Autumn Lecture 5 (Volumetric Analysis)

8/13/2019 Autumn Lecture 5 (Volumetric Analysis)

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Volumetric

Analysis

Term 1:

Week 3



Outline

• The Mole

• Avogadro’s constant

• Reacting masses

• Concentration• Titration

• Standard solutions

• Rough and accurate titres• Significant figures

• Back titration



How do we measure quantity of substance?

Element Symbol Relative atomic mass

Carbon-12 12C 12.000

Carbon C 12.011

Copper Cu 63.54

Hydrogen H 1.008

Magnesium Mg 24.312

Hydrogen = 1 g mol-1 1H 2H 3H - isotopes

Carbon-12 (12C) = 12 g mol-1

Before:

Now:



Avogadro constant

The number of atoms in exactly 12grams of isotope 12C (1 mole) was

defined as Avogadro constant.

N A = 6.02 × 10 23

Macroscopic scale Atomic scale

Amadeo Avogadro



One mole (n) is the amount of substance

which contains the same number of particles

(atoms, molecules or ions) as there are atomsin 12.00g of isotope 12C.

• The relative atomic mass in grams of all elements

contains the same number of atoms.

The Mole



Molar mass

• Molar mass numerically equals to the atomic

mass but in units of gram per mole (g mol-1).

Molar mass of a molecular substance is equals to

its molecular mass.

The mass of one mole of a substance is called

molar mass (A or Mr ).



Molar mass (or atomic weight for elements)

Atomic weight

in g/mol



• The relationship below allow us to connectthe laboratory scale with the atomic scale

using our standard dimensional analysis.

General sequence of calculation

Grams of

substance A

Molar

mass of AMoles of

substance A

Avogadro’

s Number

Elementary

units of A

(atoms,

molecules,

ions)

During conversion take care of units (grams, not kilograms)

for mass and magnitude (×1023) for elementary units!



Example of calculation:

The amount of substance n (in mole) and the number ofatoms (N) contained in it, are related to the Avogadro’s

number:

N/n = 6.02×1023 mol−1

Example: 2 g of magnesium ribbon

Number of atoms N) =

amount of substance (n) × Avogadro’s number (NA)

= 0.08226 × 6.02×1023 = 4.9523×1022 atoms

Amount of substance (n) =

mass (m) / molar mass (A)

= 2 / 24.312 = 0.08226 mol



• Stoichiometry translates as “measure of the elements”.

• Practically, stoichiometry is used for converting chemical formulas

and equations that represent individual atoms, molecules, andformula units into to the laboratory scale units which are

milligrams, grams of these substances.

Stoichiometry

aA + bB cC

Grams of

substance

A

Molar

mass of A

Moles of

substance

A

Chemical

equation

Moles of

substance

B

Molar

mass of B

Grams of

substance

B

Stoichiometry

1 g ? g ? g

Stoichiometry is a quantitative relationship among the species

involved in a chemical process in a molecular or molar amounts.



Stoichiometry

Mg(s) + O2(g) MgO(s)2 2

Magnesium ribbon with a mass of

2 g was burned in air. What is the

mass of obtained MgO?

nMg = mMg / AMg = 2 / 24.312 = 0.08226 mol of Mg

2 mol of Mg produces 2 mol of MgO. So, 0.08226 mol of

Mg will produce 0.08226 mol of MgO.

mMgO = nMgO x AMgO = 0.08226 x 40.312

= 3.31607g of MgO



Limiting and excess reagents

Excess

Limiting



Example:

H2 + O2 H2O2 2

m H2) = 20g

m O2) = 99.8gm H2O) = ?

= 20.0g ×1

2.016

= 9.92

Solution:

= 99.8g ×1

32.0

= 3.122



2H2 + O2 2 H2OInitial (mol) 9.92 3.12

Stoichiometric

coefficient

Compare 3.124.96

Limiting reagent

(completely consumed)Excess reagent

(remained unreacted)

1

Finding limiting reagent



2H2 + O2 2 H2O

Initial (mol) 9.92 3.12

Consumed (mol)

Result (mol)

-3.12-6.24

03.68 6.24

+6.24

Limiting reagent

(completely consumed)Excess reagent

(remained unreacted)

m = 6.24 mol ×18.04

1

= 112.57



A solution is a homogeneous mixture inwhich the molecules or ions of thecomponents freely intermingle.

