TITRATION CURVES

TITRATION CURVES
I- STRONG ACID vs STRONG BASE :
50 mL of 0.05 M HCl wth 0.1 M NaOH
- Initial [H 3 O + ] = 0.05 pH = 1.3
- with 10 mL NaOH
[H 3 O + ] = [50 x 0.05 – 10 x 0.1]/60 = 2.5 x 10 -2 pH = 1.6
- with 25 mL NaOH (Equivalence)
[H 3 O + ] = [OH - ] = Kw = 1.0 x 10 -7 pH = 7.0
- with 25.1 mL NaOH
[H 3 O + ] = [25.1 x 0.1– 50 x 0.05]/75.1 = 1.33 x 10 -4 pH = 10.12
Remark : The break interval of dil solutions is less than that of conc.
solutions. While in conc. solutions MO, ph ph & BTB give the same e.p.;
in dilute solution only BTB is appropriate.
II- WEAK ACID vs STRONG BASE
Equations Governing a weak acid (HA) Titration
Fraction (F)
titrated
Present
Equation
F = O HA [H + ] = Ka . Ca
O < F < 1 HA / A - [A - ]
pH = pKa + log ________
[HA]
F = 1 A - Kw
[OH - ] =
______ .Cs
Ka
F > 1 OH - / A - [OH - ] = [Excess Titrant]
• 50 mL 0.1 M HA (Ka = 1.75 x 10 -5 ) with 0.1 M NaOH
- Initial [H 3 O + ] = Ka . Ca = 1.75 x 10 -5 x 0.1 = pH = 2.88
- with 10 mL 0.1 M NaOH
Cs (10 x 0.1)/60
pH = pKa + log _____ = 4.76 + log ____________________________ pH = 4.16
Ca (50 x 0.1- 10 x 0.1)/60
- with 25 mL 0.1 M NaOH; half neutralization.
1

pH = pka pH = 4.76
- At equivalence point
All acid is converted to acetate
Kw 10 -14 50x0.1
[OH - ] = _____ ____________
.Cs =
___________ = 5 x 10 -6 pH = 8.73
Ka 1.7x10 -5 100
- with 50.1 mL NaOH
0.1 x 0.1
[OH - ] = _________________ = 1.0 x 10 -4 pH = 10.00
100.1
III- WEAK BASE vs STRONG ACID
Equations Governing a weak base (B) Titration
Fraction (F)
titrated
Present
F = O B [OH - ] = Kb . Cb
Equation
O < F < 1 B / BH + [B]
pH = (pKw - pKb) + log ________
[BH + ]
F = 1 BH + Kw
[H + ] =
________ . Cs
Kb
F > 1 BH + / H + [H + ] = [Excess Titrant]
2

NH 3 + H + + Cl - NH 4 + + Cl -
Derive the curve by titration of 50 mL 0.1 M NH 3 with 0.1 M HCl.
10
8
pH
6
2
10
Ka = 10 -9
8
pH
Ka = 10 -7
Ka = 10 -5
Ka = 10 -3
mL 0.1 M NaOH
2
Ph ph
mL 0.1 M HCl
MO
IV- KCN vs HCl K a HCN = 2.1 x 10 -9
50 mL 0.05 M KCN vs 0.1 M HCl
CN - + H 2 O HCN + OH -
- Initial
Kw 10 -14
[OH - ] =
________ . Cs =
______________ x 50 x 0.05 pH = 10.69
Ka 2.1 x 10 -9
10 x 0.1 (50 x 0.05) – (10 x 0.1) 1.5
- with 10 mL HCl; HCN = ___________ ; CN - = ____________________________ = ____
60 60 60
Buffer equation
pH = pKa + log
[CN - ] 1.5
___________ = 9 + log
[HCN] 1
- At equivalence [CN - ] = [HCN]
______
pH = 8.85
[H + ] = Ka . Ca pH = 5.08
- Excess HCl (26 mL)
(26 x 0.1) – (50 x 0.05)
[H 3 O + ] = _____________________________ = 1.32 x 10 -3 pH = 2.88
76
V- NH 4 Cl vs OH -
Derive the curve through titration of 50 mL 0.1 M NH 4 Cl with 0.1
M NaOH. Draw and what indicator is used ?
3

VI- Polyprotic acid :
Example : Maleic acid
HC.COOH
HC.COOH
H 2 M + H 2 O = H 3 O + + HM - ; K a1 = 1.2 x 10 -2
HM - + H 2 O = H 3 O + + M 2- ; K a2 = 5.96 x 10 -7
Because pK 4 & pK 1 + pK 2 9; it is feasible to titration with 2
breaks.
pH
K a2 . K b1 = Kw
HX - H + + X 2-
Maleic acid with NaOH
(6)
X 2- + H 2 O HX - + OH - mL OH - titrant
(4)
(5)
(1)
(2)
(3)
(1) [H + ] = K 1 Ca
[HX - ]
(2) pH = pK 1 + log _________
[H 2 X]
[H + ] [X 2- ]
(3) [H + ] = K 1 K 2 HX - H + + X 2- , K 2 = _______________
[HX - ]
H 2 X
[X 2- ]
(4) pH = pK 2 + log ___________
[HX - ]
[HX - ] [OH - ]
(5) Kb = __________________
[X 2- ]
(6) xss [OH - ] = Cb
[H + ] [HX - ]
H + + HX - , K 1 = _______________
[H 2 X]
4

