University of Guyana
Department of Chemistry
CHM 2105 – Analytical Chemistry 1
Assignment 2
1. A
20-­­tablet
with
reaction
After
sample
20.00 mL
of
M
saccharin
was
AgN03.
The
of
0.08181
the
solid, titration
of
2.81
0.04124
treated
is
removal
of
and
washings
required
KSCN.
Calculate
the
of
saccharin
(205.17
soluble
g/mol)
mL
of
average
number
in
tablet.
each
of
the
filtrate
M
milligrams
2. Calculate the silver ion concentration after the addition of5.00, 15.00 25.00.30.00.
35.00,
39.00, 40.00, 41.00, 45.00, and 50.00 mL of 0.05000 M AgN03 to 50.0
ml of
0.0400M KBr. Construct a titration curve from these data
pAg as a function of titrant volume.
Solution \
titration ( precipitation) reaction as AgNO3(aq) + KBr (aq) AgBr(s) + KNO3(aq)
Calculation of equvivalence point
M Ag+ x V Ag+ M Br-x V BrV Ag+ M Br-x V Br- / M Ag+
0.04 x 50/ 0.05
40 ml
Consider dissolution of AgBr as AgBr(s) Ag+(aq) + Br-(aq)
Ksp =[ Ag+ ][Br-]= 5.0 x 10-13
[ Ag+ ]= Ksp / [Br-]
Calculations for before equivalence point
plotting
1) when 5 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.005 L ) ( 0.05 mol/L)
= 0.002 - .00025
=0.00175 mol BrVolume after addition of 5 ml AgNO3 = (0.05+.005)= 0.055 L
[Br-] = 0.00175/0.055= 0.0318 M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 /0.0318
= 1.57 x 10-11 M
p Ag+ = -log [ Ag+ ]
= -log( 1.57 x 10-11)
= 10.8
2) when 15 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.015 L ) ( 0.05 mol/L)
= 0.002 - .00075
=0.00125 mol BrVolume after addition of 15 ml AgNO3 = (0.05+.015)= 0.065 L
[Br-] = 0.00125/0.065= 0.019M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 /0.019
= 2.63 x 10-11 M
p Ag+ = -log [ Ag+ ]
= -log( 2.63 x 10-11)
= 10.58
3) when 25 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.025 L ) ( 0.05 mol/L)
= 0.002 - .00125
=0.00075 mol BrVolume after addition of 25 ml AgNO3 = (0.05+.025)= 0.075 L
[Br-] = 0.00075/ 0.075= 0.01 M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 /0.01
= 5 x 10-11 M
p Ag+ = -log [ Ag+ ]
= -log( 5 x 10-11 )
= 10.3
4) when 30 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.030 L ) ( 0.05 mol/L)
= 0.002 - .0015
=0.0005 mol BrVolume after addition of 30 ml AgNO3 = (0.05+.030)= 0.080 L
[Br-] = 0.0005/ 0.080= 6.25x 10-3 M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 / 6.25x 10-3
= 8 x 10-11 M
p Ag+ = -log [ Ag+ ]
= -log( 8 x 10-11 )
= 10.09
5) when 35 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.035 L ) ( 0.05 mol/L)
= 0.002 - .00175
=0.00025 mol BrVolume after addition of 35 ml AgNO3 = (0.05+.035)= 0.085 L
[Br-] = 0.00025/ 0.085= 2.94 x 10-3 M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 / 2.94 x 10-3
= 1.7 x 10-10 M
p Ag+ = -log [ Ag+ ]
= -log( 1.7 x 10-10 )
= 9.77
6) when 39 ml 0.05 M AgNO3 added
Before eq. point , all added Ag+are consumed by Br- and excess Br- remains in the solution . We
can calculate concentration of unreacted Br- as
Moles of Br- = original moles of Br- - moles of Ag+ added
= (0.050 L ) ( 0.04 mol/L) - (0.039 L ) ( 0.05 mol/L)
= 0.002 - .00195
=0.00005 mol BrVolume after addition of 39 ml AgNO3 = (0.05+.039)= 0.089L
[Br-] = 0.00005/ 0.089= 5.62 x 10-4 M
[ Ag+ ]= Ksp / [Br-]
= 5.0 x 10-13 / 5.62 x 10-4
= 8.9 x 10-10 M
p Ag+ = -log [ Ag+ ] = -log( 8.9 x 10-10 ) = 9.05
At eq point 40 ml Ag added , there is no Ag+ . Precipitated AgBr redissolve to form Ag+.
[ Ag+ ] [Br-]= Ksp
X x X = 5.0 x 10-13
X= 7.07 x 10-7
p Ag+ = -log [ Ag+ ]= - log( 7.07 x 10-7) = 6.15
After eq. point there is only Ag+ .
Moles of Ag + = 0.001L x 0 .05 mol/L = 5 x 10 - 5
[ Ag+ ] = 5 x 10 - 5 / 0.081 L = 6.17 x 10-4 M ( total volume= 0.04 +.041 = .081 L )
p Ag+ = -log [ Ag+ ]= - log( 6.17 x 10-4) = 3.2
After 45 ml Ag + added
Moles of Ag + = 0.005 L x 0 .05 mol/L = 2.5 x 10 - 4
[ Ag+ ] = 2.5 x 10 - 4 / 0.085 L = 2.94 x 10-3 M ( total volume= 0.04 +.045 = .085 L )
p Ag+ = -log [ Ag+ ]= - log( 2.94 x 10-3) = 2.53
After 50 ml Ag + added
Moles of Ag + = 0.010 L x 0 .05 mol/L = 5 x 10 - 4
[ Ag+ ] = 5 x 10 - 4 / 0.09 L = 5.55 x 10-3 M ( total volume= 0.04 +.05 = .09 L )
p Ag+ = -log [ Ag+ ]= - log( 5.55 x 10-3) = 2.26
3. (a)
What do
solutions’?
you
understand
by
the
term
‘buffer
(b)
Write an
calculate
the
solution.
give
expression
which
hydronium
ion
Show how
this
can
be
concentration
expression
can
used
to
of
a
buffer
be
rearranged
to
of
Mg
Y2-­­
often added
the
Henderson-­Hasselbach
Why
is
a
to
a
water
specimen
that
is
to
be
titrated
for
hardness?
(b)
TI
in
a
9.76-­­g
sample of
rodenticide
to
the
trivalent
4. (a)
and
treated
Mg/EDTA
TI3+
with
solution.
MgY2-­­
+
Titration
of
equation.
small amount
The
was
oxidized
an
unmeasured
The
reaction

TIY
excess
state
of
is
-­­
Mg2+
+
the
liberated
Mg2+ required
13.34
mL
EDTA.
Calculate
percentage
of
of
0.03560
M
TI2SO4
(504.8
g/mol)
in
 
the
sample
the