1、Buffer Preparation,Biochemistry 96.3.8,Buffer,something that resists change,pH Buffer(定義),substance or mixture of substances, that permits solution to resist changes in pH upon the addition of H+ or OH-,配對的(成對的)weak acid conjugated base (weak acid salt)(HOAc OAc-) HOAc OAc-,給質子,1.What are the concen
2、tration of HOAc and OAc- in a 0.2 M acetate buffer，pH 5.00 ? The Ka for acetate is 1.7010-5(pKa = 4.77)。0.2M acetate 中含0.2 mole of acetate/litter，其中部份為HOAc，部份為OAc-，其比例及濃度可以Ka或Henderson-Hasselbalch equation(H-H方程式)解之.,Preparation of Buffer Solution,HOAc H+ + OAc-反應前 0.2 0 0反應後 0.2y 10-5 y方法(1)H+OAc-
3、產物Ka=HOAc 反應物y = OAc- 0.2 y = HOAc H = 10-5,1.7010-5 = (10-5)(y)/(0.2-y)10-5y = 3.410-61.7010-5 y2.7010-5y = 3.410-6y = 0.126OAc- = 0.126 M HOAc = 0.2-0.126=0.074 M,10-5y 代入 Ka= = 1.7010-50.2-y,方法(2) OAc- pH = pKa + logHOAc,H+OAc-Ka=HOAc 取log logKa=logH+logOAc-logHOAc-logH+=-logKa+logOAc-logHOAC,(He
4、nderson-Hasselbalch equation),OAc- pH = pKa + logHOAc,OAc- = yHOAc = 0.2 y5 = 4.77 + log y/(0.2y)0.23 = log y/(0.2y),取log 用計算機 數字+SHIFT+log鍵,y/0.2-y = 1.73.41.7y = y3.4 = 2.7yy = 0.126 OAc- = 0.126 M HOAc = 0.074 M,2.Prepare 3 L of above buffer by using solid sodium acetate trihydrate CH3COO-Na+3H2O
5、 (MW=136) and a 1 M acetic acid solution.3 L 0.126 M = 0.378 mole OAc-3 L 0.074 M = 0.222 mole HOAc0.378 mole of OAc- = Wt(g)/136 Wt(g) = 51.4 g0.222 mole = 1 M liter liter = 0.222 = 222 ml,3.Polyprotic acids 多質子酸ka1 ka2H2A H+ + HA- H+ + A-2H+HA- H+A-2 Ka1= Ka2 = H2A HA-pH = pKa1 + log HA-/H2A pH =
6、pKa2 + log A-2/HA-Which one ? Just use the one that describes the equilibrium between the species we are dealing with.,Describe the preparation of 10 L of 0.045 M potassium phosphate buffer，pH 7.5 (pKa2 = 7.2，H2PO4-HPO4=之間) 選擇pKa値較接近pH來計算pKa1=2.14 pKa2=7.2 pKa3=12.4H3PO4 H2PO4- HPO4-2 PO4-3方法很多: a。加
7、適當比例之KH2PO4及K2HPO4b。加H3PO4，再用KOH將其轉變為KH2PO4及K2HPO4c。加KH2PO4，再用KOH將其轉變為K2HPO4d。用K3PO4，再用HCl將其轉變為KH2PO4或K2HPO4,不管怎樣，先要算出此狀況下K2HPO4 及KH2PO4 之比例:HPO4= pH = pKa2 + log H2PO4- HPO4=7.5 = 7.2 + logH2PO4-HPO4= 0.3 = logH2PO4-HPO4= = 2 = 2/1H2PO4- 0.45 mole = 0.3 mole HPO4= needed 0.45 mole = 0.15 mole H2PO4
8、- needed,10L 0.045M,e. From KH2PO4 and K2HPO4 秤0.3 mole K2HPO4及0.15 mole KH2PO4 溶於10 L水中.f. From H3PO4 and KOH OH- OH-H3PO4 H2PO4- HPO4=Take 0.45 mole H3PO4，加0.45 mole KOH 使all the H3PO4 KH2PO4，再加0.3 mole KOH 使有0.3 mole 的 KH2PO4轉變為K2HPO4g. etc.,實習操作：材料：sodium acetate (MW=82.03) 1 M acetic acid acetate 之Ka=1.70 10-5(pka=4.77)定量瓶200mL去離子水儀器設備：pH meter、安全吸球1-4, 5-8, 9-12, 13-14組同學分別計算，並配製pH值為4.0、4.5、5.5及6.0的0.2 M acetate buffer 200 ml.,實驗室酸鹼度/氧化還原電位計 Laboratory pH Meter,
