Chapter 16 Worksheet 2 (ws16.2) , Calculating pH after a neutralization reaction, Understanding the shapes of acid-base titration curves Calculating the pH after a neutralization reaction
Every acid-base reaction produces another acid and another base. A neutralization reaction is a
reaction between and acid and a base that produces a weaker acid and a weaker base. Neutralization
reactions usually have very large equilibrium constants so they go to completion. Let's see why.
1. Write the balanced molecular and net ionic equations for the reaction of acetic acid with ammonia. Is this a neutralization reaction? (The table ofKa's and Kb's from ws15.1 is attached to this document.) C H"j eaoH -t N t-t"7 -7 \J I-Iy C N3 COO
C 1-t3 Cad
~,s 1:2
-
I
-{ W HLI
t
VI~v1 r.::;.( /'2.C<+IO .... ~<:'~Cli-tcl ..... \o~CQ..y;~ C ~coo J-l /<, q <;"trtJv<j'f'r
'tL-c:::.r\ .,H.Ctt (o....d }Jl-t l"'J q stQ:rvC~('" l?4":>JO!! tk.n Cf-I~CCO-).
3
Q
qc"d
2. This reaction can be thought of as occurring in 3 steps:
Step 1: Ionization of acetic acid Step 2: Ionization of ammonia Step 3: Reaction of hydronium ion with hydroxide ion Write the balanced equation for each step along with their equilibrium constants. Add the 3
reactions and calculate the equilibrium constant for the net reaction.
Step 1:
C l~:, coO
\4 -{ ~
Step 2:
\0 t-ls -t ~.-;;;?
\J
Step 3:
tk(O + + ~
~O
Net:
~3 ~Qo{J
l-. W
4
N.3
--;;;;F C
-f-lL/
~
-./ t
W3 <.ro
-+
{--1:J "::> 0
Keq =
l,'bXtG-~} co­
Keq = /.(5x(T>
(' W3(D:J-+
~.o('}. ctJ 3 coo H ~ tJ ~
~ ~ucl'\ -S+n-J~""
to
CPI<v\.
'ftc.... .
I
K~_ iD~!O'4
t--j
4trt
Keq
3.). X \0
5,Cj
I::o-e'""
t<lJe"'CJ:J
. .-
o-d-
O-fYZ
4
!
~t~ ~f::. 'T~J Q~ '"t~ fmdu~ts tkc:/j{ tlQ. rf;.CtC.Jl~ ~
Pl~+,C.Yl ( o
1
~
Acid-base titrations
An acid-base titration is an analytical technique for determining the concentration of a dissolved acid or
base and for determining the Ka for a weak acid. During a titration, a neutralization reaction is
performed by slowly adding a solution of a base (the titrant) to a solution of an acid (or vice versa).
Let's consider a titration of an acid solution whose concentration is unknown with a base whose
concentration is known. (See figure.) The base is the titrant and it is added slowly to the acid.
hase of known
concentration
acid of unknown
coneentration
When the titration begins, the base is the limiting reactant and is completely consumed (because
neutralization reactions go to completion). Excess acid and the salt produced by the reaction are now in
the flask. As the titration continues, more and more of the acid is consumed. Eventually all of the acid
(and added base) is consumed. At this point, stoichiometric amounts of acid and base have been mixed
and only a salt solution remains. This point is called the equivalence point. Given the volume of base
added to reach the equivalence point, the known molarity of the base, and the stoichiometry of the
reaction, you can calculate the number of moles of acid that was present initially. An indicator is used
to determine when the equiValence point has been reached. If the titration is continued beyond the
equivalence point, the base becomes the excess reagent and it simply accumulates (but no more salt is
produced). If the acid is weak, the Ka can be determined by monitoring the pH during the titration. As
demonstrated below, the Ka is equal to the pH half-way to the equivalence point!
5
Equivalence point calculations
Any reaction that goes to completion (not just neutralization reactions) can be performed as a titration.
Consider the generic reaction:
aA + bB
~
cC (The single arrow indicates that the reaction goes to completion.)
At the equivalence point, stoichiometric amounts of A and B have been mixed and both have been
completely consumed. Thus, at the equivalence point:
moles of A reacted
moles of B reacted
~--..... =
a
b
(a is called the stoichiometric ratio)
b
The equation can be re-written in terms of the molarity (M) and volume (V) of each reactant.
For most (but not all) acid-base reactions, the stoichiometric ratio is equal to one! The volume does
NOT have to be in liters since the volume units will cancel as long as they are the same.
If you know all but one variable, you can calculate the value of the remaining variable. Most often, you
are interested in determining one of the following:
The molarity of the acid or base being titrated. Let's assume an acid is being titrated and
a base is the titrant. In this case, you know the volume of the acid and the molarity of the
base. You measure the volume of base required to reach the equivalence point. Since you
know the stoichiometric ratio, the only unknown is the molarity of the acid.
