AB Chemistry CAB Sample

Alberta Chemistry 20-30 Sample CAB Questions

MULTIPLE CHOICE

1. Use the following information to answer the next question.

In

b

A

C

ce

sh

m

it

ce

b

w

p

M

Each molecule has a definite geometry because the atoms combine to form a molecule in order to attain a

state of minimum energy. VSEPR (valence shell electron pair repulsion) theory helps to determine the

geometry of molecular structures.

a molecule, if there are three bonding electron groups around a central atom, what is the maximum

ond angle between the bonds?

A. 180o

B. 120o

C. 109.5o

D. 90o

NS: B

120o

A

B

B

B

onsider a molecule AB3, in which A is a central atom and B is a surrounding atom. In this molecule, the

ntral atom is surrounded by three shared pairs of electrons. According to VSEPR theory, these electrons

ould be as far apart as possible, so that the electrostatic force of repulsion between them is the

inimum. In such a case, the three electron pairs arrange themselves in trigonal planar geometry, because

is the best arrangement involving the minimum possible energy. In this case, three atoms bonded to the

ntral atom, are placed at the vertices of an equilateral triangle. In this arrangement, the maximum angle

etween two bonds is 120o. The bond angle can not be less than 120o, because in that case the molecule

ill become highly energetic and thus unstable. It cannot be more than 120o, because it will contradict the

roperties of geometry.

cGraw Hill Ryerson © 2007 1

PTS: 1 DIF: Easy OBJ: Section 1.1

LOC: 20–A2.5K TOP: Diversity of matter KEY: VSEPR Theory

MSC: Knowledge


W

A

E

o

R

P

L

M

M

The electronegativity values of four elements in the same period of the periodic table are listed below:

P Q R S

3.0 2.0 3.5 4.0

These elements belong to groups 13, 15, 16, and 17, but not in that order.

hich of these elements belongs to group 16?

A. P

B. Q

C. R

D. S

NS: C

lectronegativity increases along a period from left to right. Arranged in the increasing order, the values

f electronegativity are:

2.0 < 3.0 < 3.5 < 4.0

13 15 16 17

has an electronegativity of 3.5. Therefore, R belongs to group 16.

TS: 1 DIF: Easy OBJ: Section 1.1

OC: 20–A1.3K TOP: Diversity of matter KEY: Electro negativity

SC: Knowledge

cGraw Hill Ryerson © 2007 2

3. An oxygen atom needs two additional electrons to reach the noble gas configuration of neon. Which of

the following Lewis structures represents the oxygen molecule?

McGraw Hill Ryerson © 2007 3

A.

B.

C.

D.

ANS: C

+

Each oxygen atom needs two additional electrons to reach the noble gas configuration of neon gas. Two

oxygen atoms share two valence unpaired electrons each and form O = O double bond.


LOC: 20–A2.4K TOP: Diversity of matter KEY: Lewis Structure

MSC: Knowledge


Hydrogen fluoride, HF(l), has numerous applications in the petrochemical industry and is an

essential ingredient of super acids. Its boiling point is extremely high. It is a liquid at room

temperature, while other hydrogen halides such as hydrogen chloride, HCl(g), hydrogen

bromide, HBr(g), and hydrogen iodide, HI(g), are all gases at room temperature.

Hydrogen fluoride, HF(l), has a higher boiling point than the other hydrogen halides because the

A. intermolecular hydrogen bonding in HF(l) is stronger than that in other hydrogen halides.

B. intramolecular hydrogen bonding in HF(l) is stronger than that in other hydrogen halides.

C. intermolecular hydrogen bonding in HF(l) is weaker than that in other hydrogen halides.

D. intramolecular hydrogen bonding in HF(l) is weaker than that in other hydrogen halides.

ANS: A

The boiling points of acids decrease as we proceed from HI(g) to HBr(g) to HCl(g). This is because, with

a decrease in molecular mass, London (dispersion) forces decrease. However, HF(l) has an abnormally

high boiling point and is a liquid, whereas HCl(g), HBr(g) and HI(g) are gases. This is because of the

hydrogen bonding in HF(l). Hydrogen bonding does not exist in other hydrogen halides.

