15.1 synthesis /decomposition revie15+notes,+18-19.pdf~ 9 ~ acid rain is rain with a particularly...

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Synthesis of the elements. When two pure

elements react, they combine to form a binary

compound: E + E → BC

Ex.1) Zinc metal is heated in pure oxygen:

Ex.2) Hydrogen gas reacts with solid sulfur (S8):

Ex.3) Sodium metal is heated with chlorine:

Ex.4) Synthesis of magnesium nitride

15.1 Synthesis/Decomposition Review

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Metal chlorides and oxygen. Metal chlorides react

with oxygen to form metal chlorates:

MCl + O2 → MClO3

Ex.5) sodium chloride is heated in oxygen gas:

Ex.6) formation of barium chlorate:

Ex.7) iron (III) chloride and oxygen gas:

Ex.8) synthesis of cobalt (VI) chlorate

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Metal oxides and carbon dioxide. Metal oxides

react with carbon dioxide to produce metal

carbonates:

MO + CO2 → MCO3

Ex.9) calcium oxide and carbon dioxide:

Ex.10) formation of copper (II) carbonate:

Ex.11) synthesis of potassium carbonate:

Ex.12) aluminum oxide and carbon dioxide:

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Metal oxides and water. Metal oxides react with

water to form metal hydroxides. As a result, they

are often called basic oxides.

MO + H2O → MOH

Ex.13) Magnesium oxide is added to water:

Ex.14) Sodium oxide is added to water:

Ex.15) Formation of silver hydroxide:

Ex.16) Synthesis of lead (IV) hydroxide:

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Decomposition patterns are the exact reverse of

their synthesis counterparts and are often brought

about by exposure to intense heat. Carbonates also

decompose readily with acids.

Ex.17) magnesium carbonate is heated...

Ex.18) decomposition of potassium chlorate

Ex.19) copper (II) hydroxide is heated in a crucible

Ex.20) nitrogen triiodide explosively decomposes

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Non-metal oxides and water. Many (but not all)

non-metal oxides react with water to form oxoacids.

We call them these substances acidic oxides.

NMO + H2O → Oxoacid

(All of the polyatomic acids w/ oxygen = oxoacids)

Only certain non-metal oxides are acidic. To find

them, dehydrate an existing oxoacid by removing all

hydrogens and exactly ONE oxygen atom:

Ex.1) Sulfuric acid: H2SO4 - H2O =

Ex.2) Carbonic acid: H2CO3

Ex.3) Chlorous acid:

15.2 Acidic Oxides and Acid Rain

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If a non-metal oxide can NOT be created by this

dehydration method, it is unlikely to be acidic in

water:

Ex.4) Working backwards from sulfuric and sulfurous

acids, which of the following sulfur oxides are

acidic and which one(s) is/are not?

SO SO2 SO3

As a result, not all NMO's react with water:

Ex.5) water and carbon dioxide:

Ex.6) water and carbon monoxide:

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Ex.7) synthesis of sulfuric acid:

Ex.8) nitrogen dioxide and water:

There are many "non-traditional" oxides which form

oxoacids in water. Typically, the substance in

question will contain a larger number of oxygens

and will form the oxoacid with the greatest number

of oxygen atoms.

Ex.9) Phosphorus pentoxide (P4O10) + water

Ex.10) Cl2O7 + water

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Acid rain is rain with a particularly low (acidic) pH.

“Clean” rain is naturally acidic with a pH of about

5.6, since the CO2 from respiration combines with

water to make carbonic acid, H2CO3. However, the

addition of nitrogen oxides (NOx) and sulfur dioxide

(SO2) to the atmosphere causes rain to become

even more acidic, as water combines with those

two to become STRONG acids.

Complete and balance the following:

CO2 + H2O → (pH ≈ 5.6)

NO2 + H2O → (pH ≈ 4.0)

SO2 + H2O → (pH ≈ 4.0)

H2SO3 + O2 → (pH ≈ 4.0)

SO2 is the most important gas in acid rain formation.

