Chapter Three Buffer Solution 2008.10.22 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

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

● 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

3-1.3 Composition and Type of buffer solution anti-base part (weak acid) ● Composition anti-acid part (conjugate base) Buffer pair Buffer system ● 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 ; Subnitrate NaH2PO4-Na2HPO4- Na3PO4 Z=0, A0-B- Z=+1, A+-B0 Z=-1, A--B2- Z=-2, A2--B3- 2018/11/27

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

[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

[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

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

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

(a) According to the buffer equation [A-] 1.0 Example 3-1 (a) Calculate the pH of buffer system containing 1.0mol/L HAc and 1.0mol/L NaAc. (b) What is the pH of the buffer system after the addition of 0.10 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-] 1.0 pH = pKa + log------- = 4.75 + log----- [HA] 1.0 = 4.75 2018/11/27

(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 = 1.0 + 0.10 =1.1 mol nAc- = 1.0 – 0.10 = 0.90 mol The pH of the solution becomes nB pH = pKa + log ----- nA = 4.75 + log(0.90 / 1.1) = 4.66 2018/11/27

[acid] = [NH4+] = (100×0.1)/500 = 10/500 (mol/L) Example 3-2 Put 100 ml of 0.1mol/L[HCl] into 400 ml of 0.1mol/L [NH3] solution, pH= ? (pKb = 4.75) Solution: [acid] = [NH4+] = (100×0.1)/500 = 10/500 (mol/L) [base]= [NH3] = (400 ×0.1－100 ×0.1)/500 =30/500 (mol/L) pKa (NH4+) = 14－4.75 = 9.25 [base] 30/500 pH = pKa＋log -------- = 9.25 ＋ log--------- = 9.73 [acid] 10/500 2018/11/27

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

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

--------------------------------------------------------------- Table 3-1 the relationship between capacity and concentration --------------------------------------------------------------- Buffer [Ac-](mol/L) [HAc] ratio ctotal β(molL-1pH-1) ---------------------------------------------------------------------- 1 0.1 0.1 1:1 0.2 0.115 2 0.02 0.02 1:1 0.04 0.023 ------------------------------------------------------------ 2018/11/27

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

------------------------------------------------------------ Table 3-2 The relationship between capacity and buffer ratio -------------------------------------------------------------- Buffer [Ac-](mol/L) [HAc] ratio ctotal β(mol L-1pH-1) ----------------------------------------------------------------------- 1 0.095 0.005 19:1 0.1 0.0109 2 0.09 0.01 9:1 0.1 0.0207 3 0.05 0.05 1:1 0.1 0.0576 4 0.01 0.09 1:9 0.1 0.0207 5 0.005 0.095 1:19 0.1 0.0109 ------------------------------------------------------------ 2018/11/27

So, only in the range pH = pKa±1, the buffer is Generally, we consider as follows: When the ratio ([A-] / [HA])＜1/10 or ＞ 10/1 , in other words pH＜pKa – 1 or pH＞pKa+ 1, the solution loses buffer ability . So, only in the range pH = pKa±1, the buffer is effective. This range is called buffer range. 2018/11/27

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: 3.7-5.6 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

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 = 0.05-0.2mol/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

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

[CH3COO-] pH = 4.60 = pKa + log ---------------- [CH3COOH] log ---------------- = pH – pKa [CH3COOH] = 4.60 - 4.75 = - 0. 15 ---------------- = 0.71 2018/11/27

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 0.071 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

Ways to make a buffer : Adding a conjugate base to a weak acid (NaAc + HAc) Adding a conjugate acid to a weak base (NH4Cl + NH3·H2O) Adding a strong acid to a weak base ( HCl +NH3·H2O ) Adding a strong base to a weak acid (NaOH + HAc) Example 3-3 We want to prepare a buffer solution of pH 5.10. Calculate the volume of 0.1mol/L NaOH we should add in the solution of 100ml 0.1mol/L HAc ? 2018/11/27

buffer system: HAc/Ac-, pKa = 4.75 Solution buffer system: HAc/Ac-, pKa = 4.75 HAc + OH- → Ac - + H2O Added nNaOH = produced nAc- = neutralized nHAc Suppose add x ml NaOH , nAc- = 0.1 x nHAc= 0.10×100 – 0.1 x pH = pKa + log (nAc-/ nHAc) 5.10 = 4.75 +log [0.10 x/(0.10×100 – 0.1 x)] x = 69.1 ml take 69.1 ml 0.10 mol/L NaOH solution, add to the 100 ml 0.1mol/L HAc solution, then, mixed completely. OK! You got a buffer solution of pH 5.10 . 2018/11/27

3-4.3 Preparing a Buffer in Real Life In real life, we often use a tris buffer of pH 7.60. Here is a method to prepare the buffer. (v=1L, c tris = 0.1mol/L) l. Weigh out 0.l00 mol of tris hydrochloride and dissolve it in a beaker containing about 800 ml of water. 2. Place a pH electrode in the solution and monitor the pH. 3. Add NaOH (1mol/L) until the pH is exactly 7.60. 4. Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washings to the volumetric flask. 5. Dilute to the mark and mix. 2018/11/27

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: 7.36~7.44 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

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

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-] 20 ------------ = ------- [H2CO3] 1 Thus, the ratio of [HCO3-]: [H2CO3] required to maintain a pH of 7.4 is 20 : 1 Oh! This exceeded the limit of buffer range (pKa±1). Can it maintain it? How? 2018/11/27

conjugate acid 00 conjugate base 00 anti-acid part anti-base part 共轭酸 共轭碱 H2CO3 H+ + HCO3- B- CO2 + H2O H+ conjugate acid 00 conjugate base 00 抗酸成分 抗碱成分 anti-acid part anti-base part 2018/11/27

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

Intracellular buffers: HHbO2-HbO2- , HHb-Hb-, in tissue: CO2 + H2O + Hb- == HHb + HCO3- in lung: HCO3- + HHbO2 == HbO2- + H2O + CO2 Vein transport Artery 2018/11/27

1. Explain the mechanism of buffer action for the buffer solution consisting of aqueous ammonia and ammonium chloride. 2. To 1.0 L of 0.1mol/L HAc solution is added 0.05 moles of NaOH. What is the pH of the resulting solution? Assume the volume remains at 1.0 L. Given the acid dissociation constant Ka of acetic acid =1.8×10-5 at 25℃. 2018/11/27

3. Describe how to prepare 1L of buffer solution with pH=9 3. Describe how to prepare 1L of buffer solution with pH=9.00 and ctotal=0.125 mol/L by combining a solid NH4Cl (pKa=9.25) with any necessary amount of 1.00 mol/L NaOH ? 4. A solution which contains 0.1mole of a weak monoprotic acid and 0.1 mole of the sodium salt of the acid is found to have a pH of 5.2. Calculate the dissociation constant of the acid. 2018/11/27

Summary 5. To 100ml of 0.6mol/L [1/3H3PO4] solution is added 1.2 g of NaOH. Then dilute to 200ml and mixed completely. Calculate: ① pH=? ② mOsm/L=? ③ 37℃ Π=? Summary 2018/11/27