Solution stoichiometry involves calculation of the volume of one solution required to react with another solution.

1. 1
Solution Stoichiometry: Reacting
Volumes
• Solution stoichiometry involves calculation of the
volume of one solution required to react with
another solution.
• Solution stoichiometry uses the definition molarity
to determine volumes of reagents needed to
perform the experiment.
• In liquids, it is necessary to convert volume into
moles using molarity.
Moles = Molarity x Volume

2. 2
Example: Reacting Volumes
• A 0.0823 M solution of sodium chloride (NaCl) is
added to 21.40 mL of 0.962 M silver nitrate
(AgNO3) solution forming a precipitate of silver
chloride. Calculate the minimum volume of the
sodium chloride solution required for complete
reaction.
Solution;
• First come up with the balanced chemical equation
and interpret it using the mole method.

3. 3
Example: Reacting Volumes
• Mole interpretation: 1 mol NaCl reacts with 1 mol
AgNO3 to produce 1 mol AgCl
• Calculate the number of moles of the reactant with
known molarity and volume using the relationship:
Moles = Molarity x Volume
• The reactant with known volume and molarity is
silver nitrate (21.40 mL of 0.962 M of AgNO3).

4. 4
Example: Reacting Volumes
• Calculate the number of moles of the second reactant
using the stoichiometric coefficients of the balanced
equation.
– The balanced chemical equation tells us that 1 mol
NaCl reacts with 1 mol AgNO3.
– Therefore the number of moles of NaCl required to
react with 0.0206 mol AgNO3 will be 0.0206 mol.
• Calculate the volume of the second reactant using the
equation:

5. 5
Practice: Reacting Volumes
1. Sodium hydroxide (NaOH) reacts with sulphuric acid
(H2SO4) according to the following equation:
2NaOH (aq) + H2SO4 (aq) → 2H2O (l) + Na2SO4 (aq)
Find the volume of 0.112 M NaOH solution required to
completely react with 10.0 mL of H2SO4 solution.
2. Calcium hydroxide is sometimes used in water treatment
plants to clarify water for residential use. Calculate the
volume of 0.0250 mol/L calcium hydroxide solution that
can be completely reacted with 25.0 mL of 0.125 mol/L
aluminum sulfate solution.
Al2(SO4) (aq) + 3 Ca(OH)2 (aq) → 2 Al(OH)3 (s) + 3 CaSO4 (s)

6. 6
Volumetric Analysis
• Volumetric analysis is the analytical technique for
determining the amount of a certain substance in the
solution by measuring the volume of a reacting solution.
• In the volumetric analysis, the concentration of a
solution of unknown concentration is worked out from
the volume of a solution of a known concentration.
• Technique involves working out the amount of one
reactant from the volume of the solution and its conc’.
Then the amount worked out is converted to the amount
of the other reactant using the stoichiometric
relationship from the balanced chemical equation.

7. 7
Volumetric Analysis: Example
• 25.00 mL of phosphoric acid (H3PO4) solution required
27.55 mL of a 0.155 M potassium hydroxide (KOH)
solution for neutralization.
Calculate the molarity of the phosphoric acid
The equation for the reaction is:
H3PO4 (aq) + 3 KOH (aq) → K3PO4 (aq) + 3 H2O (l)
Solution;
• From equation: 1 mol H3PO4 reacts completely with 3 mol
KOH

8. 8
Volumetric Analysis: Example
• Calculate the number of moles of the reactant with
known molarity and volume (i.e. KOH):
• Using the stoichiometric coefficients of the balanced
equation, calculate the number of moles of H3PO4
required to completely react with 4.27 x 10-3 mol KOH.
Thus;

9. 9
Volumetric Analysis: Example
• Therefore 1.42 x 10-3 mol H3PO4 were present in
25.00 mL of H3PO4
• Calculate the molarity of the H3PO4 using the
equation:

10. 10
Volumetric Analysis: Practice
• Ammonium sulfate is manufactured by reacting
sulfuric acid with ammonia according to the
reaction equation:
H2SO4 (aq) + 2 NH3 (aq) → (NH4)2SO4 (aq)
Find the concentration of sulfuric acid required to
react with 24.4 mL of a 2.20 mol /L ammonia
solution if 50.0 mL of sulfuric acid is used.

11. • Titration: the process of analyzing composition by measuring
the volume of one solution needed to completely react with
another solution.
This is a special case of a Limiting Reagent!
Volumetric Analysis (Titrimetry)

12. • In a titration, a solution of accurately known conc’ is
added gradually to another solution of unknown
concentration until the chemical reaction between the
two solutions is complete.
• Analyte: the solution of unknown concentration but
known volume.
• Titrant: the solution of known concentration
Analyte + Titrant → Products
❖ Add titrant until all of the analyte has reacted
( then need to detect excess of titrant).
Titration

13. 13
• Equivalence point – the point where the amount of titrant
added is enough to completely neutralize analyte solution.
• Indicator – substance that changes colour at (or near) the
equivalence point (i.e. when excess titrant has been added).
• End point: the point where the indicator changes its colour.
Slowly add base to
unknown acid UNTIL
the indicator
changes color
E.g. for reaction of an acid with a base.
Titration

14. Titration Setup

15. 15
Titration Calculations
1. Find the number of moles of titrant added to reach
the endpoint.
2. Determine the moles of analyte that must have
been present (use stoichiometric coefficients).
3. Determine the concentration of analyte that must
have been present (use the volume of analyte).