Sulfur dioxide can be formed by volcanic eruptions,

and nitrogen oxides by lightning strikes. These

naturally occuring processes provide key nutrients

for life (ex. nitrogen in plant life). However, both are

being formed in excessive ammounts due to hu

industrial processes. This includes electricity

generation (main), factories, and motor vehicles.

These can come back to the surface either through

wet deposition (precipitation) or

(particles simply stick to the

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being formed in excessive ammounts due to hu

industrial processes. This includes electricity



(precipitation) or dry deposition

(particles simply stick to the ground and plants)





being formed in excessive ammounts due to human



dry deposition

ground and plants)

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Because of this, rain becomes too acidic and starts

to contaminate water (lakes, oceans, rivers), soil,

and forests, harming both animals and humans. In

addition, acid rain damages buildings and

monuments, especially those made of marble or

limestone: CaCO3 + HX → CaX + CO2 + H2O.

The effect of acid deposition is not localized to

the area where the sulfur/nitrogen was released;

the acid rain spreads around the world.

Mountainous regions tend to receive the most acid

rain simply because of their altitude (air cools as it

rises, causing precipitation).

Governments have the power to work together and

fix this problem. The Clean Air Act Amendments of

1990 have reduced SO2 emissions by 10 million tons

per year (65%). Using lime (CaO, calcium oxide), SO2

can be captured as it exits a power plant, a process

called flue-gas desulfurization:

CaO(s) + SO2(g) → CaSO3(s) → CaSO4 (inert)

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Catalytic converters in cars serve a similar purpose.

Incomplete combustion of fuel (hydrocarbons)

produces nitrogen compounds, NOx, CO, and

unburnt CH compounds. A fine mesh of precious

metals within the converter captures these

molecules, allowing them to react with oxygen as it

becomes available. Once fully oxidized, they are

released:

2 NOx(g) → O2(g) + N2(g)

CO(g) → CO2(g)

CxHy(g) → CO2(g) + H2O(g)

This is why we have emissions tests for vehicles.

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A solution is any homogenous mixture. Generally

speaking, we will refer to them as substances

(solute) dissolved in water (solvent). The

concentration of these solutions is most commonly

measured as their molarity:

Molarity, M = mol (of solute)

total volume (liters)

Molarity is one of several different methods

(molality, ppm, ppt, ppb, % v/v, % m/v, % m/m) of

measuring concentration and is expressed

differently on the AP and IB exams:

(AP) (IB)

Molarity, M = mol L-1

= mol dm-3

mol/L = mol/dm3

Other units: 1 mL = 1 cm3

1 L = 1 dm3

15.3 Solution Concentration and Prep.

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Use the molarity equation to solve each of the

following problems. M = mol/L

Ex.1) Determine the molarity of 1.43 mol glucose which

has been dissolved to form 6.23 L of solution

Ex.2) Calculate the molarity of 431.5 mg NaCl (Mr =

58.443) dissolved in 235 cm3 of solution:

Ex.3) What mass of potassium iodide, in grams, would be

necessary to create 3.50 L of a 0.250 mol dm-3

solution?

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The concentration of specific ions within a solution is

fairly easy. Just calculate the overall molarity of the

substance in question and multiply that value by the

number of ions into which it dissociates:

1.00 M NaK2PO4 → 1 x Na+ 2 x K

+ 1 x PO4

3-

[Na+] = 1.00 M [K

+] = 2.00 M [PO4

3-] = 1.00 M

Ex.4) 0.300 M Sr(NO3)2 [Sr2+

] = [NO3-] =

Ex.5) 0.750 mol L-1

iron (III) sulfate solution:

Ex.6) 0.80 mol of sodium carbonate in 500.0 mL solution:

Ex.7) 0.900 M ammonium dichromate

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Concentrated solutions may also be diluted by the

addition of water or another solvent via the

equation below:

C1V1 = C2V2

"C" is used to denote any unit of concentration,

while "V" is any unit of volume. Be consistent on

both sides of the equation! The difference between

V1 and V2 is the amount ofwater which has been (or

needs to be) added to produce the dilution.