H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F----- H⎯F-----


LOC: 20–A2.7K TOP: Diversity of matter KEY: Intermolecular Bonding

MSC: Knowledge

5. Hydrogen bonds are found in all of the following compounds: glycerol, C3H5(OH)3(l), water, H2O(l),

ethanol, C2H5OH(l), and ethylene glycol, C2H4(OH)2(l). Which of the compounds listed would have the

highest boiling point?

A. water

B. ethylene

C. ethanol

D. glycerol

ANS: D


Glycerol has the highest boiling point due to the presence of three –OH groups.

CH3 – OH CH – OH CH3 – OH

H | O – CH3H | O – CHH | O – CH3

These –OH groups exhibit very strong hydrogen bonding. This results in a high boiling point.

PTS: 1 DIF: Difficult OBJ: Section 1.1

LOC: 20–A2.7K TOP: Diversity of matter KEY: Intermolecular Bonding

MSC: Knowledge

6. Which of the following compounds is the least ionic in character?

A. MgCl2(s)

B. BeCl2(s)

C. BaCl2(s)

D. CaCl2(s)

ANS: B

Ionic character increases as the difference in electronegativity between the bonded atoms increases.

Electronegativity values decrease as you move down the periodic table.

Be > Mg > Ca > Ba

Therefore, BaCl2(s) is the most ionic in character and BeCl2(s) is the least ionic in character.

PTS: 1 DIF: Average OBJ: Section 1.1

LOC: 20–A1.4K TOP: Diversity of matter KEY: Ionic Bond

MSC: Knowledge

7. Which statement(s) is(are) true?

I. The degree of electronegativity depends predominantly on the size of a cation and an anion.

II. The degree of covalence will be the highest if the cation is smaller when compared to the anion’s size.

III. The larger the size of the cation or anion, the stronger the ionic bond.

A. I and II are true.

B. I and III are true.



C. II and III are true.

D. All statements are false.

ANS: A

The smaller size of a cation distorts the electron cloud of the anion and pulls the electron density towards

itself. Thus, it introduces some covalent character is the ionic bond. The smaller the size of cation, the

greater is the degree of covalence.



MSC: Knowledge

8. An ionic compound contains two ions, X and Y. If the charge on X is +3 and that on Y is –2, which of

the following elements could be X and Y respectively?

A. Ga, N

B. Ca, S

C. Na, S

D. Al, O

ANS: D

From the given options, only gallium, Ga, and aluminium, Al, can produce ions with a charge of +3.

Between nitrogen, N, and oxygen, O, only oxygen can produce an ion with a charge of –2. Therefore,

aluminium and oxygen are X and Y and the chemical formula of this compound is Al2O3.


LOC: 20–A1.5K TOP: Diversity of matter KEY: Ionic bond

MSC: Knowledge

9. A compound consists of four carbon atoms, ten hydrogen atoms and an oxygen atom. If there is a

C–O–C bond in the compound, what is the molecular formula of this compound?

A. C4H9OH

B. C2H5OC2H5

C. CH3COC2H7

D. C2H5COCH5

ANS: B

The molecular formula of the given compound will be C2H5OC2H5. Its structural formula will be

HHHH||||

HCCOCCH||||HHHH

−−−−−−


LOC: 20–A2.2K TOP: Diversity of matter KEY: Molecular formula

MSC: Knowledge

10. Which of the following molecules is non-polar?

A. H2O(g)

B. CO2(g)

C. NH3(g)

D. CH2Cl2(g)

ANS: B

Carbon dioxide is a linear molecule in which both the C = O bonds are oriented at an angle of 180o with

respect to each other. The bonds are polar since the difference in electronegativity between carbon and

oxygen is 0.8, but due to linear geometry of CO2(g), the polarity of the two C = O bonds cancel each other

out. Therefore, CO2(g) is a non-polar.


LOC: 20–A2.9K TOP: Diversity of matter KEY: Polarity

MSC: Knowledge


11. Use the following information to answer the next two questions.

N H H

H

F H F F

H C H H H

Cl C Cl Cl Cl

Structural formulae of four compounds are given below: I. II. III. IV.

Which of these compounds is polar?

A. I

B. II

C. III

D. IV

ANS: A N

H H

H

NH3(g) is a polar compound because it is pyramidal and asymmetrical so the dipole vectors do not cancel.