The volume of titrant required to reach the equivalence point. Again assume that you are
titrating an acid with a base. In this case you know both the volume and molarity of the acid
and the molarity of the base so the only unknown is the volume of base required to reach the
equivalence point.
Do the following 2 problems on your own. We will NOT do them in class.
1. It required 10.00 mL of 0.250 M NaOH to titrate 50.00 mL ofHCl (to the equivalence point).
What was the original concentration of the HCI? (Do the same problem but change NaOH to
Ca(OHh. Do it again but react H2S04 with NaOH. The stoichiometric ratios are no longer 1 to
I! Since NaOH is a strong base, both protons in sulfuric acid will be titrated.)
2. What volume of 0.100 M NaOH are required to reach the equivalence point if you titrate
25.00 mL of 0.500 M HCl? (Do the problem again but change the numbers or change the acid or
base as above.)
6
r [:\ 'SrolC~\&~1-{'1 ~
I
I•
:: l
A l'.I<. (-I- C c. Cei W. \z.
--)-
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+ t-l't..G
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~(JQO\CDL)';:.(z..5(;'(!O-3f'G1
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-
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A
-- /V\ ~~
Vti
~
'L
'(
J.
WC\ot~
+- N.Gl
~ ~Cl + HLo /VI A V ~;; fv' fS U Vj
CI : I
""Stut CL'O>-l<1- r
1) Understanding the shape of a titration curve
There are four phases of a titration:
Phase 1:
Phase 2:
Phase 3:
Phase 4:
Before the titration begins Before the equivalence point At the equivalence point After the equivalence point Label each titration curve below with the reaction that is occurring and the species present at each stage
of the titration. For titrations involving a weak acid or weak base, label the region in which the solution
is a buffer and the point where the pH = pKa•
Titration of 25.00 mL of 0.100 M HCI with 0.100 M NaOH (strong acid, strong base):
c;oPyriQh.t..\l:).. T.~~. .!v1<;<:7~aw~.f1,ill. .<::?'!lll?al1.i~.•..II1~, . '=~!!!1.i.~i.?11 reqllir~r~!!~pr()ljucti()n . ()r.. ~is
13
II
b.,Jt<,~'7 5
10
(f:..l\C)L.
9
pH
8
7
N<lO
I-l
r;.l)
Nc.Gl CI'\~. . t.fthI')e,I+ ~
ptJ'-:"7)
Equivalence -.__
point
[), l ¥\ (...fe(
-:., r" -:. I
40
10 50
Volume of NaOH added (mL)
the following questions on the next page:
Calculate the initial pH. c~"7'" Me: i
t
I ~. ""0,,, q .xC}:>'7"> l-Ic (
· PH = 7 at t h e eqmva
. 1enc omt.
. ?
\\1"'
<'.\""'" ~)(c...~").
It-v'1 I(I'::.'~ "
Why IS
0....0
N-:;.,o 1'\
3. Why does the pH rise owly at first, very rapid! near the equivalence point, and~er the
equivalence point?
~(JfI
4. Why does it require 25.00 mL ofNaOH to reach the equivalence point?
~~S5
7
Titration of25.00 mL of 0.100 M CH3COOH with 0.100 M NaOH (weak acid, strong base):
Copyright
© The McGraw-Hili
Companies,
Inc,
Permission
required_ _for., reproduction
or disl
"-, _
"" ___ -.. ____ " ,'W" __ ,__
_,
._
•.
_
_
_ ' ,.", .. _
_________
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14
13
~---Ex<:<55 t-hO{.(-;J
12
II
(tll ",d
lO
c u~coer )
L, kq~
9
pH
8
7
111tl~
Nc1(W~(OO
6
CSq '7t<
I
I
2
:.
iiI
(fl'-i.=,It..(f
'-" p
I
,
~I+ ~ pH >7 )
I
I
I
f
i
o
40
20
30
Volume of NaOH added (mL)
50
Answer the following questions on the next page:
1.
2.
3.
4.
5.
6.
Calculate the initial pH (Ka for acetic acid is 1.8 x 10.5).
Why is pH > 7 at the equivalence point?
Calculate the pH at the equivalence point.
The pH rises slowly between 5 mL and 20 mL. Why?
Calculate the pH at 12.5 mL (half-way to the equivalence point).
After the equivalence point, the curve is identical to the strong acid/strong base titration. Why?
C~4J -:
if'{=:J..
t.~XIQ-")(a..tOO);:'
1.-"(.{)((0-3 ;(.~I s,. 1+ I'" <J~(J t!-t'j p~,:;(~
').5.CCiIML.. o+: O.(COV\ ci:J.,COGH {1.s'()cJn.l at
50,-;(<:'
"3.
<: W}C66l..l
.....
.l..
0.0500
<-- -o.c;~a
c:
0
~ ~a(j U.
_'"
00.:)'-'...)
