Transcript 1 Stoichiometry Revision DEA1

Chapter: 1 Stoichiometric Relationships
Subtopics for revision:
 States of matter
 Compounds and homogenous and heterogeneous mixtures
 Separating mixtures
 Avogadro's number, moles, molar mass and mass
 Empirical and molecular formula
 Calculating mass of product and limiting reagents
 Calculations involving volumes of gases
 Gas laws and ideal gas equation
 Real gas behaviour
 Solutions and concentration
 Titration
States of matter
States of Matter
Solids
Liquids
Gases
Distance between
particles
Close Together
Close (ish)
Far Apart
Arrangement
Regular
Random
Random
Shape
Fixed
Not Fixed – Takes
container shape
Not fixed – Fills
container
Volume
Fixed
Fixed
Not Fixed
Movement
Vibrate
Move around each
other
Move in all directions
Speed of Movement
Slowest
Faster
Fastest
Energy
Lowest
Higher
Highest
Forces of attraction
Strongest
Weaker
Weakest
Sublimation is the direct inter conversion of solid to gas without going through the liquid
state. Examples of this are:
Iodine, Carbon dioxide and Ammonium chloride (all at atmospheric pressure)
Deposition is the direct conversion of gas to solid without going through the liquid state.
Examples of this are:
Snow and Frost
NOTE! Evaporation and boiling are not the same thing! Evaporation occurs only at the
surface and takes place at temperatures below the boiling point.
Q. Where do you think boiling takes place? Why does boiling a liquid produce bubbles?
Boiling happens when particles leave throughout the body of the liquid – which is why
bubbles occur.
Boiling occurs at a specific temperature, determined by when the vapour pressure
reaches the external pressure.
Nature of Science
Compounds and Mixtures
Separating Mixtures
Summary of differences between compounds and mixtures
Homogeneous and Heterogeneous Mixtures
Homogeneous
• Example - a solution e.g. saltwater
• No individual particles can be seen throughout the solution
• Concentration is the same throughout – For example this can be checked by
evaporating several 1cm3 samples of Sodium chloride. The same amount of salt
should be formed with each sample.
• Consists of one phase
Heterogeneous
• Examples – Sand in a beaker of water, water and oil
• Individual particles can be seen throughout (different components can be
distinguished from each other)
• Does not have a uniform composition
• Consists of separate phases
• Can be separated by mechanical or physical means e.g. filtering, magnets etc…
Table of properties relating to separation techniques
Avogadro’s Number, moles, molar mass and mass
Different types of mass
• Relative atomic mass, Ar:
Used when referring to the mixture of naturally occurring
isotopes in an element and is found in the periodic table
Definition
• The weighted mean mass of an atom of an element
compared with 1/12 of the mass of an atom of carbon-12.
• E.g. The weighted average of silver is 107.87. This means
on average silver has 107.87 times the mass of 1/12th of a
carbon atom.
• NOTE: The are no units as quantities are relative.
If you compare masses with each other in relative terms you can
see the relationship:
• Oxygen has a mass
16 times greater
than an atom of
Hydrogen
• Sulphur has a mass
32 times greater
than an atom of
Hydrogen
• Sulphur has a mass
2 times greater than
Oxygen
A Mole
• A ‘mole’ represents a large number (just like dozen represents
