------------- [CH3COOH]
log = pH – pKa [CH3COOH] = = = 0.71 2018/11/27

21 Such a ratio could be established by dissolving 0.71
mol of sodium acetate and 1.0 mol of acetic acid in one liter of water. or mol NaAc and 0.10 mol HAc in the same volume. and so on. As long as the ratio of the concentrations is 0.71 (and the concentrations are not too small), the solution will be buffered at close to pH 4.60. 2018/11/27

22 Intracellular buffers: HHbO2-HbO2- , (氧合血红蛋白) HHb- Hb-, (血红蛋白)
3-5 Buffer Systems in the Body (The application of buffer solution to medical science) The pH of blood: ~7.44 Intracellular buffers: HHbO2-HbO2- , (氧合血红蛋白) HHb- Hb-, (血红蛋白) H2CO3-HCO3-, H2PO4--HPO42- extracellular buffers (in plasma) : H2CO3-HCO3-, H2PO4--HPO42- , HnP - Hn-1P- 2018/11/27

23 HCO3- + H+ H2CO3 H2CO3 + OH- H2O + HCO3- H2CO3 H+ + HCO3-
when an acid is added HCO3- + H+ H2CO3 when a base is added H2CO3 + OH- H2O + HCO3- What is the ratio of [HCO3-] : [H2CO3] required to maintain a pH of 7.4 in the bloodstream? ( given that the Ka for H2CO3 in blood is 8.0×10-7 ) 2018/11/27

24 Solution: For a solution whose pH is 7.4,
[HCO3-] pH = pKa + log [H2CO3] 7.4 = - log (8.0 ×10-7) + log solve , we got [HCO3-] = [H2CO3] Thus, the ratio of [HCO3-]: [H2CO3] required to maintain a pH of is 20 : 1 Oh! This exceeded the limit of buffer range (pKa±1). Can it maintain it? How? 2018/11/27

25 H2CO3 ===== H+ + HCO3- OH - H+ Lung === CO2 +H2O kidney 2018/11/27


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