Ex.8) 50.0 mL of a stock 16.0 M HNO3 is diluted to a final

volume of 750.0 mL. Calculate [HNO3] at this point.

Ex.9) Given 2.5 L of a 1.3 M KCl solution, what is the new

molarity of the solution if 1.5 L of water is added?

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Ex.10) 350.0 mL of a 0.500 mol L-1

sodium hydroxide

solution is left out over Spring Break. Upon

returning, you find that 78.4 mL of the solution has

evaporated. Calculate the [NaOH] at this point:

When a small sample (or aliquot) is taken from a

larger solution, the concentration of the aliquot

matches the concentration of the parent solution.

Ex.11) 67.5 g of sodium phosphate (Mr = 163.94) is

dissolved to make 1.60 L of solution. How many mL

of this solution are required to create 1.00 L of a

0.100 M solution by dilution?

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Ex.12) 10.0 g NaOH (Mr = 40) is used to produce 0.500 L of

solution "A". A 100.0 mL aliquout of solution A is

diluted to 500.0 mL to form solution "B". How

many mg of NaOH are contained by 10.0 mL of

solution B?

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Precipitation reactions occur when two or more

dissolved reactants form at least one insoluble

product. For example:

KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)

The solid produced by these reactions is known as a

precipitate and can be collected, cleaned, and

purified via filtration. The mass of the precipitate

can be used to study the contents of the original

sample in a process known as gravimetric analysis.

This process is very similar to combustion analysis

and requires you to use the mass of the products to

"track" backwards to the mass(es) of the

reactant(s). You were introduced to this concept a

few weeks ago, in combustion analysis.

You WILL need to understand solubility concepts.

Review 10.3 if you're a little "fuzzy" on this topic.

15.4 Gravimetric Analysis

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Gravimetric analysis is commonly used to determine

the purity of unknown samples. The precipitate is

collected and then used to calculate the amount of a

specific reactant within the sample:

Ex.1) You are given a 50.00-g mixture of barium chloride

(Mr = 208.234) and inert solid. After being dissolved

in water, an excess of silver nitrate (Mr = 169.875) is

added. A white precipitate is collected and, after

drying, found to have a mass of 17.48 g. Determine

the purity of the sample in terms of the chloride:

Give the balanced molecular equation for this rxn

and identify the white precipitate being described:

Using DA, work backwards from the mass of the

precipite to the mass of BaCl2 which produced it:

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Determine the purity of the sample. Percent purity

is calculated as (mass of analyte / total mass of

sample) x 100:

Ex.2) A sample to be tested in lab contains a mixture of

barium chloride and sodium chloride. 3.725 g of

this mixture is dissolved in water and titrated with

an excess of potassium sulfate. A precipitate with a

mass of 2.734 g is collected and dried.

Give the balanced molecular equation for this rxn

and mark the precipitate which will be formed:

Determine the mass % of both chlorides within the

mixture:

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Mass percentages or DA can be used to "pull" the

mass of an analyte directly from a precipitate.

Ex.3) A 3.00 g sample of an alloy containing lead and tin is

dissolved in nitric acid. An excess of sulfuric acid is

then added to the mixture, resulting in the

formation of 2.37 g of a precipitate.

Identify the precipitate which is being formed:

"Pull" the mass of the metal from the precipitate.

Determine the mass percentages of tin and lead in

this alloy:

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Sometimes, you don't know the identify of the

substance being tested. In these instances, a pure

sample is used and the molar mass of the unknown

can be calculated.

Ex.4) A metal nitrate is known to have a general formula

of M(NO3)4. A 2.000-g sample of this nitrate is

dissolved in water and slowly titrated with a 0.2100

mol dm-3

solution of sodium phosphate.

Precipitation is complete after the addition of 42.91

mL of the phosphate solution.

Give the balanced molecular equation for this rxn.

Substitute "M" for the unknown metal.

How many moles of sodium phosphate were used?

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How many moles of the unknown nitrate were

present in the sample?

