Buffer Preparation

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 OHH3PO4 H2PO4OHHPO4=
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K2HPO4 g. etc.
實驗室酸鹼度/氧化還原電位計 Laboratory pH Meter
方法(2) pH = pKa + log
[OAc-] [HOAc]
(Henderson-Hasselbalch equation) [H+][OAc-] Ka= [HOAc] 取log logKa=log[H+]+log[OAc-]-log[HOAc] -log[H+]=-logKa+log[OAc-]-log[HOAC]
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
pKa2=7.2 pKa3=12.4
pKa1=2.14
H3PO4
實習操作： 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.
配對的(成對的)
weak acid conjugated base (weak acid salt)
(HOAc
OAc-)
給質子
HOAc
OAc-
Preparation of Buffer Solution
1.What are the concentration of HOAc and OAc- in a 0.2 M acetate buffer，pH 5.00 ? The ， Ka for acetate is 1.70×10× 5(pKa = 4.77)。 。 0.2M acetate 中含 mole of acetate/litter， 中含0.2 ， 其中部份為HOAc，部份為 其中部份為 ，部份為OAc-，其比例 ， 及濃度可以Ka或 及濃度可以 或Henderson-Hasselbalch equation(H-H方程式 解之 方程式)解之 方程式 解之.
3.Polyprotic acids 多質子酸 ka1 H2A [H+][HA-] Ka1= [H2A] Ka2 = [HA-] H+ + HAka2 H+ + A-2 [H+]+[A-2]
pH = pKa1 + log [HA-]/[H2A] pH = pKa2 + log [A-2]/[HA-] Which one ? Just use the one that describes the equilibrium between the species we are dealing with.
2.Prepare 3 L of above buffer by using solid sodium acetate trihydrate ﹝CH3COO-Na+‧3H2O﹞ (MW=136) and a 1 M acetic acid solution. 3 L × 0.126 M = 0.378 mole OAc3 L × 0.074 M = 0.222 mole HOAc 0.378 mole of OAc- = Wt(g)/136 Wt(g) = 51.4 g 0.222 mole = 1 M liter liter = 0.222 = 222 ml
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-
不管怎樣，先要算出此狀況下K2HPO4 及KH2PO4 之比例: HPO4= pH = pKa2 + log H2PO4HPO4= 7.5 = 7.2 + log H2PO4HPO4= 0.3 = log H2PO4HPO4= 10L ×0.045M = 2 = 2/1 H2PO4-
∴ ⅔ × 0.45 mole = 0.3 mole .15 mole H2PO4needed
H2PO4-
HPO4-2
PO4-3
方法很多: 方法很多 a。加適當比例之 。加適當比例之KH2PO4及K2HPO4 b。加H3PO4，再用KOH將其轉變為 。 再用 將其轉變為KH2PO4及K2HPO4 將其轉變為 c。加KH2PO4，再用 。 再用KOH將其轉變為 2HPO4 將其轉變為K 將其轉變為 d。用K3PO4，再用 。 再用HCl將其轉變為 將其轉變為KH2PO4或K2HPO4 將其轉變為
[OAc-] pH = pKa + log [HOAc] [OAc-] = y [HOAc] = 0.2－ y 5 = 4.77 + log y/(0.2 y) y/(0.2－y) 0.23 = log y/(0.2－y) 取log 用計算機 數字+SHIFT+log鍵 數字 鍵
y/0.2-y = 1.7 3.4－1.7y = y 3.4 = 2.7y y = 0.126 [OAc-] = 0.126 M [HOAc] = 0.074 M
HOAc 反應前 0.2 反應後 0.2－y
方法(1) [H+][OAc-] Ka= [HOAc] y = [OAc-] 0.2 – y = [HOAc]
H+ + OAc0 0 10-5 y
產物 反應物
∴ [H] = 10-5
[10-5][y] 代入 Ka= [0.2-y] 1.70×10-5 = (10-5)(y)/(0.2-y) 10-5y = 3.4×10-6－1.70×10-5 y 2.70×10-5y = 3.4×10-6 y = 0.126 [OAc-] = 0.126 M [HOAc] = 0.2-0.126=0.074 M = 1.70×10-5 ×