• The solvent is the medium intowhich the solutes are mixed ordissolved.

• A solute is any substancedissolved in solvent.

Concentration is the ratio of the amount ofsolute to the amount of solution.

Solution and concentration



Molarity

• Chemists usually measure the concentration of solution

as the number of moles of solute per cubic decimetre ofsolution. Concentration in mol dm-3 is also called

molarity.1 cubic decimetre (dm3 ) = 1000 cm3 = 1 litre

Example: 11.7 g of NaCl was

dissolved in 500 ml of water. What is

the concentration of solution?

n = m/A = 11.7/58.5 = 0.2 mole of NaCl

M = n/V = 0.2/0.5 = 0.4 mol dm-3



• Laboratory chemicals are usually purchased in

concentrated form and must be diluted withwater (made less concentrated ) before beingused.

• During dilution, the number of moles of solute

remains constant.

Dilution

× =

×

Moles of solute in

dilute solution

Moles of solute in

concentrated solution

× = ×



Reacting masses

What if we don’t know the

concentration of HCl?



Titration

• Titration (volumetric analysis) is a common laboratory

method of quantitative chemical analysis that is used todetermine the unknown concentration of a known

reactant.

Meniscus

39.3 mL

NaOH, 1 M (titrant)

HCl, 25 mL, 1 M (analyte)

+ pH indicator

TITRANT = STANDARD

SOLUTION (concentration& volume are known)

ANALYTE = TITRAND

(unknown concentration)

pH indicator

(colour of the solution changes

depending on the pH)



Neutralization

reactants products

product of reaction

pH indicator

substance which changes the colour of the solution

depending on the concentration of H+ ions ( pH)



pH indicators

Indicator Color on Acidic

Side

Color on Basic

Side

Methyl Violet Yellow Violet Bromophenol

BlueYellow Blue

Methyl Orange Red Yellow

Methyl Red Red Yellow

Litmus Red Blue

Bromothymol

BlueYellow Blue

Phenolphthalein Colorless Pink

Alizarin Yellow Yellow Red

A common way to detect the equivalence point in an acid-base

titration is to add an pH indicator, a compound that changes color asan acidic solution becomes basic (equivalence point), or vice versa.



• Read the bottom of the meniscus, while your

eye is level with it. The meniscus can be seen

more easily against a white background.

Titration

Number of titration Rough Accurate 1 Accurate 2 Accurate 3

Initial burette reading (cm3) 10.0 10.0 10.0 20.0

Final burette reading (cm3) 35.3 35.0 35.1 45.0

Titre (cm3), Vin - Vfin 25.3 25.0 25.1 24.9



Number of titration Rough Accurate 1 Accurate 2 Accurate 3

Initial burette reading (cm3) 10.0 10.0 10.0 20.0

Final burette reading (cm3) 35.3 35.0 35.1 45.0

Titre (cm3), Vin - Vfin 25.3 25.0 25.1 24.9

Rough and accurate titres

• Rough titration is the trial process (first titration)in order to determine the approximate end point

of the reaction.

• Accurate titration is the subsequent titration

performed with the high accuracy when the endpoint of the reaction has been already

established.

+ +

3 =25.0 ml



ObservationVolume of 1.0

M NaOH (aq)

added/ mL

Colour of

indicator

0 Colorless

5 Colorless

10 Colorless

15 Colorless20 Colorless

21 Colorless

22 Colorless

23 Colorless

24 Colorless

25 Pink

26 Pink

27 Pink

28 Pink

When the two solutions just

react and neither is in excess,we have found the

equivalence point of the

titration.

=

× = .

+ → +

()

=

=

.

. = −





Acid-base titration

http://www.youtube.com/watch?v=8UiuE7Xx5l8&feature=related



Standard solution

• Standard solution is a solution containing a

precisely known concentration of an element

or a substance.• Standard solutions are usually prepared

from solid reagents.

• Standard solutions are used to determinethe concentrations of other substances, such

as analyte solutions in titrations.



Back titration

• Back titration is useful if the endpoint of

the reverse titration is easier to identify

than the endpoint of the normal titration.

• Instead of titrating the analyte, a known

excess of standard reagent

(intermediate) is added to the solution,

and the excess is titrated.



Back titrations are used when:

• one of the reactants is volatile, forexample ammonia.

