Volumetric Analysis
Titrimetric Methods
a la~ge and p, ~wful
itlim fl ic mt/yods i11clud
Tdur ·that
t ration
that
~mup of ( 11anllfat11 e J /'{), e-
,u · b1 • :f on ,n ·;a urtn · th ·· ainount of a reag •nt of kno~vncon ·en-
j·. consun1. d
b,v an an. 1/1 re
_,
~
itrations ar wid -y Lsed in analy .ical chemistry a dete mine acids, bas s oxidantsJ reductants me l ions, proteinsi and man othe species. Titr tions are a ed
on a reac i·on between the ana lyte and a standard rea ent knovvn as th _titrant. The
1
reac.ionisof nown andreproducible stoichion1e ry.
Volumetric ti.trimetry involv • rnecr u in,g the vohune qfa soluhon of known
t.Oncentration 1}urt i needed to ttact e- ·enliall).. --·o n111J t Ir vi-11- the a·nc:11 t:· _
.
:
Types
of Titrimetric Methods
tI J l l l •
Complexometric tra ons
SO E TERMS USED I
VOLUMETRIC TITR METRY
A standard solu on: is a reagent of exactly known concentra on that
is used in a trimetric analysis
Titra on: is a process in which a standard reagent is added to a
solu on of an analyte un l the reac on between the analyte and
reagent is judged to be complete.
Back- tra on : is a process in which the excess of a standard solu on used to
consume an analyte is determined by tra on with a second standard solu on.
Back- tra ons are o en required when the rate of reac on between the analyte
and reagent is slow or when the standard solu on lacks stability.
-
amp,I ~ the amount of ph.o phate in a
rnea ured x
of . tandard H r itrat
to the forn1ati n
f in olub e i
il er nitra
,e
i
Iv
then b
am.p] can be deteil.'n1in d by ad •in
o a olu1• •on of the ~ 1nple \.Vhich leads
pho p at .
k-titrated w·th a
ta:ndaird
oluhon of pota
1um
thio - anate:
, . , re th
phat
am.oun
ion plu th
,
f
am,
:ii:s
qui - alent to th
nt of thiocyanate u .ed for the back- ti ration.
. i l . e.1.~
nitr it
heinicallc-
,
aJllou:nt
o·
ph ·
-
nc oiut ·ta 1it atio11 i· a·the, .ti 1point · ea-h·, wlle1 i1e an1ount
f add d titran . , ·~11 -•ca l · .qu ..-. l n· to • ea1nount of anal:: t ·~ n f ar -1 or
•·. a1.npl th . .:qui.- ,_ t·n· ,e . ti.nth.· t tration f s d1um chlorid . ·1th ·.il er n1.tr t
·ccur- after exa. tl · l • ol of ~il.er ion has been added for •.•ach ~nole of chi iide ion
,.e·.· q Ii al
11
1
, i
1
1
··n ·h
-.a
p e
1.
•
1
End point: is the point in a tra on when a physical change occurs that is
associated with the condi on of chemical equivalence. In volumetric methods, the
tra on error Et is given by
Et = Vep - Veq
where (Vep) is the actual volume of reagent required to reach the end point and
(Veq) is the theore cal volume to reach the equivalence point.
Indicators: are organic materials (almost weak acid or weak base) are o en added
to the analyte solu on to produce an observable physical change (the end point)
at or near the equivalence point.
.th b ha ior of an ac •d-type indi ,_ ator
Hin
acid
I r
In
+ H,- O -
+
H O ;;:::::=
+ HO +
In-
b·- se-
c,
+
r
+
OH -
T -equilibrium- on tan, e pre -ion or the d'" ociation of an cid-ty -_ e indicaor tak - th fonn
[H 3Q + ] [In- ]
K ;;;; - - - - a
[Hin]
(14-1)
R, a I an·,...,,itn·g le d. to
I( 4 -2)
Th hurna e , i, not • y -, 1 . 11 e t -oJor differences in oiution on 'llntng
a mn:i tu e ot Hin ,'tnd In-~ ·p rticula 1 \.vhen th -_ ratio [HI 'l/[Jn - J i • gr at ! an
u •u1 10 or s all -r tha ah. ut O I. on -· _,quentl , the co]or ·ban. ·e det _ted b , an
I
1
a •1· e ob; r, r occu . within a li1niti d ran° _ of cono trc; tio rati
fro1n bout
IO to about 0 . 1 . .At a1 eater or n1.al r :ratio ~ th • ~01or app ars e e 1 • ally cons ant
o the ye and is independen t of th ratio. As a re ult, we can rit that th a
ag
indica or, Hin e hibic. its pure a id oior hen
[Hin ]
[In-]
an d it -. ·as.e. color when
[ Hln ]
1
.0
----..:::::::::-
[In - ]
S -o
fort - -e
ac-d color,.
r
3 o
-+-
J
1 OK.a
an.ct in the same way for the f . I] b.a· e co1o •.
To ob ain th indicator pH range.- we take the negati re logarithms. of th two
.
e pr · ion :
pH(· cid color) = - log ( 1OKa) = pKa
+
1
pH(ba ic , olor) - - log .O. lKa) = p¥ - 1
indicator pH range
== pKa
+ 1
( 14 -3)
1L
6
6
6=
-+::E---:C 3 <=> ---+-
-
C>C> -
Ph.ph Indicator for acid – base tra on
TABLE 1 4 - 1
Some Important Acid/B.ase Indicators
ComDJ.on Nan1.e
Tb.y mol blu
Transition. Range, pH
P ,Ka *
Color Changet
Indicator 'fype't
1.2-:2 ..8
J. .65§
8 .96§
R- Y
Y-B
l.