LOC: 20–A2.9K TOP: Diversity of matter KEY: Polarity

MSC: Knowledge

12. Which of the following atoms does not form hydrogen bonds?

A. O

B. N

C. F


D. Cl

ANS: D

Chlorine, Cl, does not form hydrogen bonds because of its large size. When chlorine, Cl, bonds to a

hydrogen atom, its lower electronegativity, and large atomic size, creates a dipole–dipole interaction that

is not strong enough to produce hydrogen bonding.


LOC: 20–A2.7K TOP: Diversity of matter KEY: Hydrogen Bond

MSC: Knowledge

13. Consider the Lewis structure for the following compound, A2.

A A

If there are two electrons in the inner shell of element A, in which of the following groups of the periodic

table does element A belong?

A. 3

B. 11

C. 13

D. 15

ANS: D

From the above Lewis structure, there are five electrons in the valence shell of element A. Atomic

number of an element = total number of electrons = number of valence electrons + number of electrons in

inner shells = 5 + 2 = 7. Nitrogen has the atomic number of 7 and it corresponds to group 15.



MSC: Knowledge


14. Which of the following Lewis structures represents the ethene (ethylene) molecule? H H H C C H

A.

H H H C C H

B.

H C C H H H

C.

H H H C C H

D.

ANS: A H C C H

H H

In this structure, all of the carbon and hydrogen atoms have a completed valence shell.



MSC: Knowledge

15. In a diatomic molecule of an element, each atom has one lone pair and the molecule contains a triple

bond. Which of the following elements does it correspond to?

A. Sulfur

B. Oxygen

C. Nitrogen

D. Fluorine

ANS: C


A A

N ≡ N

It is a molecule of nitrogen atom. A nitrogen atom has five valence electrons. Each nitrogen atom shares

three electrons with the other and two electrons are left unshared. Hence, there exists a triple bond in the

molecule.

PTS: 1 DIF: Easy OBJ: 1.1


MSC: Knowledge

16. What is the chemical name for CaSO4(s)?

A. Calcium sulfur tetraoxide

B. Calcium sulfate

C. Calcium sulfite

D. Calcium tetraoxysulfur

ANS: B

The chemical name of CaSO4 is calcium sulfate.

⇌ + sulfateCalcium

4CaSO(cation)

ionCaclium

2Ca +

(anion)ionSulfate

24SO −



MSC: Knowledge

17. What is the chemical name for P4O10(s)?

A. Tetraphosphorus dioxide

B. Tetraphosphorus oxide

C. Phosphorus(IV) decaoxide

D. Tetraphosphorus decaoxide

ANS: D

The name of the compound (P4O10) is tetraphosphorus decaoxide.



LOC: 20–A2.1K TOP: Diversity of matter KEY: Covalent Bond

MSC: Knowledge

18. What of the following is the name of the negative ion, also called the anion, in nitrous acid?

A. nitrite ion

B. nitrate ion

C. pernitrite ion

D. nitride ion

ANS: A

HNO2 dissociates in the following way

HNO2(aq) ⇌ anion

2cation

NOH −+ +

The name of NO2– is ‘nitrite ion’.



MSC: Knowledge

19. A compound contains an aluminium ion and some chloride ions. If the compound is neutral, how

many chlorine ions will this molecule contain?

A. 1

B. 2

C. 3

D. 4

ANS: C

An aluminium atom releases three electrons to form an Al3+ ion.

Al ⇌ Al3+ + 3e–

A chlorine atom accepts an electron to form Cl– ion

Cl + e– ⇌ Cl–

So when these two atoms combine, three chlorine atoms accept three electrons from an aluminium atom,

to form AlCl3.

Al ⇌ Al3+ + 3e–



3Cl+ 3e– ⇌ 3Cl–

Al + 3Cl+ 3e– ⇌ Al3+ + 3Cl– + 3e–

Al+ 3Cl⇌ Al3+ + 3Cl–

In the above reaction, the charge has been balanced on both sides. In other words, Al3+ and Cl–1 combine

to form AlCl3. So, there are three chlorine ions in the molecule.



MSC: Knowledge

20. Element X has two valence electrons and element Y has 7 valence electrons. Which of the following

compounds would you expect to form?

A. XY2

B. X2Y

C. X7Y2

D. X2Y7

ANS: A

The number of valence electrons in the element X is 2. It means it is a metal because an element having

less than four valence electrons, generally exhibit metallic character. Moreover, these electrons can also

be easily donated to attain the nearest noble gas configuration.