• H 3 PO 4 :
K 1 = 7 x 10 -3 ; K 2 = 7.5 x 10 -8 ; K 3 = 4 x 10 -13
Two breaks belong to K 1 & K 2 while K 3 is too weak to be titrated.
• H 2 SO 4 :
Gives only one break since pK 4
A = Phosphoric acid
B = Oxalic acid
C = Sulphuric acid
pH
A
C
B
A
B
C
VII- Polyequivalent bases
Ba (OH) 2 with HCl
mL OH -
(1)
pH
(2)
(3)
(4)
(5)
(6)
mL H + titrant
(1) [OH - ] = Kb Cb
(2) pOH = pKb 1 + log
[Ba(OH) + ]
______________
[Ba(OH) 2 ]
5

(3) pOH = 1 / 2 (pKb 1 + pKb 2 )
Ba(OH) 2 Ba(OH) + + OH -
Ba(OH) + Ba 2+ + OH -
[Ba 2+ ]
(4) pOH = pKb 2 + log _______________
[Ba(OH) + ]
(5) Ka =
(6) [H + ] = Ca
[H + ] [Ba(OH) + ]
_____________________
[Ba 2+ ]
Example : Sodium carbonate (Na 2 CO 3 )
Ba 2+ + H 2 O Ba(OH + ) + H +
V 1 = 1 / 2 V 2
2-
pure CO 3
V 1 > 1 / 2 V 2 CO 2- 3 + OH -
V 1 < 1 / 2 V 2 CO 2- -
3 + HCO 3
pH CO 3 2- /HCO 3
-
ph ph
MO HCO 3 - /CO 2
V 1 V 2 ml H +
6

• BUFFERS; Keeping the pH nearly constant
A buffer is defined as a solution that resists change in pH when a
small amount of an acid or base is added or when the solution is diluted.
A buffer solution consists of a mixture of a weak acid and its conjugate
base OR a weak base and its conjugate acid.
- Consider Acetic acid HOAc (Ca) and its salt NaOAc (Cs)
NAOAc Na + + OAc -
HOAc H + + OAc -
[HOAc]
[H + ] = Ka _______________
[OAc - ]
pH = Ka + log
and generally
[OAc - ]
_______________
[HOAc]
pH = pKa + log
[Ionized]
____________________
[HOAc]
i.e. The pH of a buffer is determined by the ratio of the conjugate acidbase
pair concentration.
The above equation is known as Henderson-Hasselbalch
equation. Similarly. For weak base and its salt;
[Salt]
pOH = pKb + log __________
[Base]
Example : Calculate the pH of a buffer prepared by adding 10 mL of
0.1 M HOAc to 20 mL of 0.1 M NaOAc
pKa = 4.76
Cs
pH = pKa + log _________
Ca
0.067 mmol/mL
= 4.76 + log ________________________ = 5.06
0.033 mmol/mL
since for
HOAc: 0.1 x 10 mL = M HOAc x 30 [HOAc] = 0.033 mmol/mL
OAc - : 0.1 x 20 mL = M OAc- x 30 [OAc - ] = 0.067 mmol/mL
7