3 .46§
R- Y
R-0
Y- B
2
2
l
2
1
l
8 .0- 9.6
Methyl yellov,,..Methyl o :-ange
B romocre:sol green
Methyl red
Bromocresol purp]e
Bromothymol blue
2.9-4- .0
3.1-4.4
3.8- 5.
4.2----6.3
5.2---0.8
6,2-7.6
6.8-8.4
4 .66§
5.00§
6.1 2§
7.10§
R-Y
Y- P
Y- B
Y-R
Phenol red
Cresol purple
Pheno]phthal in
7.6-9.2
8.3-1 0.0
Y-P
Thy:molphthalein
Ali arin.., e low GG
9.3-1 0.5
10--12
C- B
C- Y
7.81§
C- R
1
i
1
1
2
13B STANDARD SOLUTIONS
Slandard solutions play a cenlral rule in all 1i1rin1elric mcthocls of analysis. Therefore. we need to consider the desirahle properties for soch solulions. hosv lhcy arc
prepared, and hosv !heir concenlralions are expressed. The ideal standard solulion
for a tilri,netric n1ethod will
I. be sufficien1ly s!able so lhat it is necessary 10 de1cnnine irs concentration only once:
2. reacl rapidly with the analyle so 1hat the lirue required between additions of
reagent is n1inimized:
3. react n1ore or less complelely 1vith the analyle so 1ha1 satisractory end poinls are
realized; and
4. undergo n seleclive rencrion wilh the analyle lhat can be <lescribcd bv• :r balanced equation.
A primary standard: is an ultrapure compound that serves as the
reference material for a trimetric method of analysis.
A secondary standard: is a compound whose purity has been
established by chemical analysis and that serves as the reference
material for a trimetric method of analysis.
I
c:e>1r11 c..
~
~
c3I
c:::t:I Y
F11 c::I ,c31
r
~t:~
-
e>F
c31a-~..~.><.
..-. -
k.
-
....._.,,...c:J;,
c:::>
I
t 1
I
ell
i=3I
L...1
~
c::> "t='
,c3I
lr7
<::)
Cl·
The requireIDents of a titration are as follows:
8t .Bt'.~I\.Vell_
.~r·:r:)~::.- : aefi.hed/ana•
Jt~\'iri~~.o,t:·kriowrf."teactiorF
j: •~~~,-·•·§i .#?tc.~~~~@f.
,
·
·
T
o~.
t
'
·
·
·
i
s:
;
,
'
.
~et€!(IBi
-\·
betweert\'the·:_: ana1 ·: "te~!-ancr·: the:.•titrant)i lli-.thet>
-
.
:·: 4.::,-:·r here··should be a marked·change in some property of:the solutton-·when:·:
/-·>~(}hJ·- re)it(idii ·is complete. Thi~--~ ay ~e ~ ch.ange _in c9lo~_of·the·solution·: :" _
_-_\/_.;_ :·:or·m :)Join~.·eiec.tricaf\n~ :othbt f,hj,sic·a f·prciperty··of the:· soiutidn~ in-.the_:·•
_:-:·:::··:_-\·titratiqn_·:of._c1~etic acid:wi~ ~odiuni hydro~i~e~ tli~re _
is _a m_arked:increase·
•·:'/ :/ -' iii:the:·pB"of"the..solution \vlleri : $ e.-'i~actiori }s· cbmpiete. A"color change.
:;·:_.: ·;;>· is·:_usually b:r;ought aboufby)~ddi_tiorf_o f an·_1ndic.atof, wh9se· color is de~
y:·_> P.~n4ent o_n .th~ properties or' ili¢_'. s.9~µti9_n,: for exampi_e; 'the plI. •
. . .•
The J)Oi~t at ~hich etjrii;hlJrit st~iChioriletric all1ount of titrant is.
t··_:-·•.: :_-;· --~_
dded is _called the.eqmvai~ii_
c~ p_oint. Tue .-point at which the reaction is
-_·__. _· -:- obs~rv~d to be copiplete: is: cajl_~d.th~:·end point,· that is, whert ·a change ·
_;.:·:/ :..•·:·in some ·property of the solut.id~-)s·det~-cted. Th~ en.d pomt ·shoulc(·coin~ ·_.
::-:< •.·•_.·c14e·:.\vith_"the equiyalence ·pojntor l:>e it a~eproduc1ble interval from it.
\ 6.'( ThlreactionShould be
Tha~ is,the equilibriuril of the re- .
· i_--.:··:::--a6tion·:·shouid be· far to th~··nght.s6 "that a suffic1ently_shdr]J_'.challge will .
••.·: : ...··.·occur at _the ~nd p·oin~ to -obtain.•tqe desired:accuracy. If.the equilibrium ·
:\_;,,_\._/does_nqt lie f~ to the_right,·~en tjiere will ~e gradual change lI1 tlle prbp~ •
(·i\:.i):_:: erty:• .-#J.arkmg the end poirit (e♦g., · pH) .: fillct · this willbe:·difficult .to detect ·
i)\.:.-: . -:--•·._:· _.--.,.·: j · ·. --: .··. ·.. .. .. . .•:.. . ·:: . ' ·.·.--·._. · ·:·. .::··.- -: ·. . .- ...• . --·.·. . •.• . ... ·•, ' •
.. . . ;:. prec1s~ y... , ... . ... . . .
.. .- ·. . . . -.·.- · · .•· .- . . · . · ·.· ··,
· ··
an
K
~i
qha.Iititativi.