X –- 2e– ⇌ X2+

The number of valence electrons in the element Y is 7. It can not release 7 electrons to attain

electropositive character. However, it can gain one electron to attain the nearest noble gas configuration.

So, it is a nonmetal.

Y + e-- ⇌ Y–

When one cation and two such anions combine, the compound formed is

X2+ + 2Y– ⇌ XY2



MSC: Knowledge

21. Which of the following sets of bonds are correctly ordered from least to most polar?


A. C–Fe < C–H < C–O < C–F

B. C–H < C–Fe < C–F < C–O

C. C–H < C–Fe < C–O < C–F

D. C–F < C–O < C–Fe < C–H

ANS: C

In the above given bonds, one element is the same (i.e., carbon) and the electronegativity of other

elements increases in the order H < Fe < O < F. So, polar character of the bond increases in the order

C–H < C–Fe < C–O < C–F.


LOC: 20–A2.10K TOP: Diversity of matter KEY: Electronegativity

MSC: Knowledge

22. The boiling points of water, H2O(l), ammonia, NH3(g), and hydrogen fluoride, HF(l), are higher than

expected due to:

A. strong London (dispersion) forces.

B. strong hydrogen bonding.

C. strong gravitational forces.

D. strong Van der waal’s forces.

ANS: B

Water, H2O(l), ammonia, NH3(g), and hydrogen fluoride, HF(l), contain hydrogen atoms bonded to highly

electronegative elements (N, O, F). In case of such molecules, hydrogen bonding occurs. Because of the

presence of strong hydrogen bonding, these molecules require more energy to move into the vapour phase

from the liquid phase. Therefore, their boiling points are higher than the expected values.


LOC: 20–A2.8K TOP: Diversity of matter KEY: Intermolecular forces

MSC: Knowledge

23. Which of the following statements correctly reflects the correlation between boiling point and the

intermolecular forces of attraction?

A. The stronger the intermolecular forces of attraction, the lower the boiling point.

B. The stronger the intermolecular forces of attraction, the higher the boiling point.

C. The weaker the intermolecular forces of attraction, the higher the boiling point.

D. There is no relationship between intermolecular forces of attraction and boiling point.

ANS: C

Boiling point is strongly related to intermolecular forces of attraction. With an increase in the

intermolecular forces of attraction, more energy is required for the molecules to go into the gas phase and

results in higher boiling point.


LOC: 20–A2.8K TOP: Diversity of matter KEY: Intermolecular forces

MSC: Knowledge

24. Which of the following compounds is the most polar?

A. H2O(l)

B. H2S(g)

C. HF(l)

D. HCl(g)

ANS: C

All the compounds given above contain one element same that is hydrogen and its electronegativity is

2.20. The electronegativity of other elements, i.e., Cl, F, O, and S is 3.16, 3.98, 3.44, and 2.56,

respectively. The most electronegative element out of these is fluorine. So, the electronegativity

difference in HF is the most, 0.78. Therefore, HF(g) will be the most polar.


LOC: 20–A2.10K TOP: Diversity of matter KEY: Electronegativity

MSC: Knowledge


The temperature of the Pahoehoe lava can be estimated by observing its color. The results of such

estimation agree significantly with the measured temperatures of lava flows, which are invariably

greater than 1000 degrees Celsius.

Which of the following cannot be a temperature of a burning particle of the Pahoehoe lava?

A. 1010 K

B. 1276 K

C. 1300 K


D. 1350 K

ANS: A

Lava flows at a temperature of more than 1000 oC, i.e., 1273 K. Hence, lava cannot have a temperature

less than 1273 K. So, 1010 K cannot be a temperature of a burning particle of the lava.


LOC: 20–B1 2K TOP: Forms of matter–gases KEY: Celsius and Kelvin scales

MSC: Knowledge

26. In a reaction following the law of combining gas volumes, 1 L of nitrogen gas would react with

i litres of hydrogen gas, to produce ii litres of ammonia gas. The statement given above is

completed by the information in row

Row i ii

A Two Three

B Three Two

C Three Four

D One One

ANS: B

Reaction between nitrogen and hydrogen takes place as

N2 (g) + 3H2 (g) 2NH3(g)

So, we can say that one litre of nitrogen gas reacts with three litres of hydrogen gas to produce two litres

of ammonia gas.