12):
• A mole is:
– 6.02 x 1023
– Given the symbol, L
– This number is called Avogadro’s number
Some Calculations
•We can use this equation
to calculate a number of
moles from a number of
particles
N
n
L
•Where:
•n = quantity in moles
•N = number of
particles
•L = Avogadro’s
Constant (6.02x1023)
1.
Example 1: You have 3.01x1022 atoms of carbon. How
many moles is this?
•n = N / L
•n(C) = 3.01x1022 / 6.02x1023
•n(C) = 0.0500 mol
2.
Example 2: You have 6.02x1024 molecules of water. How
many moles of hydrogen atoms are present?
•n = N / L
•n(H) = 2 x n(H2O)
•n(H) = 2 x 6.02x1024 / 6.02x1023
•n(H) = 20.0
3.
Example 3: How many atoms of hydrogen are there in
2.5 moles of methane (CH4)?
•N(H) = 4 x N(CH4)
•N(H) = 4 x n(CH4) x L
•N(H) = 4 x 2.5 x 6.02x1023
•N(H) = 6.02x1024
The number of particles in a molecule
1 mol O2 means one mole of O2 molecules = 6.02 x1023 molecules of O2.
In a molecule of O2 = 2 atoms of Oxygen
Therefore one mole of 02 molecules is made up of 2 lots of 6.02 x 1023 Oxygen atoms.
So 1 mole of O2 molecules = 2 moles of Oxygen atoms
Table to show the relationship between number of moles of molecules and the
number of moles of particular atoms.
Compound
Moles of molecules
Moles of O atoms
H2O
0.1
0.1
SO2
0.1
0.2
SO3
0.1
0.3
H3PO4
0.1
0.4
O3
0.5
1.5
CH3COOH
0.2
0.4
Empirical formula and Molecular Formula
Empirical formulae
Empirical formulae is simply a way of showing how many atoms are in a molecule (like
a chemical formula). For example, CaO, CaCO3, H20 and KMnO4 are all empirical
formulae. Here’s how to work them out:
A classic exam question:
Find the simplest formula of 2.24g of iron reacting
with 0.96g of oxygen.
Step 1: Divide both masses by the relative atomic mass:
For iron 2.24/56 = 0.04
For oxygen 0.96/16 = 0.06
Step 2: Write this as a ratio and simplify:
0.04:0.06 is equivalent to 2:3
Step 3: Write the formula:
2 iron atoms for 3 oxygen atoms means the formula is Fe2O3
Calculating empirical formulae
Empirical formula examples:
Using mass
Using
percentages
Using water
of
crystalisation
to determine
empirical
formula of a
hydrated salt
Using
combustion
to determine
hydrocarbon
empirical
formula
Determining the identity of an element in an empirical formula from percentage
by mass data
Determining the number of atoms of an element in a molecule given its molar mass and
the percentage by mass of the element
Percentage composition by mass and Empirical formula
• It is common for the composition data to be given in the form of
percentage by mass.
• We can use these figures in the same way to deduce the ratio of atoms
present.
• Percentage data effectively gives us the mass present of each atom in
100g of the compound.
Tip: Just substitute percentages for mass of the element in grams. Use the
same way as before to work out the empirical formula…
Types of formulae
The empirical formula is the formula that represents the
simplest ratio of the atoms of each
element present in a compound.
For example, the empirical formula of hydrogen peroxide is HO – the ratio
of hydrogen to oxygen is 1:1.
The molecular formula gives the actual number of atoms of
each element in one molecule of the compound.
The molecular formula of hydrogen peroxide is H2O2 – there are two atoms of
hydrogen and two atoms of oxygen in each molecule.
Examples
 Oxygen
 Molecular: O2
 Empirical: O
 Water