Using the mass and moles of the unknown nitrate,

determine its molar mass.

Remove the mass of the nitrates and determine the

molar mass (and ID) of the unknown metal.

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With a knowledge of solubility rules, net-ionic

equations can be written to simplify the normal (full

or molecular equations) we normally write.

In the context of this chapter, a net-ionic equation

would only include ions which form a precipitate.

The ions which do not form a precipitate remain

dissolved throughout the reaction and are called

spectator ions. For example:

The balanced molecular equation for the

precipitation of lead (II) iodide. This is what we

would normally write when doing stoichiometry:

Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)

By showing the dissociation of soluble particles, we

create a complete ionic equation:

Pb2+

+ 2NO3- + 2K

+ + 2I

- → PbI2 + 2K

+ + 2NO3

-

15.5 Writing Net Ionic Equations

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Since K+ and NO3

- appear on both sides, they are

considered spectator ions. Remove them and

balance the remaining particles to create the net

ionic equation:

Pb2+

+ 2NO3- + 2K

+ + 2I

- → PbI2 + 2K

+ + 2NO3

-

Net Ionic: Pb2+

(aq) + 2I- (aq) → PbI2 (s)

For examples 1-3, create the complete ionic and net

ionic equations from the provided molecular

equations. Give states and balance net ionics:

Ex.1) Al(NO3)3( ) + NaOH( ) → NaNO3( ) + Al(OH)3( )

Show dissociated ions (complete ionic):

Remove spectator ions and balance (net ionic):

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Ex.2) Ca(NO3)2( ) + K2SO4( ) → CaSO4( ) + KNO3( )



Ex.3) Ni(NO3)2( ) + LiCl( ) →



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Write and balance net ionic equations for each of

the following reactions. A faster, more efficient way

of creating net ionics can be used in which lines are

drawn through the center of particles which will

dissociate. I typically include charges as well. This

removes the need to write the complete ionic.

Ex.4) (NH4)2S + Ni(NO3)2 → NH4NO3 + NiS

Ex.5) K2CO3 + MgI2 →

Ex.6) silver nitrate and potassium chloride

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Ex.7) strontium acetate and ammonium phosphate

Ex.8) ? → Hg2Cr2O7 + NaClO3

Ex.9) zinc chlorite and calcium bromide

Ex.10) ? → potassium nitrate and iron (I) sulfide

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Acids and bases are most commonly explained by

Bronsted-Lowry definitions:

Acids give H+ ex. HNO3 → H

+ + NO3

-

Bases accept H+ ex. CO3

2- + H

+ → HCO3

2-

H+

is known as a proton in acid/base reactions.

Ex.1) Identify the acids and bases in each of the reactions

described below:

a.) H2SO4 + 2 NH2- → SO4

2- + 2 NH3

b.) Na2CO3 + 2 HCl → 2 NaCl + H2CO3

c.) HNO3 + NaOH → H2O + NaNO3

15.6 Acids, Bases, and Net Ionics

~ 31 ~

One of the most common versions of an acid/base

reaction is called neutralization. This is a double

replacement reaction which occurs between a metal

hydroxide (M-OH) base and an acid, forming water

and a salt. FYI - salts are a general term for ionics.

H-A + M-OH → H2O + M-A

(Acid) (Base) (Water) (Salt)

Ex.2) HCl + NaOH →

Ex.3) Ba(OH)2 + H2SO4 →

Ex.4) H3PO4 + Mg(OH)2 →

You can also work backwards from the salt and

water to identify the acid/base combination used...

Ex.5) → Ca(NO3)2 + H2O

Ex.6) → H2O + KF

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The strength of an acid depends on the weakness of

the H-A bond. In acids, this bond is practically ionic,

making it possible for H+ to dissociate from the

parent molecule:

H-A(?) → H+(aq) + A

-(aq)

Weaker H-A bonds dissociate more readily in water,

a process known as ionization. Acids which ionize

easily (nearly 100%) are said to be strong acids (SA).

Weak acids (WA) dissociate to a lesser extent.