LOC: 20–B1.3K TOP: Forms of matter–gases KEY: Law of combining volumes

MSC: Knowledge


Andy took a sample each of milk, silt in water, tomato juice, and wine and tested them for the

presence of suspended particles. He was able to detect the presence of suspended particles in all but

one of the samples.



Which of the above substances will have uniform composition and appearance?

A. Milk

B. Silt in water

C. Tomato juice

D. Wine

ANS: D

Uniform appearance and composition is the property of a homogeneous mixture. Heterogeneous mixtures

consist of visibly different substances and phases. Of the given substances, milk, silt in water, and tomato

juice are examples of suspensions. A suspension is a heterogeneous mixture in which particle sizes are

large; visible and will settle when left undisturbed. These three substances will not have uniform

appearance and composition. Wine is a solution of alcohol and water. All solutions are homogeneous

mixtures and will have uniform appearance and composition. The particles present in homogeneous

mixtures cannot be detected even with microscope.


LOC: 20–C1.1K TOP: Acids and Bases KEY: Homogeneous mixtures

MSC: Knowledge

28. Which of the following compounds is not an electrolyte?

A. C2H5OH(aq)

B. H2SO4(aq)

C. HCl(aq)

D. CuSO4(aq)

ANS: A

C2H5OH(aq) is a molecular compound that will dissolve in water, but, as a molecular compound, does not

ionize, and is therefore not an electrolyte.


LOC: 20–C1.4k TOP: Acid and Bases KEY: Electrolyte

MSC: Knowledge


The table shows four experiments, A, B, C, and D, in which a constant mass of sodium chloride,

NaCl(s), is dissolved in the same amount of solvent at different temperatures. The time taken (in

seconds) for the NaCl(s) to dissolve was 10 s, 15 s, 18 s, and 25 s.

Experiment Temperature (oC)

A 37

B 40

C 25

D 58

Which experiment would have taken 18 s to dissolve the NaCl(s) completely?

A. A

B. C

C. B

D. D

ANS: A

Solubility of a compound is directly proportional to the temperature at which salt is dissolved. Therefore,

the higher the temperature, the faster the dissolution. The events, arranged in ascending order of the time

taken to dissolve the salt in water, are C < A < B < D. Therefore, event A took 18 s.


LOC: 20–C1.9k TOP: Acid and Bases KEY: Solubility

MSC: Knowledge


The white washing of walls gives shiny finish to them. Slaked lime is the main ingredient used

for white washing walls. It can be prepared as follows.

CaO(s) + H2O(l) Ca(OH)2(s)

The calcium oxide, CaO(s), must dissolve in water for the reaction to occur.

An increase in pressure will have which of the following effects on the reaction of calcium oxide with

water: McGraw Hill Ryerson © 2007 18

A. Solubility of calcium oxide will increase.

B. Solubility of calcium oxide will decrease.

C. There will be no effect on the solubility of calcium oxide.

D. The product formed will be different.

ANS: C

The effect of pressure is observed only in case of gases. Since the reaction given above does not include

any gaseous component, the increase in pressure will have no effect on the solubility of calcium oxide.


LOC: 20–C1.9K TOP: Acids and Bases KEY: Solubility

MSC: Knowledge


NH4ClO4(s) is used as an oxidizer to provide the required thrust for the lift off of the space shuttle. NH4ClO4(s) reacts with dinitrogen trioxide, to form an acid solution along with water and nitrogen gas.

What is the name of the acid formed in this process?

A. Chlorous acid

B. Hypochlorous acid

C. Perchloric acid

D. Chloric acid

ANS: C

The following reaction takes place

2 NH4ClO4 + N2O3 ⇌ 2N2 + 2 HClO4 + 3H2O

The acid obtained in the process is HClO4. It contains ClO4– polyatomic ion and its name is perchlorate

ion. According to the nomenclature of acids, the acid obtained in the above process is perchloric acid.


LOC: 20–C2.1K TOP: Acids and Bases KEY: Mono/polyprotic acid

MSC: Knowledge



A scientist treats dilute sulfuric acid with barium chlorite to produce an acid. The reaction

is shown below:

Ba(ClO2)2(aq)+ H2SO4(aq) BaSO4(s) + 2HOClO(aq)

Name the acid formed in the reaction.

A. chloric acid

B. hypochlorous acid

C. chlorous acid

D. hypobromous acid

ANS: C

The acid obtained in the process is HOClO(aq). It contains ClO2–(aq) polyatomic ion. This ion is named

chlorite. According to the nomenclature of acids, the suffix ‘ite’ is changed to ‘ous’. Therefore the acid

obtained is chlorous acid.