 Molecular: C2H6
 Empirical: CH3
Molecular: C6H12O6
Empirical: CH2O
Sodium Chloride

 Molecular: H2O
 Empirical: H2O
 Ethane
Glucose



Molecular: n/a*
Empirical: NaCl
Copper Sulphate



Molecular: n/a*
Empirical: CuSO4
*Why do these two not have an empirical formula?
Determining molecular formulas
 Formulas that show all the atoms present in a molecule, are
called molecular formula.
 For example, CH2 is an empirical formula. There is no
molecule that exists with just one atom of carbon and two
atoms of hydrogen, but there are many molecules with
multiples of this ration, such as C2H4, C3H6, and so on.
Calculating molecular formulae
The molecular formula can be found by dividing the Mr by the relative mass of the empirical formula.
Example: What is the molecular formula of hydrogen peroxide given that
its empirical formula is HO and the Mr is 34?
1. Determine relative mass of empirical formula:
empirical formula mass = H + O = 1.0 + 16.0 = 17
2. Divide Mr by mass of empirical formula to get a multiple:
multiple =
relative molecular mass
mass of empirical formula
3. Multiply empirical formula by multiple:
HO × 2 = H2O2
=
34
17
= 2
Determining Molecular Formulas From
Empirical Formulas
Your unknown molecule is found to have Mr = 30.06. Empirical formula
is CH3. Determine the molecular formula of the unknown molecule.
Solution:
• Determine empirical formula mass:
Mr(CH3) = 12.00 + (3 x 1.01) = 15.03
• Divide molecular mass by empirical formula mass to tell you the
multiplier for the empirical formula:
Multiplier = 30.06 / 15.03 = 2
• Multiply the empirical formula by the multiplier:
Molecular formula = CH3 x 2 = C2H6
Calculating mass of product and limiting reagent
Calculating the mass of a product
E.g. what mass of magnesium oxide is produced when 60g of
magnesium is burned in air?
Step 1: READ the equation:
2Mg + O2
2MgO
IGNORE the oxygen
in step 2 – the
question doesn’t ask
for it
Step 2: WORK OUT the relative formula masses (Mr):
2Mg = 2 x 24 = 48
2MgO = 2 x (24+16) = 80
Step 3: LEARN and APPLY the following 3 points:
1) 48g of Mg makes 80g of MgO
2) 1g of Mg makes 80/48 = 1.66g of MgO
3) 60g of Mg makes 1.66 x 60 = 100g of MgO
1) When water is electrolysed it breaks down into hydrogen and oxygen:
2H2O
2H2 + O2
What mass of hydrogen is produced by the electrolysis of 6g of water?
Work out Mr:
1.
2H2O = 2 x ((2x1)+16) = 36
2H2 = 2x2 = 4
36g of water produces 4g of hydrogen
2. So 1g of water produces 4/36 = 0.11g of hydrogen
3. 6g of water will produce (4/36) x 6 = 0.66g of hydrogen
2) What mass of calcium oxide is produced when 10g of calcium burns?
2Ca + O2
Mr: 2Ca = 2x40 = 80
2CaO
2CaO = 2 x (40+16) = 112
80g produces 112g so 10g produces (112/80) x 10 = 14g of CaO
3) What mass of aluminium is produced from 100g of aluminium oxide?
2Al2O3
4Al + 3O2
Mr: 2Al2O3 = 2x((2x27)+(3x16)) = 204
4Al = 4x27 = 108
204g produces 108g so 100g produces (108/204) x 100 = 52.9g of Al2O3
Finding how much reactant need to make04/08/2016
product
The Limiting Reagent

In a reaction, there is we can describe reactants as being ‘limiting’ or in ‘excess’


Limiting – this is the reactant that runs out
Excess – the reaction will not run out of this
2 H2 + O2  2 H2O

For example, if you have 2.0 mol H2 and 2.0 mol O2



To determine this, divide the molar quantity of each reactant by its coefficient
in the equation. The smallest number is the limiting reactant:



H2 is the limiting reactant – it will run out
O2 is present in excess – there is more than enough
H2: 2.0 / 2 = 1.0 – smallest therefore limiting
O2: 2.0 / 1 = 2.0
You should use the limiting reactant when doing all further calculations
including:


Determining amounts of products formed
Determining amounts of other reactants used
Limiting reagent and reactant in excess
Percentage purity
The theoretical yield is determined by the limiting
reactant
Percentage yield
=
actual yield (in g)
theoretical yield
X100

The theoretical yield, which is usually expressed in grams or moles,
refers to the maximum amount of product obtainable, assuming
100% of the limiting reactant is converted to product.

Identification of the limiting reactant depends on the mole ratios in
the balanced equation for the reaction. Hence grams of reactants
need to be converted to moles.
Determining Theoretical Yield


The theoretical yield is the amount of product you expect to make if all of
your limiting reactant fully reacts.
It can be calculated by following these steps:
Calculate moles of all reactants
Determine limiting reactant
Use mole ratios to calculate moles of product expected
Convert moles of product to mass/volume/solution etc.

All this does is string together all the calculations you have already met.
Calculations involving volumes of gases
How do the number of particles in the two flasks
compare?

The number of particles in the two flasks are the
same.

Even though bromine molecules are larger and
heavier, this is not relevant because of the nature of
gaseous state. We assume that the volume of
individual gas molecule are zero.

Particles in a gas are widely spaced out with
negligible forces between them. Most gas space is
EMPTY. We call this an IDEAL GAS.

Therefore gas volume is determined only by the
number of particles and the temperature and
pressure.
Avogadro’s law
In 1811 the Italian scientist Amedeo Avogadro developed a
theory about the volume of gases.
Avogadro’s law:
Equal volumes of different gases at the same pressure
and temperature will contain equal numbers of particles.
For example, if there are 2 moles of O2 in 50 cm3 of oxygen
gas, then there will be 2 moles of N2 in 50 cm3 of nitrogen
gas and 2 moles of CO2 in 50 cm3 of carbon dioxide gas at
the same temperature and pressure.
Using this principle, the volume that a gas occupies will
depend on the number of moles of the gas.
Alternatively, it can be stated that equal numbers of particles of all gases, when
measured at the same temperature and pressure, occupy equal volumes.