• an acid or a base is an insoluble salt, forexample calcium carbonate

• a particular reaction is too slow

• direct titration would involve a weak acid- weak base titration (the end-point of

this type of direct titration is very difficultto observe)



Example: Back Titration to Determine

the Amount of an Insoluble Salt

0.125g sample of chalk was placed in a 250mL beaker

and 50.00mL 0.200M HCl was added. The excess HCl

was then titrated with 0.250M NaOH. The average

NaOH titre was 32.12mL. Determine the mass of

calcium carbonate present in a chalk.

Step 1: Determine the amount of HCl in excess from the



Step 1: Determine the amount of HCl in excess from the

titration results

a) Write the equation for the titration:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

acid + base → salt + water

b) Calculate the moles, n, of NaOH(aq) that reacted in the

titration:n = M x V

M = 0.250 mol L-1

V = 32.12 mL = 32.12 x 10-3 L

n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol

c) From the balanced chemical equation, 1 mole NaOH reacts

with 1 mole of HCl. So, 8.03 x 10-3 mol NaOH reacted with 8.03 x

10-3 moles HCl. The amount of HCl that was added to the chalk

in excess was 8.03 x 10-3 mol.



Step 2: Determine the amount of calcium carbonate in chalk

a) Calculate the total moles of HCl originally added to the chalk:

n(HCl total added) = M x VM = 0.200 mol L-1

V = 50.00 mL = 50.00 x 10-3 L

n(HCl total added) = 0.200 x 50.00 x 10-3 = 0.010 mol

b) Calculate the moles of HCl that reacted with the calcium

carbonate in the chalk.

n(HCl titrated) + n(HCl reacted with calcium carbonate) =

n(HCl total added)

n(HCl total added) = 0.010 mol

n(HCl titrated) = 8.03 x 10-3 mol

8.03 x 10-3 + n(HCl reacted with calcium carbonate) = 0.010

n(HCl reacted with calcium carbonate) = 0.010 - 8.03 x 10-3 =

1.97 x 10-3 mol

) f



c) Write the balanced chemical equation for the reaction

between calcium carbonate in the chalk and the HCl(aq).

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

d) From the balanced chemical equation, calculate the moles of

CaCO3 that reacted with HCl.

From the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1

mol HCl reacts with ½ mol CaCO3. So, 1.97 x 10-3 mol HCl had

reacted with ½ x 1.97 x 10-3 = 9.85 x 10-4 mol CaCO3 in the

chalk.

e) Calculate the mass of calcium carbonate in the chalk.

n = mass / An = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl)

MM(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol-1

mass = n x A = 9.85 x 10-4 x 100.09 = 0.099g

The mass of calcium carbonate in the chalk was 0.099g.



Reliability

Chemistry involves the interpretation ofquantitative measurements, usually made as apart of experiment.

Each measurement has 4 aspects:

1) Object of measurement

2) Value

3) Units

4) Reliability

Example:

The mass of iron was 4.05 grams.



• The accuracy of a measurement is indicated bythe number of digits used to represent it.

• Digits that result from a measurement such thatonly the digit farthest to the right is not knownwith certainty are called significant figures.

• The number of significant figures in ameasurement is equal to the number of digitsknown for sure plus one that is estimated.

Significant figures

This digit has some uncertainty This digit has some uncertainty

These two digits are known for sure These three digits are known for sure



When are zeros significant?

Rules for determining number ofsignificant figures:

• All nonzero digits are significant.

• Zeros between nonzero digits are

significant.

• In a number with no decimal point, zeros

at the end of the number (tracing zeros)

are not necessarily significant.• If a number contains a decimal point,

zeros at the beginning (leading zeros)

are not significant, but zeros at the end

of the number are significant.

125.3

Example:

402

33,000

0.00340

Wh d i ifi fi ?



Why we need significant figures?

33.100 mL

20.256 mL

0.0005 mL

0.3 mL

+

+

+

53.6565 mL

Analytical pipette

Measuring cylinder

Significant figures express the accuracy of experiment.

53.7 mLrounding



What to write in log book

Lab report template

• Changes from

procedure (if any)• Results

• Observations

• Calculations

Complete lab report (long)

• Risk assessment

• Short intro• Procedure

• Results

• Observations

• Calculations

• Short conclusion

Documents

Autumn Lecture 5 (Volumetric Analysis)