LOC: 20–C2.1K TOP: Acids and Bases KEY: Mono/Polyprotic acid

MSC: Knowledge


A student compares the ionization of monoprotic acids with that of polyprotic acids, and records the

following observations.

Acid Ka1 Ka2 Ka3

ethanoic acid 1.76 × 10–5

tartaric acid 6.0 × 10–4 1.5 × 10–5

hydrosulfuric acid 1.3 × 10–7 7.1 × 10–15

phosphoric acid 7.6 × 10–3 6.2 × 10–8 2.2 × 10–13

When the above mentioned acids are titrated with NaOH; which of the following acids will have the

highest pH value at the final equivalence point in the titration curve?

A. ethanoic acid

B. tartaric acid

C. hydrosulfuric acid


D. phosphoric acid

ANS: D

Equivalence points

pH

Volume of base

Phosphoric acid, H3PO4(aq), has three ionizable hydrogen atoms, while tartaric acid,

C2H4O2(COOH)2(aq), and hydrosulfuric acid,H2S(aq), have two ionizable hydrogen atoms, and ethanoic

acid, CH3COOH(aq), has one ionization hydrogen atom. In case of phosphoric acid we will have the

maximum number (three) of ionizations and, therefore, the maximum number of equivalence points. So,

out of above acids, phosphoric acid will have equivalence point at the highest pH value, as shown above

in the titration curve. In the case of other acids, the number of equivalence points will be fewer and they

will reach equivalence with the NaOH(aq) at a lower pH value.


LOC: 20–C2.11K TOP: Acids and Bases KEY: Mono/Polyprotic acids

MSC: Knowledge

34. Strong acids increase the concentration of H3O+(aq) and OH–(aq) ion in aqueous solutions to a large

extent. While weak acids and bases increase the concentration of the respective ions to a small extent. The

relative strength of an acid or a base is generally determined by the dissociation constant Ka or Kb.

Which of the given acids is the strongest acid?

A. HCN(aq)

B. HF(aq)

C. CH3COOH(aq)

D. HNO2(aq)

ANS: B

As per the information, the strength of an acid is directly proportional to the dissociation constant of the

acid (Ka). From the given data McGraw Hill Ryerson © 2007 21

4 × 10–7 < 1.8 × 10–5 < 4.5 × 10–4 < 6.7 × 10–4 < 6.7 × 10–4

HCN < CH3 COOH < HNO2 < HF

∴ HF is the strongest acid.


LOC: 20–C2.10k TOP: Acid and Bases KEY: Strong and weak acids

MSC: Knowledge


Acid dissociation constant of some acids

Acids pKa

hydrofluoric acid 3.3

methanoic acid 3.8

ethanoic acid 4.8

hydrogen sulfide 7.1

Which of the above mentioned acids will give H+ ions most easily?

A. Hydrogen sulfide

B. Ethanoic acid

C. Methanoic acid

D. Hydrofluoric acid

ANS: D

pKa values are related to Ka (dissociation constant) values as follows

pKa = –log10Ka

The stronger the acid, more will be the dissociation constant Ka and lesser will be the pKa values. Now

from the above data we can say that hydrofluoric acid has the smallest pKa values. Hence, it is the

strongest acid out of these and will give H+ ions very easily.


LOC: 20–C2.11K TOP: Acids and Bases KEY: Strong acids and bases

MSC: Knowledge



When HCl(aq), a strong acid, and NaOH(aq), a strong base, are mixed together, a

significant quantity of heat is produced. The products are sodium chloride, NaCl(aq)

and water, H2O(l).The reaction that takes place is:

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

This reaction is a i reaction and ii ions are the spectator ions. The statement given above is

completed by the information in row

Row i ii

A hydrolysis Cl–(aq) and H+(aq)

B neutralization H+(aq) and OH–(aq)

C esterification Na+(aq) and OH–(aq)

D neutralization Na+(aq) and Cl–(aq)

ANS: D

This reaction is called a neutralization reaction as it involves the combination of hydronium ions,

H3O+(aq) and hydroxide ions (OH–) to form water. HCl(aq) and NaOH(aq) ionize and dissociate into ions

in solution according to the following equation:

H3O+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) Na+(aq) + Cl–(aq) + 2H2O(l)

In this reaction Na+(aq) and Cl–(aq) ions remain unchanged during the reaction and they see other ions

react. So, Na+(aq) and Cl–(aq) ions are the spectator ions.