The volume occupied by a gas is known as the MOLAR VOLUME
This can be used in a similar way to using MOLAR MASS but these
calculations are easier because gases have the same molar volume
under the same conditions
The Molar Volume of an Ideal Gas

At standard temperature and pressure (STP):

Molar Volume of Gas
= 22.7 dm3 mol-1
= 2.27x10-2 m3 mol-1

T = 273K (0oC)

P = 1.01x105 Pa (100kPa)
Molar volumes of gases
If the temperature and pressure are fixed at convenient
standard values, the molar volume of a gas can be
determined.
Standard temperature is 273 K and pressure is 100 kPa.
At standard temperature and pressure, 1 mole of any gas
occupies a volume of 22.7 dm3. This is the molar volume.
Example: what volume does 5 moles of CO2 occupy?
volume occupied = no. moles × molar volume
= 5 × 22.7
= 113.5 dm3
Example calculation:
What volume of H2 gas is produced when 0.050 mol Li reacts with
excess Hydrochloric acid at S.T.P.?
Solution:
 Write balanced equation

2 Li(s) + 2 HCl(aq)  2 LiCl(aq) + H2(g)

Work out the mole ratio:
2 moles Li to 1 mole H2
0.050 mol Li reacts to form 0.025 mol of H2




Plug into the formula
V = 0.025 x 22.7 = 0.567 dm3
Another example… with limiting reagents
At STP, 30 cm3 ethane reacts with 60 cm3 oxygen. Which reactant is
present in excess, how much remains after the reaction and what volume
of CO2 is produced?
Solution:

2 C2H6(g) + 10 O2(g)  6 H2O(l) + 4 CO2(g)

Limiting reactant:


C2H6: 30/2 = 15
O2: 60/10 = 6

therefore O2 limiting, ‘6’ will be the number used in all further calculations as there is enough
O2 for ‘6’ of the reaction
Volume CO2 produced:

C2H6 remaining:




V(CO2) = 6 x 4 = 24 cm3
V(C2H6 used) = 6 x 2 = 12 cm3
V(C2H6 remaining) = 30 – 12 = 18 cm3 EXCESS AFTER REACTION
Note: there is no need to convert to moles as they are all in a ratio of moles
as you would divide by 22.7 to get to moles, do your sums and then multiply
by 22.7 to get back to volumes… So why bother?
Gas laws and ideal gas equation
Look at the diagrams – if the pressure, what happens to the
volume?
Pressure of a gas in inversely
proportional to its volume.
The product of pressure and
volume is a constant.
BOYLE’S LAW
What happens when particles of a gas are heated?

There is an increase in the average kinetic energy of the particles.
If the pressure is held constant, what happens to volume?

There is a proportional increase in the volume
How can you describe the relationship between temperature and volume?
CHARLES’
LAW
What happens when particles of a gas are heated?

There is an increase in the average kinetic energy of the particles.
If the volume is held constant, what happens to pressure?

There is a proportional increase in the pressure
How can you describe the relationship between temperature and pressure?
GAY-LUSSAC’S
LAW
Q: Why do pressurised cans often carry a warning to be
stored in a cool place?
Higher temperatures in
a fixed volume will
mean an increase in
pressure.
This can cause the can
to explode!
These three gas laws can be summarised as follows:
BOYLE’S LAW
CHARLES’ LAW
GAY-LUSSAC’S LAW
http://www.youtube.com/watch?v=bftkRnTcFj8
NOTE:Temperature
must be in Kelvin (K)
Ideal gas equation: converting units
It is very important when using the ideal gas equation that
the values are in the correct units.
The units of pressure, volume or temperature often need
to be converted before using the formula.
Pressure
to convert kPa to Pa:
× 1000
Volume
to convert dm3 to m3:
to convert cm3 to m3:
÷ 1000 (103)
÷ 1 000 000 (106)
Temperature
to convert °C to Kelvin: + 273
CHARLES’ LAW
HINT: Look at what information you have… What
equation can you use?
Molar Volume of a Perfect Gas



We learnt about the molar volume of gases last lesson….how can
they be the same?
The distance between particles is much bigger than the size of the
particles….so particle size makes very little difference:
The blue particle is twice the size of the red particle, but the blue
particles are not taking up twice the amount of space.
10 units
10 units

In reality, the relative distance between the molecules is much, much
greater than this.
The Ideal Gas Equation

The volume a gas takes up is determined by:





Pressure
Temperature
Moles of gas
The value of the constant is directly proportional to the fixed mass of gas,
or the number of moles, n
This combines to form the ideal gas equation
PV = nRT

Where:





P = pressure in Pa
V = volume in m3
n = moles of gas
R = gas constant, 8.31 J K-1 mol-1
T = temperature in K
Calculating the Mr of gases
Using the ideal gas equation
Ideal Gas Assumptions

Particles occupy no volume

Particles have zero intermolecular forces

These are not always valid, particularly at:


Low temperature
High pressure
Study the equation to make some
predictions:
PV = nRT
How would does temperature change if you decrease pressure
at fixed volume?
If you decrease pressure, temperature will decrease.