- · .. . , •• • •• -
-. '
• • •
••
• •
•
•
• •
•
•
•
•
• •
• •
•
•
•
•
•• ••
'
•
•
- •
'
•
• •'
-
• -
;, •
•
•
:
•
•
t
• • :. ' .
•'
·: •
•
•'
: •
•
_,
Titer:
titer is the weight of ~~lyte that is chemically equivalent to 1 mL of the titra~t,
usually expressed in milligrams. For example, if a potassium dichromate solution
has a titer of 1.267 mg Fe, each milliliter potassium dichromate will react with
1.267 mg iron, and the weight of iron titrated is optained by simply multiplying
the volume of titrant used by the titer. The titer can be expressed in terms of ~y
form of the analyte desired, for example, milligr~s FeO or FeiO3 .
Titer = milligrams a.1""1.alyt:e that react
witl-i. 1 rnL o f tit1-anr_
A + T _______ Product
Titer of A = Molarity of T(mmole / ml) x mole ra o(mmole A / mmole T) x M.WT (A)
Example S.37
A standard solution of potassium dichromate contains 5.442 g/L. What is its titer
in terms of milligrams Fe3O4?
Solution
The iron is titrated as Fe2+ and each Cr2O12- will react with 6Fe2 + or the iron from
2Fe3O4:
6Fe2 + + Cr2O72 - + 14H+ ~ .6Fe3 + + 2Cr3 + + 7H20
The molarity of the K2 Cr2O7, solution is
5.442 g/L
294.19 g/mol
g/L
= 0.01850 mol/L
Therefore the titer is
0 _01850(_mm_o_l_K_2_C_r2_0_1)
mL
x _2 (
1
mmol F~304 ) X 231 _54(_m_g_F_e_3_O_4_)
mmol K2Cr2O1
mmol Fe3O4
= 8.567 mg Fe3O4/mL K2Cr2 O1
titr
e
...
•
""
on curu
p
ssoc ted
...
Cl
•
pat of pH
s
ti r
or b se
t
determ n·ng endpo·nts
e k
A
- orb .
e
l
■
mount o ti r nt
t
-uc
rnd d""
C
ong (
rves are
Ill
■
SOCJa 10n
pie e )
s e ul or
canst nts of
0
1
RCID-BHSE TITRHTIONS
~tron~ ~ci~ versus ~tron~ ~ase-me fasq litrntions
In the case of a strong acid versus a strong base, both the titrant and the analyte are completely ionized. An example is the titration of hydrochloric acid with
sodium hydroxide:
(8.1)
•
Example 8.1
Calculate the pH at 0, 10, 90, 100, and 110% titration for the titration of 50.0 mL
of 0.100 M HCI with 0.100 M NaOH.
Solution
At 0% pH= -log 0.100 = 1.00
At 10%, 5.0 mL NaOH is added. We start with 0.100 M X 50.0 mL
H+. Calculate the concentration of H+ after adding the NaOH:
=
5.00 mmol
mmol H+ at start
= 5.00 mmol H+
mmol OH- added = 0.100 M X 5.0 mL = 0.500 mmol OHmmol H+ left
= 4.50 mmol H+ in 55.0 mL
[H+] = 4.50 mmol/55.0 mL = 0.0818 M
pH = -log 0.0818 = 1.09
_;At 90%
mmol H+ at start
mmol OH- added= 0.100 M
II1J11ol H+ left
= 5.00 mrnol H+
x 45.0 mL = 4.50 mmol OH= 0.50 mmol H+ in 95.0 mL
[H+] = 0.00526 M
pH = - log 0.00526 = 2.28
At 100%: All the H+ has been reacted with OH-, and we have a 0.0500 M solution of NaCl. Therefore, the pH is 7.00.
At 110%: We now have asolution consisting of NaCl and excess addedNaOH.
mmol OH- = 0.100 M X 5.00 mL =0.50 mmol OH- in 105 mL
[OH-] = 0.00476 M
pOH = -log 0.00476 =2.32; pH= 11.68
14
12
.p---
r
10
'
Phenolphthalein
transition range
a
6
' >
◄
.
4___,
2
L--61"
0
4-0
20
0
60
100
BO
140
120
____ mL NaOH
Fig_ 8 _l _
C>f
0_ 1
...lkf
'T:i.t:ra.tie>ri
:E-3.:C::1
c::-....ir""V"e:
""V'"e:r:S""L:I.S
C)_
1
1 0 0 mI.....,
r---:l".a..~:E----:1:-
fe>r
__.lkf
la~le ~.l
Equations Governing a Strong-Acid (HX) or Strong-Base (BOH) Titration
Strong Base
Strong Acid
Fraction F
Titrated
Present
F=O
HX
O<F<l
F=l
F>l
x-
HXIXOH_/X_
Equation
Present
[H+] =[HX]
[H+] = [remaining HX]
[H+] =vfw (Eq. 7.13)
[OH-] =[excess titrant]
BOH
BOHfB+
B+
H+ffi+
Equation
[OH-] =[BOH]
[OH-] =[remaining BOH]
[H+] =vfw [Eq. 7.13)
[H+] =[excess titrant]
EXAMPLE 14-1
·Generate the hypoth, tical titration cuTv,e fo the titration of 50.00 rnL of 0.0500
M HCl with 0. ~000 M . 0H.