16. 16
Titrations can be used in the analysis of
• Acid-base reactions
• Redox reactions
H2SO4 + 2NaOH 2H2O + Na2SO4
5Fe2+ + MnO4
- + 8H+ Mn2+ + 5Fe3+ + 4H2O
• Precipitation reactions
Ag+ (aq) + Cl- (aq) AgCl (s)
2Ag+ (aq) + CrO4
2- (aq) Ag2CrO4 (s)

17. An acid-base titration (titrating an acid with base).
Start of titration
Excess of acid
Point of
neutralization
Slight excess
of base
At equivalence point : moles H+ = moles OH-

18. 18
What volume of a 1.420 M NaOH solution is required
to titrate 25.00 mL of a 4.50 M H2SO4 solution?
WRITE THE CHEMICAL EQUATION!
volume acid moles acid moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
M
acid
rxn
coef.
M
base
n(H2SO4) = C(M) x V n(H2SO4) = 0.025 L x 4.5 mol/L = 0.1125 mol
From stoichiometry
2 mol NaOH = 1 mol H2SO4
 n(NaOH) = 2 x n(H2SO4) = 2 x 0.1125
= 0.225 mol
C (M)
n
 V =
1.420 M
0.225 mol
 V = = 158 mL NaOH

19. A Redox Titration

20. 20
16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize
25.00 mL of an acidic FeSO4 solution. What is the molarity
of the iron solution?
WRITE THE CHEMICAL EQUATION!
volume red moles red Moles oxid Volume oxid
M
acid
rxn
coef.
M
base
n(MnO4
-) = C(M) x V n(MnO4
-) = 0.01642 L x 0.1327M = 0.0022 mol
From stoichiometry
5 mols Fe2+ = 1 mol MnO4
-
 n(Fe2+) = 5 x n(MnO4
-) = 5 x 0.0022
= 0.011 mol
V
n
 C (M) =
0.025 L
0.011mol
= = 0.4358 M FeSO4
5Fe2+ + MnO4
- + 8H+ Mn2+ + 5Fe3+ + 4H2O
16.42 mL = 0.01642 L 25.00 mL = 0.02500 L

21. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for
titration to an equivalence point. What is the concentration of the
NaOH?
Part A: Standardize a solution of NaOH — i.e., accurately
determine its concentration.
Step 1: Calculate amount of H2C2O4
90.04 g/mol
1.065 g
 n =
M
m
 n = = 0.0118 mol
Step 2: Calculate amount of NaOH required
H2C2O4(aq) + 2 NaOH(aq) ---> Na2C2O4(aq) + 2 H2O(liq)
acid base
From stoichiometry
1 mol H2C2O4 = 2 mols NaOH
 n(NaOH) = 2 x n(H2C2O4) = 2 x 0.0118
= 0.0236 mol NaOH
Step 3: Calculate concentration of NaOH
V
n
 C (M) =
0.03562 L
0.0236 mol
= = 0.663 M NaOH

22. Apples contain malic acid, C4H6O5. 76.80 g of apple requires
34.56 mL of 0.663 M NaOH for titration. What is weight % of
malic acid?
Part B: Use standardized NaOH to determine an amount
of acid
Step 1: Calculate amount of NaOH used
 n = C. V = 0.0229 mol
Step 2: Calculate amount of malic acid titrated
C4H6O5(aq) + 2 NaOH(aq) ---> Na2C4H4O5(aq) + 2 H2O(l)
From stoichiometry
1 mol C4H6O5 = 2 mols NaOH
 n(C4H6O5) = 1/2 x n(NaOH) = 1/2 x 0.0229
= 0.0115 mol
Step 3: Calculate the mass of malic acid titrated
= 1.54 g
 n = 0.663 M x 0.03456 L
 m = n.M = 0.0115 mol x 134 g/mol

23. Apples contain malic acid, C4H6O5. 76.80 g of apple requires
34.56 mL of 0.663 M NaOH for titration. What is weight % of
malic acid?
Step 4: Calculate weight percentage of malic acid
= 2.01 %
m (apple)
m (malic acid)
 % malic acid = x 100%
76.80 g
1.54 g
 % malic acid = x 100%

24. aA + bB products
Remember that both A and a are on top and both B and b
are on the bottom
A general expression
Consider a general
reaction:
Therefore
From stoichiometry:
Amount (mol) A
Amount (mol) B
a
b
=
But we also know that: Amount of A in mols, nA = CAVA
Amount of B in mols, nB = CBVB
[A] VA
[B] VB
a
b
=
CA VA
CB VB
a
b
=