How would volume change if you heat something at fixed
pressure?
If you increase temperature, volume will increase.

How would pressure change if you decrease the volume at fixed
temperature?
If you decrease volume, pressure will increase.


Use of the ideal gas equation enables us to calculate how
systems respond to changes in pressure, volume, and
temperature, and to calculate molar mass.

Gas density can also be derived from using the ideal gas
equation by applying the relationship:

When substituting values into the ideal gas equation, pay
attention to the SI units:
PV = nRT
Example calculations:

1.048 g of unknown gas A, occupies 846 cm3 at 500K
What is it’s molecular mass?
 n( A) 

PV

RT
846
1000 1000  0.02056
8.31 500
101,000 
Mm(A) = mass / moles = 1.048 / 0.02056 = 50.96 g/mol
More calculations

The volume of an ideal gas at 54.0 °C is increased from 3.00 dm3 to 6.00
dm3. At what temperature, in °C, will the gas have at the original pressure?

Use a modified version of the ideal gas equation:
PV
T
1
1

1

PV
T
2
2
2
Since original and final pressure should be the same, we can remove this
from the equation as they cancel out:
54.0 converted to
Kelvin by adding 273
V V
T T
1
2
1
2
3.00 / 327.0 = 6.00 / T2
T2 = (6.00 x 327.0 / 3.00) = 654 K = 381oC
Key Points

The Ideal Gas equation:
PV = nRT

Also:
PV
T
1
1

1

PV
T
2
2
2
Provided that:


Molecules have zero volume
Molecules experience no attraction to each other
Real gas behaviour
An ideal gas is defined as one
that obeys the ideal gas law
PV=nRT under all conditions. E.g.
For one mole of gas, the
relationship PV/RT should be
equal to 1.
However, this just theory, there is
no such thing as an ideal gas. All
gases that exist are know as
REAL GASES, and these will
deviate from ideal behaviour.
Graph showing variation from 1 of real 1 molar gases at
different conditions
Q: What conclusions can we draw from the graph and the deviation in relation
to pressure and temperature?
• The gas behaves most like an ideal gas at low
pressure and shows greatest deviation at high
pressure.
•The gas behaves most like an ideal gas at high
temperature and shows greatest deviation at low
temperature.
BUT WHY??? WE NEED TO QUESTION THE
VALIDITY OF THE TWO ASSUMPTIONS WE
MAKE ABOUT IDEAL GASES…
1. The volume of the gas particles in negligible
2. There are no attractive forces between particles. (Hence an ideal gas can never
liquefy!)
Assumptions
 The volume of the gas particles in negligible
=n
Assumption - Volume
 At low pressure, e.g. 1x105 Pa (STP), the volume occupied by the
particles of a typical gas is only 0.05% of the total volume =
negligible so VALID ASSUMPTION…

But if you increase the pressure to 5x105 Pa, the volume of the
particles is about 20% of the total volume = not so negligible…

Therefore, the volume of a real gas at high pressure is larger than
that predicted by the ideal gas law and PV/nRT > 1
Assumption
 There are no attractive forces between
particles.
=n
Assumption – Interactive Forces
 At low pressure, e.g. 1x105 Pa (STP), the particles are so widely
spaced apart, interactive forces are highly unlikely = Valid
assumption!

But if you increase the pressure to 3x107 Pa, the particles are much
closer, so interactive forces strengthen. This will cause a reduction
of pressure of the gas.

Therefore, the pressure of a real gas when at high pressure is
smaller than that predicted by the ideal gas law and PV/nRT < 1
Assumption
 There are no attractive forces between
particles.
=n
Temperature
 At low temperature, the lower kinetic energy of the particles
increase the inter-particle forces.
 Therefore, the pressure of a real gas when at low temperature is
smaller than that predicted by the ideal gas law and PV/nRT < 1
CONCLUSION: Real gases deviate from ideal behaviour when either
or both assumptions are not valid.
Assumptions
1. The volume of the gas particles in negligible
2. There are no attractive forces between particles. (Hence an ideal gas can
never liquefy!)
Solutions and concentration
Concentration

The amount of solute dissolved in a unit of solution.

