Analytical Chemistry
Lecture No. 4
Date :16 /12/ 2012
Dr. Mohammed Hamed
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Buffers solutions: - buffers are solutions that resist changing pH when small
amounts of acid or base are added they resist changing pH by neutralizing
added acid or base
The components of a buffer solution :- It is a mixture of weak acid and its
salt:- CH3COOH is weak acid , CH3COONa is Salt of weak acid
The Mechanisms of buffer solution to remain stable:
- If base is added (OH-):
It will be buffered by the following reaction
CH3COOH + OH- --------- CH3COO- + H2O
So the pH will not change. the weak acid present in the buffer mixture can
neutralize added base.
- If acid (H+) is added:
it will be buffered by the following reaction,
CH3COO- + H+ ------------ CH3COOH
The pH will not alter significantly because the CH3COOH formed is a weak
acid.
Two important characteristics of a buffer include:
1. pH Range
• The pH range is the range of pH values over which a buffer system works
effectively.
• The pH range of an acid or base can be seen as the region of little pH change
vs. volume of base/acid added on a titration curve
• It is best to choose an acid with a pKa close to the desired pH.
2. Buffering Capacity
• The buffering capacity of a buffered solution represents the amount of acid or
base that can be absorbed by the solution without a significant change in pH.
• A buffer with a large capacity contains large concentrations of buffering
components. Therefore the capacity of a buffered solution is determined by the
magnitudes of [HA] and [A-].
Many organic acids are weak acids. Weak acids do not completely
dissociate in water. The dissociation of an organic acid is described by the
following reaction:
AH-------------------------- H + + Aweak acid
conjugate base
CH3COOH----------------------------CH3COO- + H+
weak acid
conjugate base
H2PO4-------------------H3PO4 + H+
weak base
conjugate acid
One can analyze the strength of a weak acid. This means that the amount of
hydrogen ion released can be determined. To do this, one can use the following
expression:
Ka=[H+] [A-] / [HA]
where Ka is defined as the acid dissociation constant.
The larger the value of Ka, the stronger the acid is. Because Ka values vary over
a wide range, they are usually expressed using a logarithmic scale:
pKa = - log Ka
The hydrogen ion is one of the most important ions in biological systems. The
concentration of this ion affects most cellular processes. For example, the
structure and function of most biological macromolecules and the rates of most
+
biochemical reactions are strongly affected by [H ]. The pH scale has been
devised as a convenient method of expressing hydrogen ion concentration. pH
has been defined as the negative logarithm of the hydrogen ion concentration:
+
pH = - log [H ]
The Henderson-Hasselbalch equation provides a convenient way to think
about buffers and pH:
pH = pKa + log[A-] / [HA]
If one were examining the dissociation of acetic acid, the HendersonHasselbalch relationship, then, would be:
pH = pKa + log[CH3COO-] / [CH3COOH]
The Henderson-Hasselbalch equation can be used to determine if an aqueous
solution of a conjugate acid/base pair is functioning as a buffer. If the
concentration of the weak acid is equal to that of its conjugate base, the ratio of
these two components is one. When this is the case, the Henderson-Hasselbalch
equation reduces to
pH = pKa
because the log(1) is equal to zero.
When the pH of the solution is equal to the pKa of the ionizing group, the
solution is functioning at maximum buffering capacity (best buffer). An aqueous
solution of a conjugate acid/base pair functions as a good buffer when the ratio
of the conjugate base to weak acid ranges from 1:9 to 9:1. Substituting these
ratios into the Henderson-Hasselbalch equation, one finds that this aqueous
solution functions as a good buffer when the pH of the solution is within
approximately one pH unit of the ionizing group’s pKa.
pH = pKa + 1
because the log (1/9) is -0.999 and the log of (9/1) is +0.999.
Using the Henderson-Hasselbalch Equation
Example :- How would you prepare 10mL of a 0.01M phosphate buffer, pH
7.40, from stock solutions of 0.10M KH2PO4 and 0.25M K2HPO4? pKa of
KH2PO4 = 7.20. The following approach may be helpful in solving this type of
buffer problem in which both the conjugate acid and base are added separately.
Please note that the numbers are not rounded off until the very end and are
rounded based on the limits of the pipets, cylinders, etc. that are required to
accurately measure the calculated volumes.
-
1. Use the Henderson Hasselbalch equation to find the ratio of A to HA.
-
pH = pKa + log [A ] / [HA]
-
7.40 = 7.20 + log [A ] / [HA]
-
0.20 = log [A ] / [HA]
-
1.584893192 = [A ] / [HA]*
-
-
*Since [A ] / [HA] = 1.584893192, we can say that [A ] / [HA] =
1.584893192/ 1. In this case [A ] = 1.584893192; [HA] = 1.
2. Calculate the decimal fraction (part/whole) of each buffer component.
-
A = 1.584893192 / (1.000 + 1.584893192) = 1.584893192 / 2.584893192=
0.61313682
HA = 1.000 / 2.584893192= 0.38686318
3. Find the molarity (M) of each component in the buffer by simply
multiplying the molarity of the buffer by the decimal fraction of each
component.