Remarks :
- The ionization of acid is suppressed by salt addition and, therefore could
be neglected.
- We can use mmole instead of molarity, the ratio is the same.
- The buffer can be obtained by mixing an excess of weak acid with
strong base or excess salt with strong acid.
- If the solution is diluted, ratio remains constant and the pH does not
change, or in other words dilution does not change the ratio of the
buffering species.
- Addition of small amount of strong acid is converted to weak acid
through reaction with OAc -
H + + OAc - HOAc
with a small change of [A - ]/[HA] and pH. Meanwhile, the same amount
of acid added to unbuffered (NaCl) solution leads to marked pH decrease.
The amount of acid/base that can be added without causing a large
change in pH is governed by the BUFFER CAPACITY of this solution.
Such capacity increases with concentration of the buffering species. The
higher their concentrations, the more acid/base it can tolerate.
The buffer capacity (buffer intensity, buffer index, buffer value) of
a solution is defined as
= d C BOH / d pH = -d C HA / d pH
where d C BOH & d C HA represent the number of moles per liter of strong
base or acid, respectively, needed to bring about a pH change of d pH.
The is a positive number, the larger it is, the more resistant the solution
is to pH change.
For weak acid/conjugate base buffer solution of greater than 0.001
M, the buffer capacity is approximated by (as derived by Vanslyke).
C HA . C A-
= 2.303 ___________________
C HA + C A-
i.e. for 0.1 mol/L HA and 0.1 mol/LA -
0.1 x 0.1
= 2.303 ___________________ = 0.05 mol/L per pH
0.1 + 0.1
If we add NaOH until it becomes 0.005 mol/L the change in pH
d pH = d C BOH / = 0.005/005 = 0.1 = pH
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In addition to concentration effect, the is governed also by HA/A -
ratio.
It is maximum when the ratio is unity, i.e. when pH = pKa. This
corresponds to the midpoint of a titration of a weak acid. In general,
buffer capacity is satisfactory over pH range of pKa ± 1.
Example :
A buffer solution is 0.2 M HOAc & NaOAc calculate change in pH
upon adding 1.0 mL of 0.1 M HCl to 10 mL of this solution.
Solution :
Initially pH = pKa since [HOAc] = [OAc - ]
pH = 4.76
After acid addition
HOAc = 2.0 + 0.1 = 2.1 mmole
OAc - = 2.0 – 0.1 = 1.9 mmole
1.9/11
pH = 4.76 + log _________ = 4.71
2.1/11
Compared to the addition in unbuffered solution the concentration
[H + ] 10 -2 M and pH would be 2.0.
The above phenomenon (i.e. resistance to pH change) is true as
long as we do not exceed the amount of the buffer reserve.
Problem : Consider to 0.1 M HA and NaA mixture added in increment is
0.01 M NaOH, pKa = 4.76 calculate .
- HA + OH - = A - + H 2 O
NaOH 0.1 0.1
0.01 (0.1-0.01) 0.1 + 0.01
[Cs] 0.1 + 0.01
pH = pKa + log ________ = 4.76 + log _______________ 4.85
[Ca] 0.1 - 0.01
pH = 4.85 – 4.76 = 0.09
0.01
= ___________ = 0.11
0.09
NaOH 0 0.01 0.02 0.03 0.04 0.05 0.06
PH 4.76 4.85 4.94 5.03 5.13 5.24 5.36
- 0.11 0.11 0.11 0.10 0.09 0.08
9

Remarks :
(1) is not fixed, it depends on base addition.
(2) more base decreases , since acid reserve decreases.
(3) is maximum when [Cs] / [Ca] 1
(4) more efficient capacity () as components concentrations are
increased.
Calculate the pH of 0.1 M HA & 0.1 M NaA -
(a) before dilution
(b) after 50 times dilution.
(c) 10,000 times dilution (pKa = 4.76)
[Cs]
(a) pH = pKa + log _________ = 4.76
[Ca]
(b) mM HA = 0.1/50 & mM A - = 0.1/50
& since CV = C` V`
0.1/50
pH = pKa + log ___________ = 4.76
0.1/50
[H 3 O] + [A - ] [H 3 O + ] (0.1 x 10 -4 + [H 3 O + ])
(c) Ka = ______________ ; 10 -4 = ____________________________________
[HA] (0.1 x 10 -4 – [H 3 O + ])
since A - + H 2 O HA + OH - ; Cs = 0.1 x 10 -4 + [A - ]
& HA + H 2 O A - + H 3 O + ; Ca = 0.1 x 10 -4 – [H 3 O + ]
The above quadratic equation is solved for [H 3 O + ], pH 5.0.
Comment : While 50 times dilution does not change pH of buffer, 10 4
fold dilution increases the pH by 0.24.
Bates expressed this effect quantitatively in terms of DILUTION
VALUE which is the change in pH on dilution: positive dilution value;
pH rises on dilution : negative dilution value; pH decreases on dilution.
Remark : Temperature change affects pH of basic buffer more than pH
of acid buffer. Why ?
10

Buffers in Drugs and Biological system :
(1) Salicylic acid in soft
glass bottles is
influenced by glass
giving Na salicylate
forming buffer
system.
pH
5
4
3
2
1
HA + A -
HCl
M
10 -2 10 -4 10 -6
Dilution effect on 3 solutions
(2) Ephedrine base + ephedrine HCl act as buffer i.e. the drug itself may
act as its own buffer. Nevertheless, air CO 2 and also glass alkalinity
may disturb buffer effect and the need to add buffer is a must to
maintain the system within a certain pH.
(3) Buffered aspirin is to maintain pH of gastric juice at relatively high
value to avoid aspirin hydrolysis.
(4) Blood pH is maintained at 7.2-7.4
- Inside the cells are KH 2 PO 4 , K 2 HPO 4 & hemoglobin/ oxyhemoglobin.
- Outside the cells are plasma protein, NaH 2 PO 4 (acid salt) and
Na 2 HPO 4 (basic salt) & CO 2 /HCO - 3 . The main buffer is
CO 2 /HCO - 3 .
[HCO - 3 ]
_____________
pH = 6.1 + log
[CO 2 ]
At pH 7.4, [HCO - 3 ] / [CO 2 ] is 20
> 20 alkalosis; < 20 acidosis and each reflects specific clinical
symptoms. For example, diabetes may give rise to acidosis which can be
fatal.
HA
11