Buffers solutions that resist pH changes base addition ofacid containacidic componentHAHA+ OH -  H 2 O + A - basic component A-A- + H +  HA conjugate.

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Buffers solutions that resist pH changes base addition ofacid containacidic componentHAHA+ OH -  H 2 O + A - basic component A-A- + H +  HA conjugate pairweak acid + conjugate base weak base + conjugate acid Which of the following are buffer systems? a) KH 2 PO 4 /H 3 PO 4 b) NaClO 4 /HClO 4 c) KCl/HCl d) KCHOO/CHOOH

Henderson-Hasselbalch Equation pH =pKapKa + log[A-][A-] [HA] [A - ] =[HA]pH =pKapKa adjust pH by changing [A-][A-] [HA] best buffering [A-][A-]  [HA] 10:1maximum pH = pKapKa  1 If you want a buffer at pH = 8.60 a) HA b) HB c) HC K a = 2.7 x K a = 4.4 x K a = 2.6 x = 2.5 x = log[C - ] [HC] [C - ] [HC] = 1.02 HC  H+H+ + C -

Calculate the pH of a buffer1.0 M CHOOH 1.0 M KCHOO K a = 1.8 x HAHA A-A- pH =pKapKa + log [A-][A-] [HA] = log [1.00] [1.00] = 3.74 What is the pH after addition0.1 mole of HCl to 1.0 L HCl  H + + Cl - H+H+ acidreacts with baseCHOO - + H +  CHOOH [CHOOH][CHOO - ][H + ] I C E x+x x0.9 + xx = log (0.9) (1.1) = 3.65 in H 2 O pH = 1

Preparation of a buffer of specific pH pH =pKapKa + log [A-][A-] [HA] “phosphate buffer”pH = 7.4 H 3 PO 4  H 2 PO H + K a1 = 7.5 x H 2 PO 4 -  HPO H + K a2 = 6.2 x HPO 4 2-  PO H + K a3 = 4.8 x =3.98 x dissolveNaH 2 PO 4 andNa 2 HPO 4 in water 7.4 =-log (6.2 x )+ log [HPO 4 2- ] [H 2 PO 4 - ] 0.19 =log [HPO 4 2- ] [H 2 PO 4 - ] [HPO 4 2- ] [H 2 PO 4 - ] 1.55 = pK a = 7.21

Buffers 1. Mix weak acid Mix weak base + salt of conjugate base + salt of conjugate acid 2. Partial neutralization weak base there must be an excess of weak acid(weak base) assume that all of the strong base reacts forming stoichiometric amount of A - leaving unreacted HA of weak acidwith strong base with strong acid

Which of the following solutions will be buffered? a) 100 mL 1.0 M HNO mL 1.0 M NaOH b) 100 mL 1.0 M NH 3 c) 100 mL 1.0 M HNO 3 d) 100 mL 1.0 M CHOOH mL 1.0 M HCl + 50 mL 1.0 M NaOH + 50 mL 2.0 M KOH

Calculate the pH of a buffer prepared by mixing: 40.0 mL of 1.0 M C 2 H 5 OOH 60.0 mL of 0.1 M NaOH K a = 1.3 x mol C 2 H 5 OOH = 0.04 L= 0.04 molx 1.0 mol L mol OH - = 0.06 L= molx 0.1 mol L C 2 H 5 OOH+ OH -  C 2 H 5 OO - + H 2 O mol C 2 H 5 OOH = = mol C 2 H 5 OO - = volume = L [C 2 H 5 OOH] =0.34 M[C 2 H 5 OO - ] =0.06 M

Calculate the pH of a buffer prepared by mixing: 40.0 mL of 1.0 M C 2 H 5 OOH 60.0 mL of 0.1 M NaOH K a = 1.3 x [C 2 H 5 OOH] =0.34 M[C 2 H 5 OO - ] =0.06 M pH = pKapKa + log[C 2 H 5 COO - ] [C 2 H 5 OOH] pK a =- log (1.3 x ) pH = log = < < 5.89 (0.1 – 10)