MA- = 0.01M x 0.61313682 = 0.006131368M
MHA = 0.01M x 0.38686318 = 0.003868632M
4. Calculate the moles of each component in the buffer. Moles = Molarity x
Liters of buffer
-5
molesA- = 0.006131368M x 0.01L = 6.131 x 10 moles
-5
molesHA = 0.003868632M x 0.01L = 3.869 x 10 moles
5. Calculate the volume of each stock solution required to make the buffer
Liters of stock = moles of the buffer component / Molarity of the stock
-5
-4
LA- = 6.131 x 10 moles / 0.25 M = 2.452 x 10 L = 245μL
-5
-4
LHA = 3.869 x 10 moles / 0.10 M = 3.869 x 10 L = 387μL
6. To prepare this buffer, one would use appropriately-sized pipets to measure
and transfer each component to a 10mL volumetric flask and bring the solution
to volume with dH2O.
Hydrolysis of Salt
Salt : an ionic compound, the product of an acid base neutralization reaction:
acid + base ----------------- water + a salt
HCl + NaOH --------------- H2O + NaCl
One way to predict the behavior of a salt in water. A salt may be thought of as
the product of a reaction between an acid and a base.
1. When the parent acid and base are strong, a water solution of the salt is
neutral.
2. When a salt is formed from the reaction between a weak acid and a strong
base, a water solution of that salt will be basic.
3. When a salt is formed from a strong acid and a weak base, a water solution
of the salt will be acidic.
4. When a salt is formed from a weak acid and a weak base, a water solution
of the salt could be acidic, basic or neutral depending on the nature of the
ions.
1-Hydrolysis of the Salt of Strong Acid and Strong Base
The strength of the acid is determined by how far the equilibrium lies to the
right. Qualitatively, this may be judged by the Ka of the acid. A large
Ka indicates a strong acid; a small Ka indicates a weak acid. Strong acids, such
as HCl, have Ka values in the vicinity of infinity. This implies that the
dissociation of HCl is virtually complete, and the equilibrium lies completely to
the right, therefore, the concentration of the acid equals the concentration of
hydronium ions produced. strong bases, such as NaOH, will dissociate
completely. The concentration of OH- in solution will be equal to the
concentration of the strong base.
A typical strong acid problem might be: What is the pH of a 0.010 M HCl
solution? Since HCl is a strong acid, the hydronium ion concentration will be
equal to the HCl concentration:
[H3O+] = 0.010 M
The pH can be found by taking the negative log of the hydronium ion
concentration:
pH = -log[H3O+] = -log(0.010) = 2.00
A typical strong base problem might be: What is the pH of a 0.010 M NaOH
solution? Since NaOH is a strong base, the hydroxide ion concentration will be
equal to the NaOH concentration:
[OH-] = 0.010 M
The pH can be found by first finding the pOH by taking the negative log of the
hydroxide ion concentration, and then converting the pH to pOH. To find the
pOH:
pOH = -log[OH-] = -log(0.010) = 2.00
The pH can then be calculated from the equation pH + pOH = 14:
pH = 14.00 - 2.00 = 12.00
2- Hydrolysis of the Salt of Weak Acid and Strong Base
This salt will hydrolyze to form a basic solution. The salt sodium
acetate will ionize into sodium ions an d acetate ions. The sodium ion
does n o t undergo hydrolysis b u t the acetate ion will react with free
hydrogen ions forming acetic acid.
Na + C2H3O2- + HOH <====> HC2H3O2 + Na + OH
This reaction causes a shift in th e [H ]/[OH ] by removing [H ]. This salt is
classified as a basic salt because th e acetate ion, by definition, is a weak
Bronsted-Lowry base (proton acceptor).
The solution of such a salt is basic in nature. The anion of the salt is reactive.
It reacts with water to form a weak acid and OH- ions.
A- + H2O; ↔ HA + OHWeak acid
for example, the salt CH3COONa. It ionises in water completely to give
CH3COO- and Na+ ions. CH3COO- ions react with water to form a weak acid,
CH3COOH and OH- ions.
CH3COO- + H2O ↔ CH3COOH + OHC(1-x)
Cx
Cx
Thus, OH- ion concentration increases, the solution becomes alkaline.
Applying law of mass action,
Kh = [CH3COOH][OH-]/[CH3CO-] = (Cx×Cx)/C(1-x) = (Cx2)/(1-x) ) ...... (i)
Other equations present in the solution are:
CH3COOH ↔ CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH]
H2O ↔ H+ + OH-,
Kw = [H+][OH-]
From eqs. (ii) and (iii),
log [OH-] = log Kw - log Ka + log[salt]/[acid]
-pOH = -pKw + pKa + log[salt]/[acid]
....... (iii)
...... (ii)
pKw - pOH = pKa + log[salt]/[acid]
pH = pKa + log[salt]/[acid]
Considering eq. (i) again,
Kh = cx2/(1-x)
or
Kh = Ch2/(1-h)
When h is very small, (1-h) → 1
or
h2 = Kh/C
or
h = √Kh/C
[OH-] = h × C = √(CKh) = √(C*Kw/Ka)
[H+] = Kw/[OH-]
= Kw/√(C*Kw/Ka) = √(Ka*Kw)/Kc
-log [H+] = -1/2log Kw - 1/2log Ka + 1/2log C
pH = 1/2pKw + 1/2pKa + 1/2log C
= 7 + 1/2pKa + 1/2log C.