1
Titrating a Strong Base with a Strong Acid
Ti ratio .. 1r\ri. , f. r ·,t o, b1.. . a •d, •v d •n n an l. 0rou · .a- to tho ·: for
. t1o ig acid·· . hort of :h qu ivalenc :· point th . .olution i.s ht hi-.· ba·,1c tl1e
hyd ·ox·,de ion concentration bein nume:ically r -lated to the analytical molarity of
the bas . The ,olution i~ n unal at the quivalenc . po n~ and b con1- s.aci ic rn he
• 0 •on .
•. yond the equivalence po.~
1nt· hen he hydtonium ion •one ntration ii •o he an ·•,tlca cone n ation o' th .. ,;c- s· .tro a,id
1
I
I
1
·1
I
EXAMPLE 14--2
----•
- - i i~ I
■ fi .. -••-n#fiqaz....,.._
I - , - ,- .
..............
-..-...,..,,111'1
_ _ _ _4_
i ■_,,iii•111-v~
...,,,..
..-.,...._r
¥WSI}.--:,.,
.. _
• .,______
.., ..__
11
,.,._..__
I
1"1"1-
..,~-.,-,...--
Calcula1e the pH dur~ng die tii1ratio.n of 50.00 mL of 0.0500 M -aOH with
0. 000 M HCl after th addition o." b lfollo ·n,g 10 um --_o reag nt: (a) 24.50
111
(b) 25 ..00 mL (c) 25.50 rnL.
1
(a)
At
24 .
50
mL
add
d;
[H
o +] i .
-to,icbiometric ,consid. ration- but
[
10
·.
-]
[H O +
or~ginal
mall and
be
ut_
be
from [O·H-]
1
v er y
can
UOm
can.n
aOH - no. mmol
to al volume o · •so]ution
=
so.oo x o.osoo _
~o oo +
24.
Kw/ 6.71 X 1o - - )
=
1 .00 X 10 -
1 _
=
4
so x o. :ioo
= 6 .7
o
1.49 X ·0 - 11 M
pH = -lo· ( I . .19 X 10-
i"
1noll '
C
=
=
co -
d
from
obtained
=
1;10
1
1
0.83
1
X
added
10- - M
/'(6.71 X 10- 4 )
(b) This i .. the equi • a]enc
1 .o
[H ,o + ] =
pH
A
[H o + ]
point wher
=
-
lo.eJl .00 X
~
,x
0-7
10-
=
4
[OH-]
=
1.00
x
0- 1
= 7 .00
. 5 . 0 mL add d
-
, H
25.50 X 0.100 - 50 .00 X 0.0500
75. 0
14 .---.--....---.....----- - -- - --
------,
Figure 14·4 Titrari nc1rv , for
NaOH ,jtJ1 HCI. Cur . A: 50.001nL
1
10 - - - - -
of 0.0)00 aOH
l Cl. Cun,eB· 0.00 mL of O. 05
aOH vHh 0,.0100 MHCl.
M
::r
Q.
4
IO
J
20
_5
o!ume HCJ, mL
0
·
with 0.1000
•
I 14C I TITRATION CURVES FOR WEAK ACIDS
Pour distinctly different types of calculations are needed to derive a titration curve
for a weak acid (or a weak base):
l . At the beginning, the solution contains only a weak acid or a weak base, and the
pH is calculated from the concentration of that solute and its dissociation
constant.
2. Afler various incre111enls of titranl have been added (in quanli lies up lo, bu! not
including, an equ ivalent amount), the solution consists of a series of buffers.
The pH of each buffer can be calculated from the analytical concentrations of
the conjugate base or acid and the res:idual concentrations of tJ1e \.Veak acid or
base.
3. At the equiva lence point, the solution contains only the conjugate of the weak
acid or base being titrated (that is, a salt), and the pH is calculated from the concentration of this product.
4. Beyond the equivalence point, the excess of strong acid or base titrant represses
the acidic or basic character of the reaction product to such an extent rhar the
pH is governed largely by the concentration of the excess titrant.
Weali Reid versus Strong ease- -R Hit
l.ess Straightforward
C>::.E-J: -
:E-3:C>.A.c -+- N"a.
:E-3'::zC>
N"a..
C>.A.c -
la~le ~-~
Equations Governing a Weak-Acid (HA) or Weak-Base (B) 1itration
Weak Base
Weak Acid
Fraction F
Titrated
Present
F=O
HA
0 <F< 1
HA/A-
Equation
Equation
Present
B
.
CB
pH= (pKw - pKb) + log -c (Eq. 7.56)
BH+
[OW] =
[OH-]
F
(Eq. 7.32)
= [excess titrant]
[W] =
[H+]
~
= [excess titrant]
(Eq. 7.39)
Example 8.2
Calculate the pH at 0, 10.0, 25.0, 50.0, and 60.0 mL titrant in the titration of
50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.
