5.3: Acid-Base Titrations – Analyzing With
Volume
A titration is used to determine the concentration of an unknown solution.
P. 326
Because HCl is strong acid the
H3O concentration is 1:1
So…
pH = -log(0.10) = 1.00
OH is what neutralizes and it reacts with the H3O+ ions present so it will take
more to neutralize the H3O+ from the HCl then from the HCN.
Criteria for Titration
Titrations are one of the most powerful tools that chemists have.
During a titration you use a solution of known concentration to
determine and volume need to react to calculate the number of
moles.
Once you have the number of moles needed you can use that
information to determine the concentration of the unknown
solution.
Criteria for Titration
Titration Accuracy
A standard solution is a solution of KNOWN
concentration.
Titration Procedure
1.
2.
Titration Calculations
You only take the two trials that are the closest
together….or 3 if they are all close together.
Titration Calculations
Once you have calculate the average volume
needed you can use the known concentration of
the “titrant” to calculate the number of mols (n)
n=C●V
Once you have number of mols of “titrant” you
an use the MOLE RATIO to calculate the number
of mols of UNKNOWN SAMPLE.
At least one of the two solutions; sample or titrant must be strong.
The endpoint is when the colour change is seen.
The equivalence point is when the mols of titrant and sample are EQUAL.
23.67 + 23.59 = 47.26 ÷ 2 = 23.63 ml
Standard Solutions
1. The solution can be created from dissolving a known
amount of pure solid.
2. The solution is created (diluted) from another solution of
known concentration.
No if its not pure….yes if its kept pure and not exposed to the air.
A volumetric Pipette
No if its not pure….yes if its kept pure and not exposed to the air.
A volumetric Pipette
A volumetric Pipette
A burette
An appropriate indicator changes colour to signal this point.
V = 0.02500 L
V = 0.02815 L
C = 0.1072 mol/L
C=?
0.09520 mol/L
C = n/V
n=C•V
n = (0.1072 mol/l)(0.02500 L)
( )
C = n/V
C = (0.00268 mol) ÷ (0.02815 L)
C = 0.0952042629 mol/L
1
n = 0.00268 mol
1
Titration is very much like stoichiometry from grade 11.
n = 0.00268 mol
Pg 332: Practice Problem 5.3.1
#’s
1, 2
Steps to Solve Titrations
m
n
M
n
C
V
( )
1
1
Steps to Solve Titrations
m
n
M
n
C
V
Again….but with the new info from other species….different volume
and or concentration usually.
V = 0.04623 L
C = 0.203 mol/L
V = 0.0250 L
0.188 mol/L
C=?
C = n/V
n=C•V
n = (0.203 mol/l)(0.04623 L)
n = 0.00938469 mol
( )
1
2
C = n/V
C = (0.004692 mol) ÷ (0.0250 L)
C = 0.0952042629 mol/L
n = 0.187682 mol
Pg 334: Practice Problem 5.3.2
#’s
1, 2, 3
Calculating Volume in Titrations
m
n
M
n
C
V
Again….but with the new info from
other species….different volume and
or concentration usually.
C = 0.423 mol/L
V = 0.025 L
C = 0.158113883 mol/L
V=?
C = n/V
n=C•V
n = (0.423 mol/l)(0.025 L)
n = 0.010575 mol
( )
1
2
[OH-] = 10-pOH
[OH-] = 10-0.500
[OH-] = 0.3167766 mol/L
[Sr(OH)2] = [OH-](1/2)
[Sr(OH)2] = [0.3167766](1/2)
[Sr(OH)2] = 0.158113883
n = 0.005875 mol
C = 0.423 mol/L
V = 0.025 L
C = 0.158113883 mol/L
V=?
0.0334 L
C = n/V
n=C•V
n = (0.423 mol/l)(0.025 L)
n = 0.010575 mol
( )
1
2
C = n/V
V = n/C
V = (0.005875 mol) ÷ (0.158113883 mol/L)
V = 0.0334410863 L
n = 0.005875 mol
Pg 336: Practice Problem 5.3.3
#’s
1, 2, 3
Calculating Molar Mass in Titrations
Again….but with the new info from other
species….different volume and or concentration
usually.
m
n
M
n
C
V
Only thing that’s different is if the acid or
base is MONOPROTIC (1) or DIPROTIC
(2)….this effects balancing ratio.