Acetic acid (vinegar, weak acid):

HC2H3O2 → H+(aq) + C2H3O2

-

[] = 1.00M (1.3%) [] = 0.013 M [] = 0.013 M

Nitric acid (very strong acid)

HNO3 → H+(aq) + NO3

-

[] = 1.00M (100%) [] = 1.00 M [] = 1.00 M

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Strong bases dissociate 100% in water to provide

hydroxide ion, as shown below:

M-OH(?) → OH-(aq) + M

+(aq)

Group 1 MOH (metal hydroxides) are VERY soluble

in water, and thus make very strong bases. Other

MOH are less soluble or only become so in acids.

Strong Acids

HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4

Strong Bases

Group 1 MOHs - LiOH, NaOH, KOH, RbOH, CsOH

w/ acids only: Ba(OH)2 , Ca(OH)2, Sr(OH)2

In net ionic equations, strong acids/bases are shown

in their fully-dissociated states. Weak acids/bases

are NOT shown as being dissociated. MEMORIZE

the list above.

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When writing net ionics for acid/base reactions,

show SA/SBs as fully-dissociated. WA/WBs are not

shown as being dissociated.

Ex.7) Convert the equations below into net ionic format

using the "line" method from the prior chapter:

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

Al(OH)3(s) + 3 HBr(aq) → 3 H2O(l) + AlBr3 (aq)

(HAc is used to represent acetic acid in most

reactions. The Ac represents the acetate poly.)

HAc(aq) + KOH(aq) → H2O(l) + KAc (aq)

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Complete the following molecular equations and

provide their net ionic counterparts:

Ex.8) HClO4( ) + Mn(OH)2( ) →

Ex.9) KOH( ) + HClO3( ) →

Ex.10) HCl( ) + NH4OH( ) →

Ex.11) Na2CO3( ) + HNO3( ) →

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Write the molecular and net ionic equations for each

of the following descriptions:

Ex.12) hydrocyanic acid and sodium hydroxide

Ex.13) silver hydroxide and sulfuric acid

Ex.14) acetic acid and sodium hydrogen carbonate

~ 37 ~

Particle-Level Diagrams (PLDs) are used to show

how molecules and ions interact quantitatively at

the atomic level and have recently become a

favorite question type on both the AP and IB exams.

First, PLDs can be used to show how water

molecules interact with one another. Remember,

water is polar and contains partial + (around

hydrogens) and partial - charge (around oxygen).

Ex. 1) Indicate the polar regions and associated charges on

the water molecule shown below. In the blank

space, show how 3 water molecules would interact

using dashed lines to represent hydrogen bonds.

15.7 PLDs in Solution

3 x H2O

~ 38 ~

When soluble ionic substances dissolve in water,

they dissociate to form separate ions surrounded by

water molecules. Show 4 water molecules

interacting with each of the ions shown below:

Ex.2)

When drawing PLDs in solution, water is usually

omitted from the diagram. As a result, its presence

is implied by the spaces shown between the ions.

Knowing this, we can easily draw PLDs of ionic

compounds dissociating in water:

Ex.3)

Ca2+

(aq)

Cl-(aq)

Ca2+ Cl

-

NaCl(aq)

Li2S(aq)

~ 39 ~

Polyatomics break apart from their attached ions,

but do not themselves dissociate:

Ex.5)

Soluble ionics are shown as dissociated, insoluble

products become "piles" at the bottom of the PLD:

Ex.6)

K2SO4(aq)

(NH4)2Cr2O7(aq)

NaOH( )

Cu(OH)2( )

PbS( )

BaSO4( )

~ 40 ~

Acids are generally soluble, but only strong acids can

be shown as dissociated:

Ex.8)

Bases follow general ionic solubility rules. Some are

soluble and strong, others are insoluble and weak.