3- Hydrolysis of the Salt of a Strong Acid and a Wea k Base
The solution of such a salt is acidic in nature. The cation of the salt which has
come from weak base is reactive. It reacts with water to form a weak base and
H+ ions.
B+ + H2O ↔ BOH + H+
Weak base
Consider, for example, NH4Cl. It ionises in water completely into NH4 and Cl
ions. ions react with water to form a weak base (NH4OH) and H+ ions.
NH+4 + H2O ↔ NH4OH + H+
C(1-x)
Cx
Cx
Thus, hydrogen ion concentration increases and the solution becomes acidic.
Applying law of mass action,
Kh = [Hx ][NH4 OH]/[NH4+ ]=(Cx.Cx)/C(1-x) = (x2 C)/((1-x))
...... (i)
where C is the concentration of salt and x the degree of hydrolysis.
Other equilibria which exist in solution are
NH4OH ↔ NH+4 + OH-,
H2O ↔ H+ + OH-,
Kb = [NH+4][OH-]/[NH4Oh]
Kb = [H+][H-]
.... (ii)
..... (iii)
From eqs. (II) and (iii)
Kw/Kb =[H+ ][NH4 OH]/[NH4+ ] =Kh
.... (iv)
[H+] = [H+ ][NH4+]/[NH4OH] = Kw/Kb ×[NH4+ ]/[NH4 OH]
log [H+] = log Kw - log Kb + log[salt]/[base]
-pH = -pKw + pKb + log[salt]/[base]
pKw - pH = pKb + log[salt]/[base]
pOH = pKb + log[salt]/[base]
Relation between Hydrolysis constant and Degree of hydrolysis
The extent to which hydrolysis proceeds is expressed as the degree
of hydrolysis and is defined as the fraction of one mole of the salt that is
hydrolysed when the equilibrium has been attained. It is generally expressed
as h or x.
h = (Amount of salt hydrolysed)/(Total salt taken)
Considering again eq. (i),
Kh = x2C/(1-x) or
Kh = h2C/(1-h)
When h is very small (1-h) → 1,
H2 = Kh × 1/c
or h = √(Kh/C)
= √(Kw/Kb * C)
[H+] = h × C = √(C*Kh)/Kb
log [H+] = 1/2 log Kw + 1 1/2log C - 1/2log Kb
pH = 1/2pKw - 1/2 log C - 1/2 pKb
= 7 - 1/2 pKb - 1/2log C
4-Hydrolysis of the Salt of Weak Acid and a Wea k Base
The ions of this type of salt will hydrolyze to form a solution that may
be acidic, basic, or neutral, depending on th e strength (ionization
constant) of th e weak acid an d weak base that formed th e salt. If the
ionization constant of the weak acid is greater
Maximum hydrolysis occurs in the case of such a salt as both the cation and
anion are reactive and react with water to produce H+ and OFT ions. The
solution is generally neutral but it can be either slightly acidic or slightly
alkaline if both the reactions take place with slightly different rates. Consider,
for example, the salt CH3COONH4. It gives CH3COO- and ions in solution.
Both react with water.
Other equilibria which exist in solution are:
CH3COOH ↔ CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH]
NH4OH ↔ NH+4 + OH-,
H2O ↔ H+ + OH-,
Kb = [NH+4] [OH-]/[NH4OH]
Kw = [H+][OH-]
..... (i)
..... (ii)
..... (iii)
From Eqs. (i), (ii) and (iii),
Kh = Kw/Ka.Kb = [CH3COOH][NH4OH]/[CH3COO-][NH+4]
Let C be the concentration and h be the degree of hydrolysis
.... (iv)
Kh = h2/(1-h)2
When h is small, (1-h) → 1.
Kh = h2
h = √Kh = √Kw/Ka*Kb
[H+] Ka × h
= Ka × √Kw/Ka*Kb
= √Kw * Ka/Kb
-log [H+] = -1/2log Ka - 1/2log Kw + 1/2log Kb
pH = 1/2pKa + 1/2pKw - 1/2pKb
= 7 + 1/2pKa - 1/2pKb
When pKa = pKb, pH = 7, i.e., solution will be neutral in nature.
When pKa > pKb. The solution will be alkaline as the acid will be slightly
weaker than base and pH value will be more than 7. In case pK a < pKb, the
solution will be acidic as the acid is relatively stronger than base and pH will be
less than 7.