Solution
At 0 mL, we have a solution of only 0.100 M HOAc:
(x)(x)
0.100 -
X
=
[H+] =
1.75 X 10- 5
x =
1.32 X
10- 3 M
pH= 2.88
At 10.0 mL, we started with 0.100 M X 50.0 mL = 5.00 mmol HOAc; part has reacted with OH- and has been converted to OAc-:
mmol HOAc at start
mmol OH- added= 0.100
M
mmol HOAc left
=
=
5.00 mmol HOAc
X 10.0 mL
1.00 mmol OH= mmol OAc- formed in 60.0 mL
= 4.00 mmol HOAc in 60.0 mL
We have a buffer. Since volumes cancel, use millimoles:
pH
= pKa +
pH
=
[OAc-]
log - - [HOAc]
1.00
4.76 + log - 4.00
=
4.16
At 25.0 mL, one-half the HOAc has been converted to OAc-, so pH= pKa:
mmol HOAc at start
= 5.00 mmol HOAc
mmol OH-= 0.100 M X 25.0 mL = 2.50 mrnol OAc- formed
mmol HOAc left
= 2.50 mrnol HOAc
pH
2.50
2.50
= 4.76 + log - - = 4.76
At 50.0 mL, all the HOAc has been converted to OAc- (5.00 mmol in 100 mL, or
0.0500 M):
1.0 X 10- 14
1.75 X 10 _5 X 0.0500 = 5.35 X 10-6 M
pOH = 5.27
pH= 8.73
At 60.0 m.L, we have a solution of NaOAc and excess added NaOH. The hydrolysis of the acetate is negligible in the presence of added OH- . So the pH is determined by the concentration of excess OH-:
mmol OH-= 0.100 M X 10.0 mL = 1.00 mmol in 110 mL
[OH-] = 0.00909 M
pOH = -2.04; pH,= 11.96
12
10
Phenolphthalein
~
8
:::c
Cl...
6
Methyl f""ed
------·-=.,:::..::::..::=-=::_ __
Midpoint: pH= pKa
4
2
0
L---..L.._
0
_._____.__
20
_._____,__
__.__---''----'---..______.__....,____.__
40
60
80
100
120
__.______._~
140
mL o_ 1 .lv.I NaOH
Fig_ B _S _
--r-11:.r-c:t.:f:..1<:>r.1.
1 C)Q rnI......, C)_ 1
r-,.;r-~e:>.E-:£_
c::"t..I.r-v-~
LI<:£ F-:£C>__.A_c::
re>r
'V"~TS"'-.:1.S
Q_ 1
.A,,:;£
XA
P
14 3
G · 1 ... te a cur for t ~ titration of 50.00 nIL o - 0.] 000 M ace c acid \Vith
0.100
o •ium hydroxide ..
1
Weah Dase versus Strong_ Reid
The titration of a weak base with a strong acid is completely analogous to the above
case, but the titration curves are the r~verse of those for a weak acid versus a strong
base.
12
10
a
=
c:::,_
D
6
l'Vl=thyl
r=c:I
4
2
0
0
20
4-0
60
mL 0-1
fig_ 8.8.
SC>
1 C>O
1 20
1 4-C>
LkE Hc::;1
Titration curve for 100 ITIL
0. 1 .lv.E ~~3 versus Q_ 1 .lv.E 1-ICI_
Example: Generate the hypothe cal tra on curve and calculate
the pH for the tra on 0.1M from HCl with 25ml (0.1M) Ammonia?
Kb = 1.75 x 10 -5
NH3
+
H+
+ c1-
~ NH4+
a) Before addi on HCl (V=0 ml)
-ll
I
N
O. l
o-
· ,X -
'.'l
X O-I
-
)( 1 1, : ·2
l
■.
• - 2.
2 : 10 -
11 -12
·-
2 . ..
o
m
VL
+ c1-
b) Before the equivalent point ( V = 10ml from HCl)
A er addi on 10ml from HCl , ammonium chloride NH4Cl will formed and excess
from ammonia. Which is buffer solu on and in this case we can calculate pOH and
then pH as.
f:NH..,OHI
(Nff-~I
(salt I
[bascl
= [NH4Cl] salt
= Residual of Ammonia
c) At the equivalent point:
Which is mean addi on equal volume (25m) from HCl to 25ml(0.1M) Ammonia
to form salt of weak base only NH4Cl and we can calculate the pH to the salt of
weak base as.
-
_.
I
d) Post equivalent point:
This case mean addi on from HCl, like addi on 26ml. We can calculate the pH
from the excess of HCl as.
•h X(I I - ~ X(1, 1 - 2X II,-' mnl/L
IH I -
25
+ •~
14
12
10
8
:c
Cl..
6
4
2
0
4
0
20
40
60
80
100
mL 0.1 M HCI
2. Calcula e
-.. ·o - s ,o
120
e pH
4
fig. 8.9. Titration curves for
100 mL 0.1 M weak: bases of different Kb values versus 0.1 M HCI.
140
'1L......
.--1.
• .
.55 . 00 a 1td 6 .00 [ _L o
ti
of 50.00 mL of
*(~
0. 1 . 00
a _ mon.i •.
14 38,.
O. 00 - 5 ~ 00
, 9 00 5 . .. .
5 ;, - 0
00 -. HC ·n the ti ra-
~-~
ti on o
50~ 0-1n.L ali _ ~ ot of 01.. 000 .
•. aOH L-- itr· ~~~~□ ~100 _
C~Cffl ·_ te · hepH .- he · olu_- -.o - _ - o~·:0 2s~oo 40 oo,
•·.- n a t r t e add tio·a
1
;;t
1
4s.oo - -_ ""oo. so.oo s .00
55·,.o- • and 60.
1
m
--. d: ta
Weak acid versus Weak base:
In this type of tra on the change in the pH at the equivalent point
is very small, no suitable simple indicator is used to determine the
eq.point.
NH4OH + HOAC ↔ NH4OAC + H2O
Example: Titra on 0.1M NH4OH with 50ml(0.1M) CH3COOH.
a) Before the addi on of NH4OH.
pH•
t
•½
t
pK. X
• ,2~37
4 ..74 -
+
]og
t
Jacidl
lo.g_ (10-- 1 )
0,5 .- 2:87
b) A er addi on 10ml NH4OH.
Buffer solu on from HOAC/ OAC- will formed.