The volume that is usually taken is 1dm3. The amount of solute may be expressed
in g or mol therefore the units of concentration are g dm-3 or mol dm-3.

Unlike gases, the volume of a liquid is not directly related to it’s amount. For
solutions, we express the amount through it’s CONCENTRATION.
Molarity

The number of moles of a substance dissolved in one litre (dm3) of a solution.
moles
concentration 
volume

Units: mol dm-3




Pronounced: moles per decimetre cubed
Units often abbreviated to ‘M’ (do not do this in an exam!)
Volume must be in litres (dm3) not ml or cm3
This is the most useful measure of concentration but there are others such as %
by weight, % by volume and molality.
Concentration can also be expressed in mass (g dm-3)
Parts per million (ppm)
•
•
•
Another unit of concentration
Denotes one part per 106 by mass.
Useful for very low concentrations such as found in air and water pollution
•
A concentration of 1 ppm for a substance means that each kilogram of solution contains
1 milligram of solute.
Assuming a density of 1 g dm-3, 1ppm also means each dm3 of solution contains 1 mg of
solute.
•
You can make a dilution from a more concentrated starting solution, called a stock solution,
by adding a solvent.
As a solution is diluted, the number of moles of solute remains the same, but now they are
spread over a larger volume. Hence, concentration is decreased.
Number of moles is constant. n=cV
Concentration x volume must be constant through dilution
Titration
• If we had an unlabelled bottle of HCl, how can we find out it’s
concentration using neutralisation?
• We can react an acid with a standard solution of alkali such as NaOH, and
determine the exact volumes reacting together.
• If we have the volumes of both and the concentration of one we can work
out the unknown concentration.
Titrations can be used to find:
1. Concentration of a solution
2. Mass of chemical
3. Molar Mass (Mm)
4. Formula
5. Determining percentage purity
Carrying out a titration
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Titration calculations examples
What is the concentration of an NaOH solution if 25.0 cm3
is neutralized by 23.4 cm3 0.998 mol dm-3 HCl solution?
1. Calculate no.
moles HCl:
moles = (conc. × volume) / 1000
= (0.998 × 23.4) / 1000
= 0.0234
2. Determine ratio NaOH + HCl  NaCl + H2O
of NaOH to HCl: ratio NaOH:NaCl = 1:1
3. Calculate no.
0.0234 moles HCl = 0.0234 moles NaOH
moles of NaOH:
4. Calculate conc.
of NaOH:
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conc. = (moles × 1000) / volume
= (0.0234 × 1000) / 25.0
= 0.936 mol dm-3
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Titration calculations
98 of 35
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More titration calculations
99 of 35
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Example

It is found by titration that 25.0 cm3 of an unknown solution of sulfuric acid
is just neutralised by adding 11.3 cm3 of1.00 mol dm-3 sodium hydroxide.
What is the concentration of sulfuric acid in the sample.
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O

Use:
n2C1V1 = n1C2V2
2 x C1 x 25.0 = 1 x 1.00 x 11.3
Where:
n = coefficient
C = concentration
V = volume
‘1’ refers to H2SO4
‘2’ refers to NaOH
C1 = (1 x 1.00 x 11.3) / (2 x 25.0) = 0.226 mol dm-3
More Titration Problems…
standard solution?
Solutions
Standard solutions
A standard solution is a solution of known
concentration.
Standard solutions are made by dissolving an
accurately weighed mass of solid in a known
volume of solvent using a volumetric flask.
The volumetric flask has a thin neck, which is marked
with a line so it can be filled accurately to the correct
capacity.
The standard solution can then be used to find the
concentration of a second solution with which it reacts.
This is known as volumetric analysis or titration.
103 of
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Preparing standard solutions
104 of
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Back titration

A known excess of one reagent A is allowed to react with an unknown amount of
reagent B

At the end of the reaction, the amount of A that remains unreacted is found by titration
with a standard solution

A simple subtraction calculation gives the amount of A that has reacted with B and can
be used to calculate the amount of B that has reacted
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Solution
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Remember!
Alcohol is not the
answer but it is a
solution*!
*Of ethanol, water and various other bits and pieces