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
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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:
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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.
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4. 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-
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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:
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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.
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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.
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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,
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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
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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
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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
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12. [acid] = [NH4+] = (100×0.1)/500 = 10/500 (mol/L)
Example 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] /500
pH = pKa＋log = 9.25 ＋ log = 9.73
[acid] /500
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13. 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:
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14. 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).
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15. ---------------------------------------------------------------
Table 3-1 the relationship between capacity and concentration
Buffer [Ac-](mol/L) [HAc] ratio ctotal β(molL-1pH-1) : :
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16. 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).
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17. ------------------------------------------------------------
Table 3-2 The relationship between capacity and buffer ratio
Buffer [Ac-](mol/L) [HAc] ratio ctotal β(mol L-1pH-1) : : : : :
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18. 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.
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19. 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.
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20. 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.
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21. 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.
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22. [CH3COO-] pH = 4.60 = pKa + log ---------------- [CH3COOH]
log = pH – pKa
[CH3COOH] = =
= 0.71
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23. 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.
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24. 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 Calculate the volume of 0.1mol/L NaOH we should add in the solution of 100ml 0.1mol/L HAc ?
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25. 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 = 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
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26. 3-4.3 Preparing a Buffer in Real Life
In real life, we often use a tris buffer of pH 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.
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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: ~7.44
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-
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28. 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 )
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29. 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?
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30. conjugate acid 00 conjugate base 00 anti-acid part anti-base part
共轭酸
共轭碱
H2CO 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
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31. H2CO3 ===== H+ + HCO3-
OH - H+
Lung === CO2 +H2O
kidney
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32. Intracellular buffers: HHbO2-HbO2- , HHb-Hb-, in tissue: CO2 + H2O + Hb- == HHb + HCO3- in lung: HCO3- + HHbO2 == HbO2- + H2O + CO2
Vein
transport
Artery
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33. 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℃.
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34. 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.
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35. 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
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