Cl 1,£0"0NH'~
~
i L,O'
- , ;;;::== Cll ,C,0011
+ H1()
~1!"""""'1-
+ (),J 1-
NHjOH + f-1 +
c) At the equivalent point, addi on 50ml from NH4OH, and
Ammonium acetate will formed, salt of both weak acid and weak
base.
d) Post equivalent point , a er addi on 60ml from ammonia. That is
mean another buffer solu on will form that produce all HOAC were
convert to Ammonium acetate , and the excess of ammonia.
50 x"O.I
50"+"60
60 l< 0.1 - 5tJ X" 0.1
.=so+" M
Titration of ~o~ium Carbonate=H m~rotic Base
1- Sodium carbonate is a Br¢nsted base that is used as .a primary standard for the
standardization of strong acids. It hydrolyzes in two steps:
-
[ H ,c o,3 J [ OH- J
1
2-
2[C 03]
1
values should differ by at least 10 4 to obtain good separation of the equivalence
point breaks in a case such as this.
Kb
3- where Kai and Ka 2 refer to the Ka values of H2CO3 ; HCO3 - is the conjugate acid
_,_
of C0~2
Ka2=
and H".)CO-:t is the conjugate acid of HCO-:t 2 ·[H-+-J [ HC0 3 J
[ H+] [C0,3 ]
Ka1= I<.1 =
·K 2 ·- ·
[H?C03]
-
·-
·
] [ H C 03
[HO
[C0 2 (aq)J
1 0- 7
.2 X
1
[H 3 0-+] , Co 3 -
Hco -
K~ 2 = - -- -- -- - [HC·0 3 ]
=
4_69
X
J iQ-l I
4- A titration curve for Na2CO3 versus HCI is shown in Figure 8.10 (solid line).
-----------------------.
12 ~ . -(A) Point
(B) point
10
~
a
(C) point= first eq.point
(D) point
' '\
1
Phenolphthalei n
+
I
Methyl red
4
fv'.lethyl orange
D
(E) point= second eq.point
2
0
L----'---'--.......l--...__ ----lL--.L.--L--....,___ __,._____,__
0
20
40
60
BO
mL 0.1 .M HCI
100
__.___
_,___....__ __
120
(A) point: before any addi on(at 0.0ml volume from HCl)
In this case we have only sodium carbonate c ,o 2 - and the pH is determined
3
from the hydrolysis of Bronsted base as:
-
_ ,
[HCO3] [OH-J
Kb=
1
-
[HC0 3 ] [OH-J
2-
a
[CO;-]
[C03]
-
[HC0 3 ] = [OH-]
10- 14
5 x 10- l l
pOH= 2.34
_ [OH -]2
0.1
pH = 14 – pOH = 11.7
(B) point: A er addi on 25ml HCl:
In this case a part from carbonate
buffer region is established.
.
-
co/- is converted to Hco3-and
[H .
]
K i ==
'
= (50-25) x0.1 / VT(75ml)= 2.5mmole / 75ml
= 25 x0.1 / Vt(75ml) = 2.5mmole/75ml
= 4.69 x10-11
pH= 10.32
(C) point: first eq.point( addi on 50ml HCl):
In this case all were
converted to
mono-protanted salt
from Ka1 and Ka2 as.
( pH = pKa2 )
and we calculate the pH for
= (50-50) x0.1/ Vt(100ml) = 0.0mmole/100ml
HC03 - : = 50x0.1 / Vt(100ml) = 5mmole / 100ml
[H+] == ✓K 1 K2
pH= 8.3
(D) point: A er first eq.point (addi on 75ml HCl):
point, the
Beyond the first eauivalence
- HC03 - is partially converted
...
to H,_C01(C02) and a second buffer region is established, the pH being established
by [HC03 -]/[CO2 ].
= 50x0.1 /Vt(125ml) = 5mmole / 125ml will formed
= (50 – 25 ) x0.1 M / 125ml = 2.5mmole / 125ml residual concentra on
= 2.5 mmole / 125ml will formed from
H-,CO~~ = 25 x0.1 / Vt(125ml)
.,_.
[H ] [HC013]
Kl
a =
[H2C03]
:= =
I
4.2 X 10- 7
pH = 6.37
(E) point: second eq.point( addi on 100ml HCl):
At the second equivalence point (E) the pH is determined by the extent of
dissocia on of carbonic acid, the principal species present, and [H+] is
calculated from Equa on.
Ka1=
= 50 x 0.1 / Vt(150ml)= 0.0333
[H
4.2 X 10- 7 ==
] [HC0 J
(0.1) [50/(50
[H+] == 1.177 x 10-4 M
3
100)]
1
- [H+]2
0.033
pH= 3.92
Important notes:
Phenol~hilifileffi fa u~ea lo ilelect ilie fu~I enO ~offi~ anil meiliJI oranie fa used to detect the second one.
2- Neither end point, however, is very sharp.
1-
3- phenolphthalein is colorless beyond the first end point and does not interfere .
4- The second equivalence point, which is used for accurate titrations, is normally not very
accurate with methvl orange indicator be cause of the gr~dual change in the
color of the methyl orange. This is caused by the !Z!adual decrease in the pH due to
the HC03-/C02 buffer system beyond the first end point.
If beyond the first equivalence point we were to boil the solution after each
addition of HCI to remove the CO2 from the solution, the buffer system of
HC0 /CO would be removed, leaving only HC0 in solution.