LOC: 20–C2.9K TOP: Acids and Bases KEY: Neutralization

MSC: Knowledge

37. Use the following information to answer the next question:

A student was asked to identify the three acidic compounds among the six samples listed below.

She applied the Arrhenius concept to classify the acids.

AlCl3(aq), H2SO4(aq) , CuSO4(aq) , HCl(aq) , NH3(aq) , NaOH(aq)

She classified both H2SO4(aq) and HCl(aq) as acids. She could not identify the third acid.

Which of the remaining compounds could she not select as acid?


A. AlCl3

B. CuSO4

C. NH3

D. NaOH

ANS: A

AlCl3 is an acidic compound. However, it does not follow the Arrhenius as it does not contain any proton

(or hydrogen ion). CuSO4 is a salt whereas NH3 and NaOH are basic compounds.


LOC: 20–C2.7k TOP: Acid and bases KEY: Arrhenius (modified) theory

MSC: Knowledge


The ionization equations of some acids and bases are given below:

+ H ⇌ HAcid

HCl(aq)Base2 )O(l

l

−

−

3O+(aq) + Cl–(aq)

+ H ⇌ HAcid4 (aq)NH +

Base2 )(O 3O+(aq) + NH3(aq)

+ CO ⇌ HCOAcid2 )O(H l

Base

23 (aq) 3

–(aq) + OH–(aq)

+ CH ⇌ CHAcid4 (aq)NH +

Base3 (aq)COO 3COOH(aq) + NH3(aq)

Which of the above mentioned bases are Bronsted-Lowry bases but not Arrhenius bases?

A. H2O(l) and CO22–(aq)

B. CO32–(aq) and CH3COO–(aq)

C. CO32–(aq) only

D. H2O(l), CO32–(aq), and CH3COO–(aq)

ANS: B

According to the Arrhenius concept, a base is a substance, which can donate OH– ions. Hence, a base

must be a source of OH– ions. But according to the Bronsted-Lowry theory, a base must be a proton

acceptor. H2O(l) can accept a proton. Also it can give OH– ions in the solution. Hence, it is an Arrhenius

as well as a Bronsted base, while CO32–(aq) and CH3COO–(aq) ions are only proton acceptors and not

OH–(aq) donors. Hence, these are Bronsted-Lowry bases but not Arrhenius bases.



LOC: 20–C2.8K TOP: Acids and Bases KEY: Arrhenius (modified) theory

MSC: Knowledge

39. If the pH of a solution, measured with the help of a pH meter, is 5.127, [H3O+] is

A. 5.67 × 10–5 mol/L

B. 7.46 × 10–6 mol/L

C. 5.23 × 10–7 mol/L

D. 7.92 × 10–6 mol/L

ANS: B

Reading on the pH meter = 5.127

pH of the solution = 5.127

We know that pH = –log [H3O+] or log [H3O+] = –pH

log [H3O+] = –5.127 mol/L

[H3O+] = antilog [–5.127] = 7.46 × 10–6 mol/L.


LOC: 20–C2.6K TOP: Acids and Bases KEY: Hydronium ion/pH

MSC: Knowledge


In an experiment, 30.0 mL of a solution of 0.100 mol/L HCl(aq) is titrated with a standard

0.100 mol/L NaOH(aq) solution. The initial pH of the solution is 1.000.

The pH of the titrant after adding 5.0 mL of 0.100 mol/L NaOH(aq) is

A. 2.334

B. 1.523

C. 1.225

D. 1.143

ANS: D

After the addition of 5 mL of 0.1 M NaOH solution, the volume of the solution becomes

= 30 + 5 = 35 mL


Initially the amount of H+ ions present in the titration flask = 0001

3010 ×.

= 3.0 × 10–3 M.

Similarly the amount of OH– ions present in 5 mL of solution =0001

51.0 × .

= 0.5 × 10–3M

Amount of H+ ions left after neutralization in solution

= 3.0 × 10–3 – 0.5 × 10–3

= 2.5 × 10–3M

Now concentration of H+ ions

[H+] = 35

105.2 3−× × 1 000 = 0.072 M

pH = –log (.072)

pH = 1.143.


