Transcript Volumetric Analysis

Chapter 5
Volumetric Analysis
Author: J R Reid
CONCEPT OF VOLUMETRIC
ANALYSIS
The reactants will react with the standard
solution from burette of a known
concentration called titrant.
The analyte in which the concentration to
be determined is placed in a conical flask.
The appropriate indicator will add into the
conical flaks to determine the end point,
showed by the color changing of the
indicator
ACID BASE TITRATION

Involves a neutralization reaction in which an acid
is reacted with an equivalent amount of base

Titration curve - pH of the solution versus volume
of titrant added

Titrant - usually strong standard acid or strong
base

Equivalence point - is the point of stoichiometric
equivalence between the analyte and the reagent

End point - a point when reaction is completed
CALCULATIONS
Write the balanced chemical equation for the reaction
Extract all the relevant information from the question
Check that data for consistency, for example, concentrations are
usually given in M or mol L-1 but volumes are often given in mL. You will
need to convert the mL to L for consistency. The easiest way to do this
is to multiply the volume in mL x 10-3
Calculate the moles of reactant (n) for which you have both the
volume(V) and concentration(M) : n = M x V
From the balanced chemical equation find the mole ratio known
reactant : unknown reactant
Use the mole ratio to calculate the moles of the unknown reactant
From the volume(V) of unknown reactant and its previously calculated
moles(n), calculate its concentration(M): M = n ÷ V
EXAMPLES
Question: 30 mL of 0.10 M NaOH neutralised 25.0 mL of HCl. Determine the
concentration of the acid
 Write the balanced chemical equation for the reaction
NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)
 Extract the relevant information from the question:
NaOH
V = 30 mL, M = 0.10M
HCl
V = 25.0 mL, M = ?
 Check the data for consistency
NaOH
V = 30 x 10-3 L , M = 0.10 M
HCl
V = 25.0 x 10-3 L, M = ?
 Calculate moles NaOH
n(NaOH) = M x V = 0.10 x 30 x 10-3 = 3 x 10-3 moles
 From the balanced chemical equation find the mole ratio
NaOH:HCl
1:1
 Find moles HCl
NaOH: HCl is 1:1
So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point
 Calculate concentration of HCl: M = n ÷ V
n = 3 x 10-3 mol,
V = 25.0 x 10-3 L
M(HCl) = 3 x 10-3 ÷ 25.0 x 10-3 = 0.12 M or 0.12 mol L-1
SELF-EXERCISE
1. 50mL of 0.2mol L-1 NaOH neutralised 20mL of H2SO4.
Determine the concentration of the acid (0.25 M)
2. 25.0mL of 0.05M Ba(OH)2 neutralised 40.0mL of nitric
acid. Determine the concentration of the acid. (0.0625
M)
WATER DISSOCIATION
Water is an amphoteric solvent that is having acidic
and basic properties.
H2O
H+ + OHKw= [H+][OH-]= 1.0 x 10-14
[H+]= [OH-]= 1.0 x 10-7 M
Kw=molar equilibrium constant
pH CALCULATIONS

pH= -log [H+]

pKw=pH + pOH

At 25oC, Kw= 1.0 x 10-14, then
14 = pH + pOH

When [H+]=[OH-]=1.0 x 10-7 M or pH=pOH=7, the solution is
neutral.
If [H+]>[OH-] or pH<7, acidic solution.
 If [H+]<[OH-] or pH>7, basic solution.

EXAMPLE:

Calculate
a) pKw for water ionization constant
b) pH and pOH values for 1.5 x 10-3 M hydrogen ion solution.
Solution:
a) pKw = -log Kw
= -log 1.0 x 10-14 = 14.00
b) [H+] = 1.5 x 10-3 M
pH= -log 1.5 x 10-3 = 2.82
pOH= 14.00 - 2.82 =11.18
WEAK ACID AND WEAK
BASE




Weak acid and weak bases will ionize partially
The acidity constant (Ka) is used to calculate the total H+ that has
ionized
The basicity constant (Kb) is used to calculate the total amount of
OH- ionized.
e.g: acetic acid with a concentration of C
Initial (M)
Equilibrium (M)
CH3COOH
C
C-x
H+ + CH3COO0
0
x
x
Ka= [H+][CH3COO-] = 1.75 x 10-5
[CH3COOH]
(x)(x) = 1.75 x 10-5
C-x
If x at the denominator is neglected, the general formula to
determine [H+] for weak acid (HA) with a concentration of CHA:
[H+]= √KaCHA
For weak base, e.g NH3 with basicity constant, Kb will undergo the
following reaction in water.
NH3 + H2O
NH4++ OH-
The same method can be applied to determine [OH-] for a weak base (B)
with a concentration of CB is:
[OH-] = √KbCB
EXAMPLE