5-
-
3
-
3
2
Titration of Pol~protic Rcids
1- Compouud· .1vvith two o· more acid functional group .·yi.Id multipl,e end points in a
1
titration provided that the functional group",differ ·ufficiently in .·.t.ren th as acids
For example of dipro c acids are H2A, H3PO4, Malonic acid =HOOC CH2-COOH
And
2- In order to obtain good end point breaks for titratiqn of the first
proton, ICa 1 should be at least
3-
1 04
><
.Fi:"ci2.-
ff ilii ratio i.small r ilie .rror become exce we, particularly in the 1egion of !he fir.tequivalence point
and amm rigoroustreatment of the equilibrium relation 1ip .i required.
4- Two indicators must be used to detect the equivalent points in the tra on od
dipro c acid H2A with sodium hydroxide NaOH.
5- Figure 8.12 illustrates the titration
r.11rvefor a d-inrQtic acid H--.A,
and Table 8.3 summarizes the equations governing the ~if ferent portions of the titration curve.
14-
2 eq.point
12
Buffer-
10
:::c
or:::::L
1 eq.point
a
Buffer-
6
4-
~H
=
pK01 -+- log
H
2
-~~:.J)
A/HA-
2
20
4-0
120
100
80
60
mL NaOH
Table 8.3
Equations Governing Diprotic Acid (H 2 A) Titration
Fraction F Titrated
Equation
Present
F = 0 (0%)
[H+] = V Kai CH,A (Example 7. 7)
(or Eq. 7.20; Ex. 7.17, quadratic, if -strong acid)
0 < F < 1 (>0 to <100%)
pH
= pK
01
CHA+log~
(Eq. 7.45)
H2A
(or CHA-+ [H+] and Ca A-[H+] if a strong acid)
2
F
=
[H+] = V K 1Ka2 (Eq. 7.84)
• (or Eq. 7.83 if H 2 A -strong acid)
1 (100%) (1st eq. pt.)
0
pH
1 < F < 2 (>100 to <200%)
F
=
2 (200%) (2nd eq. pt.)
F> 2 (>200%)
A2-
= pK 2
[OH-]
0
CA2-
+ log - CHA-
= ✓ K.v
•
GA2-
(Eq. 7.45, Ex. 7.16, 7.22)
(Eq. 7.32)
KA2
(or Eq. 7.29, Ex. 7.19, quadratic if A
[OH-] = [excess titrant]
2-
-strong base)
Example:
When 100.0 IllL.. of 0.10 A4 rnalonic acid is titrated with 0_ l 0 .M N aOH the follo wing titration
curve is observed:
l\llalonic acid = HOOC·~ CH 2 --COOH
abbreviated H 2 A
pH
12
10
8
6
D
4
B
2
100
50
200
150
vol. of NaOH added (rnL)
Given that
questions:
Kai
=
1.5 x
10-3 and K
Wri~e out th
B.
Calculate the pH at:
l)
2)
3)
4)
5)
=
2.0 x
10-6 for malonic acid, answer the :fol1owing
a tion and thie qui[brium x.pi-e , ion a · ociated with
A.
1· ·
32
point A pH= 1.94)
point B
point C
point D
p int E
K.t 1 and K_. 2 .
Solu on
[H +] [HOOCCH2c,o o -1 = 1.5
[HOOCC"2 COO,H ]
~
X
10-3
1
[H
+
[OOCC"2C 0 0 2 - ]
1
= Z.O
x I 0 -6
[HOOCC~COO -]
B.
P •o int A: 100 mL. of 0 ..10 M H 2 A
(Addi on 0.0ml from NaOH)
H
I
+
+
0
+x
+
E
0.10 -
X
X
X
x2
0.10 ·-
The 5% rule fai l s .
Ignor th
x
=
amount of I[
X
1 . 15 x 10-2 M by successive approximation _
] r I, a ed by 2 nd di
.o · ation (it wiH be n gligible).
t
=-J -r ...
1
"',5X'
- Q -
) : : ,·· "' 1 1 '
(A er addi on 50ml from NaOH), in this case the solu on contains both HA- and
H2A which is buffer solu on:
[HA-] = 50 x0.1 / Vt(150ml)= 0.033Molar
[H2A] = (100 – 50) x0.1 / Vt(150ml) = 0.033M
[HA-] = [ H2A]
pH = pKa1 + log HA- / H2A
pH = pKa1= 2.82
-r -p - -ie -: i ~hich i _ _oth an _cid
. specie
• )•.
an d- a .-b. ase (amph.otenc
~
(A er addi on 100ml from NaOH)
P. = P al ' P
1
2
a2
·
-·
1
'
••• . •
~
•
. ••
.
•
.
•
'1 . .
_
•.
•.·
.•.
I
.•
I
= 2 :2 • S70 = 4__-6
2
In this case all H2A was converted to HA-, which is mono-pro c salt and we can
calculate the pH from the above law or .
PCJo,i r.tt: ~
:: h .a .lfwa.y
te3
s e c o n d equ.jvaJ.eTJJoe pc,iri.t w h e r e [:FILA.-]
A er addi on 150ml from NaOH), in this case 100ml from NaOH will convert all H2A
to HA-, and the other 50ml will convert an equal amount from HA- to A-2. Which is
buffer solu on contains A-2 / HA- and we can calculate the pH as:
[HA-] formed = 100ml x 0.1 / Vt( 250ml) = 0.04M
[HA-] remaining or residual = (100 – 50) x0.1 / Vt(250ml) = 0.02M
[A-2] formed = 50 x0.1 / Vt(250ml) = 0.02M
[HA-] = [A-2]
pH = pKa2 + log [A-2] / [HA-] = 5.70
Poin.t: E: seco:nd eq_Lii-vaJ.,e nce p o i n t ; on.ly A. 2 - presen.t ( a w ,e a.k. bas~)-
:EIA.I
1.0 m
C
-x.