Ex.9)

HCl(aq)

HAc(aq)

HF(aq)

HCN(aq)

Ba(OH)2( )

Al(OH)3( )

~ 41 ~

PLDs can also be draw to show the result of

chemical reactions. Net ionics are handy here:

Ex.10) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Net Ionic:

PLD:

Ex.11) HCl(aq) + NaAc(aq) → HAc(aq) + NaCl(aq)

Net Ionic:

PLD:

~ 42 ~

Ex.12) Iron (III) chloride and sodium sulfate:

Molecular:

Net Ionic:

PLD:

Ex.13) Carbonic acid and sodium acetate:

Molecular:

Net Ionic:

PLD:

Beer's Law describes the mathematical relationship

between concentation (molarity) and absorbance

within a device known as a

A spectrophotometer uses a collimator and slit to

create a beam of light with a very specific

wavelength. This beam is allowed to pass through a

small crystal vial known as a

the analyte, or sample being studied. The

spectrophotometer then compares the intensity of

the light entering the sample against the amount

passing through (being

between these values is known as the

15.8 Beer's Law and Spectrophotometry

~ 43 ~

describes the mathematical relationship


within a device known as a spectrophotometer


create a beam of light with a very specific


small crystal vial known as a cuvette which contains

, or sample being studied. The

otometer then compares the intensity of


passing through (being transmitted). The difference

between these values is known as the absorbance

Beer's Law and Spectrophotometry

describes the mathematical relationship


spectrophotometer.



which contains

otometer then compares the intensity of


). The difference

absorbance.

Beer's Law and Spectrophotometry

~ 44 ~

Beer's law is given by:

A = εbc

A = absorbance: how much light is NOT making it

through the analyte and to the detector.

ε = molar absorptivity: also known as the molar

extinction coefficient. A constant for a substance at

a given wavelength, it measures absorbance per

unit of concentration.

b = path length: distance the light travels through

the sample, in cm. Usually 1 cm.

c = concentration: the molar concentration of the

sample being analyzed.

Ex.1) At a given frequency, what should happen to the

absorbance of a sample as the concentration of the

analyte increases? Decreases?

~ 45 ~

Ex.2) You'll often solve for ε, as it is a constant at a given

wavelength for a substance. How would you

rearrange the equation to do so?

Ex.3) Which value(s) in the equation would change if you

used a larger cuvette?

Ex.4) You must always wipe cuvettes clean before placing

them into a spectrophotometer. How would a

fingerprint influence the measured absorbance

value of a sample?

Ex.5) How would the fingerprint in #4 influence your

estimate of the analyte's concentration?

~ 46 ~

Knowing their is a proportional relationship

between A and c, you can use the known values of a

sample to estimate those of an unknown sample:

Ex.6) A 0.500 mol dm-3

sample of copper (II) sulfate is

found to have an absorbance of 0.328 at 635 nm.

Estimate the concentration of an unknown sample

with a measured absorbance of 0.980.

Another tactic is to solve for the molar absorptivity

at a given wavelength/concentration and use it to

solve for another concentration

Ex.7) A 0.010 M methanol solution is found to have an

absorbance of 0.225 at 9030 nm in a 1 cm cuvette.

If another sample of methanol has an absorbance of

0.680 under the same conditions, what is its [ ] ?

Determine the value of ε:

Solve for [ ] in second sample:

~ 47 ~

A third tactic (and most common) involves creating

multiple standard solutions with known [ ] and

measuring absorbance at a given wavelength.

Ex.8) 3 samples of copper (II) sulfate are tested at 635

nm, producing the following absorbance values:

Sample 1: 0.050 mol dm-3

Abs = 0.09


Abs = 0.20


Abs = 0.37

Graph this data w/ a best fit line:

Abs

[CuSO4] mol dm-3

~ 48 ~

Ex.9) Why is absorbance marked on the y-axis with [ ] on

the x-axis?

Ex.10) Estimate the [ ] of an unknown copper (II) sulfate

solution with abs = 0.14:

Ex.11) Would it be realistic to estimate the [ ] of an

unknown solution with abs = 0.89? Why or why

not?

15.1 synthesis /decomposition revie15+notes,+18-19.pdf~ 9 ~ acid rain is rain with a particularly...

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