C = 0.1261 mol/L
V = 0.02810 L
m = 0.328 g
M=?
92.6 g/mol
C = n/V
n=C•V
n = (0.1261 mol/l)(0.02810 L)
n = 0.00354341 mol
( )
1
1
n = m/M
M = m/n
M = 0.328g ÷ 0.00354341
M = 92.56620036 g/mol
n = 0.0035341 mol
m
n
M
V = 0.0305 L
m = 2.73 g
C = 0.1112 mol/L
V = 0.0250
0.500 L L
M
?
V == 0.0250
L
C = n/V
n=C•V
n = (0.1112 mol/l)(0.0305 L)
n = 0.00333916 mol
( )
1
2
?
Trick question, read carefully.
“A 25.0 ml sample of this….”
m
n
M
C = n/V
n = 0.0016958 mol C = 0.0016958/ 0.0250 L)
C = 0.067832 mol/L
V = 0.0305 L
m = 2.73 g
C = 0.1112 mol/L
V = 0.0250 L
M = 80.5
g/mol n = m/M
?
M = m/n
M = 2.73g ÷ 0.033916 mol
M = 80.49298266 g/mol
C = n/V
n=C●V
n = (0.067832 mol/L)(0.500 L)
n = 0.033916 mol
m
n
M
C = n/V
C = 0.0016958/0.0250 L
C = 0.067832 mol/L
Pg 338: Practice Problem 5.3.4
#’s
1, 2, 3
Calculating Percent Purity
How Math People See The World
Calculating Percent Purity
How it Really Is
Calculating Percent Purity
Just like % yield questions before.
% purity is a measure of the % of mass you have theoretically
calculated (stoich) and actual mass given.
% purity calculations will be used with titration to calculate the
actual.
mGIVEN = 0.3470 g
C = 0.1481 mol/L
mCALC = 0.312141768
??
g V = 0.02026 L
n = m/M
C = n/V
m=n•M
n=C●V
m = (0.00299848 mol)(104.1 g/mol) n = (0.1481 mol/L)(0.02026 L)
n = 0.00299848 mol
( )
1
1
n = 0.00299848 mol
mGIVEN = 0.3470 g
C = 0.1481 mol/L
mCALC = 0.312141768 g V = 0.02026 L
% Purity = Actual Mass (Calculated) x 100%
Given Mass (In Question)
% Purity = 0.312141768 g x 100%
0.3470 g
= 89.9544 %
~ 89.95 %
C = 0.131 mol/L
V = 0.03139 L
=0.0856463125 mol/L
CGIVEN =?
CCALC = ??
C = n/V
n=C●V
n = (0.131 mol/L)(0.03139 L)
n = 0.00411209 mol
( )
1
2
C = n/V
C = n/V
C = 0.0214115781 ÷ 0.2500 L
n = m/M
n = 2.70 g ÷ 126.1 g/mol
n = 0.0214115781 mol
n = 0.002056045 mol
C = 0.131 mol/L
V = 0.03139 L
n = 0.00411209 mol
CGIVEN =0.0856463125 mol/L
CCALC = 0.0822418
mol/L
??
( )
1
2
C = n/V
C = 0.002056045 mol ÷ 0.025 L
n = 0.002056045 mol
C = 0.131 mol/L
V = 0.03139 L
CGIVEN =0.0856463125 mol/L
CCALC = 0.0822418 mol/L
% Purity = Actual Concentration (Calculated) x 100%
Given Concentration (In Question)
% Purity = 0.0822418 mol/L
x 100% = 96.024916643 %
0.0856463125 mol/L
~ 96.0 %
Pg 341: Practice Problem 5.3.5
#’s
1, 2, 3
P. 344
Pg 344 – 346
#’s
1, 2, 5, 6, 9, 11, 13