E
0 . 0 3 3 .lid -
m o,1 /300 :mL
X
mpl:ion:
pOF:I
=
1 _3 :x. I o - s
0_033
- l o g [C>I-1:-]
=
:x .I 00<¾:> =
4_ 89
s _,o
>< 1- o
X
:x:2
0_033
KJK.,.,2 =
-f-X
X
=
=
o,
0
-+-x
5_0 >< 10,- 9
Check. ass
~
-+--
:x.
-
Q_04-·~
:X:.
2
0_033
-c:::: 5 <;¾'
so as
u m.p1tion g o o d
EXAMPLE 1 5-9
Con tr .c a cur · , for th tlm:ration of 25.00 mL of 0.1000 M malei acid,
= COO , with 0.1000 M aOH.
Sy.m bolizing the acid a H2
w,e can wri ,e th._ two di ociation qui.libri a a.--
HM
H .o ~ H O
-
HM-+ H -O ~H
1
~
o-
+
1
•.•
-
+ M 2-
K
l
Ka_
= 1. X 10- 2
I
== 5 ..'
1
X 10- 7
A er addi on : 0.0ml , 5ml, 24.9ml, 25ml(first eq.point), 25.01ml , 25.5ml , 49.9ml
and 50ml(second eq.point) from NaOH.
SALTS OF POLVPROTIC ACIDS-ACID OR BASE?
!
Salts of acids such as H 3 P04 may be acidic or basic. The protonated salts possess
both acidic and basic properties (H2 P04 - ,.· HP04 2 -), while the unprotonated salt is
simply a Brszsnsted base that hydrolyzes (P04 3 -).
1. Amphoteric Salts. H 2PO4 - possesses both acidic and basic properties. That
is, it is amphoteric. It ionizes as a weak acid and it also is a Br0nsted base that
hydrolyzes:
Ka 2
=
[H+][HP042 -]
= 7.5
X 10-s
(7.76)
[H2PO4-J
Kb= -
Kw
Kai
[H3PO4](QH-]
= -----[H2PO4 -1
1.00 X 10- 14
- - - - = 9.1 X 10- 13
1.1 X 10-2
(7.77)
The solution could, hence, be either alkaline or acidic, depending on which ionization is more extensive. Since Ka2 for the first ionization is nearly 105 greater than
Kb for the second ionization, the solution in this case will obviously be acidic.
An expression for the hydrogen ion concentration in a solution of an ion such
a_s
F-I2P<=>-4--
ca.ri.
be
e>hTai:r1e.<l
as
f~lle>ws_
.· .
1+
.+K
,
C:N aHA/Kal
So when:
1-
2- If
and
[HA-] / Ka1 ≤ 1
} ____is_not-_negligible
C HA _ x K a 2.
C
I+ HA .
K
a1
3- If Ka2× [HA- ] ≤ Kw
and [HA-] / Ka1 >> 1
Kw is not negligible
1 is negligible
ICa 2 + IC..v
C HA -
C HA _ X :
ICa i
EXAMPLE 1 5-6
Calculate the hydroruum ion _-oncentration of a ·1 .00 X 1o·- 3 MNa HPO4 •. oluf on
he pe tinent di o ia ion con tan '· are Ka a d K:u · which both contain
[HPO~- . Their ·: lue are K. 2 = 6.-. ·' X 10- 8 and Ka . == 4.5 X 10- 1 . Con iderin again the ·· -umpt·on .·
1 d t , quation
we fi
X
10--- 3
:x 10- • i I li h larger than 1 o that the denomiuator can be s.1m..
plified~. • 'he prod 1ct. K~1- c .a:2H . _, i ,y no mean. much arger than K however.
·et r for e u ea par iall, imp "fi d v ion of quation 15-15 .
t
a t
1 5 - 1 6
t h a t
l m O
0
) / ( 6 . 3 ?
'\1
l
1
. . · ·.
H.o+] =
• ••
I
1'
/~4.5 X 10- 13 X 1.00 X 10- 3 + 1.00 X
JO-l
.. .
· - - - - - - - - - - - - - - - = 8.1
· 1.00 X 10- 3 /(6.32 X 10-)
X
_
., ..
10 ] M
EXAMPLE 15~7
..-wwcA
n ao. oo .
f im , , ,
,32X- 10~
;but the ...
:rrw_
,·.
• ...•..• •. •.
~1,
EXAMPLE 15 - 8
Calcu]aite the hydronmum ion c ncentration of a 0.100 M NaHCO olution.
a sun1e as 'W hav earlier (page 400 that: [H. co ] << £CO,_ aq)] and that
th _ foll,o ing eq uilibria de cri be the y te1n ade•q uate ly :
w_
1
K
1
= , H _Q+ ][HCO_i"]
[C0 2 (aq)] :
== 4.,2 X 10- 7
HC03.
I
+ H.,,
-o ~ H. O + co 2 -
_
Ka2 =
==
[HO+][Co_ - ]
.·
[HC·0 3
.
4.69 X 10-n
Clearly, .CNaB !Kai in the denominator of Eq 1ation 15-15 i ll!UC~: l~rg,er ~barf··
unity· in addition, Ka2CNaHA ha . a value of •..69 X 10- 12,, which i subst~nti~lly
greater than Kw· Thu ·- - quation l 5-1 6 apphe and
• •, •• • • • •
Volumetric apparatus