Calculate the pH of 1.0 x 10-2 M acetic acid solution. Ka for acetic
acid is 1.75 x 10-5.
Solution
CH3COOH
H+ + CH3COOInitial(M)
1.00 x 10-2
0
0
Equilibrium (M)
1.00 x 10-2-x
x
x
Ka = [H+] [CH3COO-]= 1.75 x 10-5
[CH3COOH]
(x)(x)
= 1.75 x 10-5
1.00 x 10-2-x
X at the denominator can be neglected since C > 100 Ka
(x)(x) = 1.75 X 10-5
1.00 x 10-2
x = 4.18 x 10-4 M = [H+]
pH= -log 4.18 x 10-4 = 3.38
SELF-EXERCISE
Calculate the pH and pOH of a 1.00 x 10-3 M solution of acetic acid.
(answer: pH=3.88;pOH=10.12)
TITRATION OF STRONG ACID
WITH STRONG BASE
The titrant and the analyte ionize completely since strong
acid and strong base dissociate completely
e.g: titration between HCl and NaOH
HCl + NaOH
H2O + NaCl
TITRATION CURVE
STRONG ACID WITH STRONG BASE
NaOH (aq) + HCl (aq)
H2O (l) + NaCl (aq)
• Before addition of NaOH
- pH = 1.00
• When the NaOH added
- pH increase slowly at first
• Near the equivalence point (the point which equimolar
amounts of acid and base have reacted)
- the curve rises almost vertically
• Beyond the equivalence point
- pH increases slowly
CALCULATION OF pH AT EVERY
STAGE OF TITRATION
1) After addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of
0.100 M HCl
Total volume = 35.0 mL
i) Moles of NaOH in 10.0 mL
M = mol
L
= 0.1 x (10 mL/1000 mL)
= 1.00 x 10-3 mol
ii) Moles of HCl in 25.0 mL
M = mol
L
= 0.1 x (25 mL/1000 mL)
= 2.50 x 10-3 mol
Amount of HCl left after partial neutralization
= (2.50 x 10-3) - (1.00 x 10-3)
= 1.50 x 10-3 mol
Concentration of H+ ions in 35.0 mL
M = mol
L
= 1.50 x 10-3 mol
(35/1000)
= 0.0429 M
[H+] = 0.0429 M,
pH = -log 0.0429 = 1.37
2) After addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl
[H+] = [OH-] = 1.00 x 10-7
pH = 7.00
3) After addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 mL of HCl
Total volume = 60.0 mL
Moles of NaOH added
M = mol
L
= 0.1 x (35 mL/1000 mL)
= 3.50 x 10-3 mol
Moles of HCl in 25.0 mL solution = 2.50 x 10-3 mol
After complete neutralization of HCl, no of moles of NaOH left
= (3.50 x 10-3) - (2.50x10-3)
= 1.00 x 10-3 mol
Concentration of NaOH in 60.0 mL solution
M = mol
L
= 1.00 x 10-3 mol
(60/1000)
= 0.0167 M
[OH-] = 0.0167 M
pOH = -log 0.0167 = 1.78
pH = 14.00 – 1.78
= 12.22
WEAK ACID versus STRONG
BASE WITH BUFFER REGION
The titration between acetic acid and NaOH.
CH3COOH + NaOH
H2O + CH3COONa
The acetic acid which is only a few percent ionized
is neutralized to water and equivalent amount of
CH3COONa salt
•For the first part of the graph, we have an excess of sodium hydroxide. The
curve will be exactly the same as when we add hydrochloric acid to sodium
hydroxide.
•Once the acid is in excess, there will be a difference.
•Past the equivalence point - a buffer solution containing sodium acetate
and acetic acid.
As soon as the titration is started, some of the acetic acid is converted to
sodium acetate
As the titration proceeds, the pH slowly increased as the ratio
[Acetate]/[Acetic acid] changes.
At the midpoint of the titration, [Acetate]/[Acetic acid] and the pH is equal
to pKa.
At the equivalence point, we have a solution of sodium acetate
Since this is a Bronsted base, the pH at the equivalence point will be
alkaline.
The pH will depend on the concentration of sodium acetate
The greater the concentration, the higher the pH.
WEAK BASE versus STRONG
ACID WITH BUFFER REGION
The titration between hydrochloric acid and
ammonia
NH3 + HCl
NH4+ + Cl-
•At the very beginning of the curve, the pH starts by falling quite quickly as the acid is
added, but the curve very soon gets less steep (some of the NH3 is converted to
NH4+). This is because a buffer solution is being set up - composed of the excess
ammonia and the ammonium chloride being formed
•At the midpoint of the titration, [NH4+] equals [NH3]
•The equivalence point (solution of NH4Cl form) is somewhat acidic (less than pH 5),
because pure ammonium chloride isn't neutral. However, the equivalence point still
falls on the steepest bit of the curve.
•Beyond the equivalence point, the pH is determined by the concentration of H+
added in excess.