-l>
NCt<~('('Jj + f-.tz...o
CJ
-(j.ostx')
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0
o.~.s:o
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~oo
H
(
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(a 6(':]::.
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l ¢ r.l '"
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f';8 t& ...~
...1rR. ~.2
~I "il"b.]
q
V
8
I~Cl"V11 <-<;ll j, I qt-\~ r tlQ f\ rs+ d fGf C ~ JJo.o f-.( (~ QdcJ~d ~coo ~" qve)
q(/ -t"-' <->-.}qJ fG t~ ~(jlUode1--JC~ PC(Vl-i-) jCJ(j (".q~ a bu~~r,
Y.
HG~\J"<r) o~rQ·hcJ..-QII~ ) t~ Iou+~, ~vt'5 ~:J.Iv\ ~ VV1tddl~
ot rlA lov4 t~r ~(Ov1,
OVI t"~ low pH Qvd of flQ. tav~~.r ~j(o//\ t~~ 1'"':>111'+ j.vIL(""~L. loqs~.
iUc bl~L- etl Qvd) tLQ~ ('S~1- VAUCJ~ ~Cld. % (1/\ ~S~ ~l("j/AS )
e.OcJ.",
drc.J(' ,S Ahof-( pr"'CldJ~"S a (C\rc;)c;:f <::::.L-o~...Q l"'\ ,Cc(-i3 <r:la-7
G""
r~v'\
5. I
\1\
1"4.
I'"
\M /
~ ~Idd ~
d~~ <":'J t
~Gl\,,,"GO-J =-
(
+4
c
~ t~ tou+~r ~(Ol/] \
10 u+ Mr r~'jI
~
[c~CCPHJ
ff.{",
()V1 (
~ I}-u.)Ot ~ ~
e1LJ I U::J. tQ~o
r+;:. 12...5" L)
rt-e. '" ~Lf7iI
pt-:a + I<>JC I) -=
[c~ccoll.J
Mt... ,.... tLo- QqUlUC'J«~'-~e
~ c::urLR 1<'
p+) PN IS dq~""1"<;?6 ~
(d~~f-( ~q \ 10 p~UtClL)'S f,+v-crTt()J)
'.
"X<:'6'S J.joO-!-/
VII'
Titration of 25.00 mL of 0.100 M NH3with 0.100 M Hel (weak base, strong acid):
O.l(:bM
tl
t-J ,?,
Copyright © The McGraw-Hili Companies, Inc. Permission required for reproduction or di!
~
11
10
9
~
~Hj?~ N~t
8;
7t:~~-.----f>t..S~~~r-__
-_+l
6. }~..~
pH
~,
;,
Equivalence.. -,-,-­
point
5
4
;k-------F-xG~s ~C\ ~-----~~I
(+ N I-llj't)
3
2
\
o
20
10
30
40
50
Volume of He1 added (mL)
Answer the following questions on the next page:
1. Calculate the initial pH (Kb for ammonia is 1.8 x 10-5; It is coincidence the the Kb for ammonia is
about the same as the Ka for acetic acid).
2. Why is pH < 7 at the equivalence point?
3. Calculate the pH at the equivalence point.
4. What is the pH half-way to the equivalence point?
[alrl '" ft.'t"'l6-o;)(C,\~.J
V()\~ -:
A<.(dl,.C,.
3.
K0
2 5l> J
) (
~
tv tJ t{
+_
C..t'tJ
3
~ t~ -= =l \. ?-11­
'S~ ({ ~~I.I1'7" +<::i r
-:. (. )'-\)U6-
-
-'if LO't ~o
':,
(~)( (6'-'5'" A
_ILl
5'": ~(~,)(lO
_{
~\ -::: ~.1-'X(C> ~ ~(5.<7')(l6-IGj\()(lSO J tt''! - 5" .'L~
'1.
Pt-l :=. \' \:4 ~(q,2~
9
· Maleic acid is a diprotic acid (a dicarboxylic acid). Its fonnula is C4H404. Its structure is shown below.
Only hydrogens a and d ionize (do you understand why?)
o
0
II
II
H-O-C-C=C-C-O-H
a
I I
H r.J
h
.V
C
For this titration, use the following symbols: H 2A, HA-, A 2-, where A
,
C4H2 04
THration Curve of 50 mL
0.10 MMaleic Acid with 0.10 MNaOH
14' " " ' I " " - - - . . . , . - - - " r " - - - r - - - - - , - - - - - r - - - - ,
. . . . . . . . . -. . '"'. . . . ._.. .,.. .,.. . . ",. . . . ."." . . . . . "."..,"
"~-
·"':;:";·;1m.~
." lif''''.....'I'....''Ii.OI'.''....,',.u."
o~----~----~----~-----+----~----~
o
25
150
75
100
mL of NaOH added
125
150 '
What would the titration of a triprotic acid (such as phosphoric acid) look like?
10