LOC: 20–C2.5K TOP: Acids & Bases KEY: Hydronium ion/pH

MSC: Knowledge

41. In the above experiment, the initial pOH of the HCl(aq) solution is 13.000. The pOH of the solution,

after adding 30.05 mL of NaOH, will be

A. 2.192

B. 3.072

C. 4.079

D. 5.209

ANS: C

After adding 30.0 mL of the NaOH solution the acid is completely neutralized. On adding 30.05 mL of

NaOH, the additional volume of NaOH = 0.05 mL. Now amount of NaOH in 0.05 mL of

NaOH = 0001

1.0 × 0.05 = 5 × 10–6M

Total volume of solution becomes 30 + 30.05 = 60.05 mL.

Concentration of OH– ions

[OH–] = 05.60

105 6−× × 1 000

= 05.60

1005 × × 10–5


= 05.60

500 × 10–5

= 8.327 × 10–5

pOH = –log [OH–]

= –log [8.327 × 10–5]

= –[log 8.325 – 5]

= –[0.921 – 5]

= 4.079.


LOC: 20–C2.5K TOP: Acids and Bases KEY: Hydroxide ion/pOH

MSC: Knowledge


pH value of some of the common substances

Substance pH value

Battery acid 0.5

Vinegar 2.4 – 3.4

Lemon juice 2.2 – 2.4

Beer 4.0 – 5.0

Milk 6.8

Lime water 10.5

In which of the above solutions will the phenolphthalein indicator be pink?

A. Lemon juice

B. Vinegar

C. Battery acid

D. Lime water

ANS: D

Phenolphthalein is light pink in the pH range 8.3 – 10.0. It is colourless in acidic solutions (low pH); and

in basic solutions it is deeper pink. Only lime water has a pH above 8.3, and will be pink in the presence

of phenolphthalein indicator.



LOC: 20–C2.2K TOP: Acids and Bases KEY: Indicators

MSC: Knowledge

43. In which of the following solutions, will methyl orange be red?

A. Lemon juice

B. Beer

C. Milk

D. Lime water

ANS: A

Methyl orange is red in solutions with a pH less than 3.2, orange in solutions with a pH of 3.2 to 4.4 and

yellow in solutions with a pH greater than 4.4. Methyl orange is red in lemon juice which has a pH of 2.2

to 2.4.


LOC: 20–C2.2K TOP: Acids and Bases KEY: Indicators

MSC: Knowledge


A 0.20 mol/L solution of HCN(aq) is prepared. Its pH is 4.92 and [H3O+] is 1.2 × 10–5 mol/L. Students

determined the number of significant digits in the values and recorded them in the table below:

Student Number of significant digits in pH Number of significant digits in concentration

A 2 2

B 3 2

C 1 5

D 0 7

Which of the students reported the pH and concentration of hydronium ion, H3O+(aq), to the correct

number of significant digits?

A. A

B. B

C. C

D. D McGraw Hill Ryerson © 2007 28

ANS: A

The number before the decimal point is not significant in pH, so the value, 4.29 has only 2 significant

digits. In all other values, all numbers except zeroes before the number are significant, and the exponent is

not considered in counting significant digits. Therefore, the value 1.2 x 10–5 has 2 significant digits.


LOC: 20–C2.4K TOP: Acids and Bases KEY: Concentration

MSC: Knowledge

45. A student measures 100 mL of a 0.2 mol/L NaCl(aq) solution and dilutes it to 500 mL. The

concentration of NaCl(aq) in the diluted solution will be

A. 0.02 mol/L

B. 0.04 mol/L

C. 0.05 mol/L

D. 0.06 mol/L

ANS: B

Applying the molarity equation,

(concentrated) C1V1 = C2V2 (dilute)

(0.20 mol/L) (100 mL) = C2(500 mL)

C2 = 500

10020.0 × = 0.04 mol/L


LOC: 20–C1.11K TOP: Acids and Bases KEY: Dilution

MSC: Knowledge


A saturated solution of sodium chloride, NaCl(aq), is prepared with 100 g of water at a

temperature of 90 oC. The solubility of sodium chloride, NaCl(s), at different temperatures

is shown in the graph below:


The Solubility of NaCl in 100 g of Water at Varying Temperatures

60


How much NaCl(s) would be needed?

A. At least 20 g

B. At least 30 g

C. At least 40 g

D. At least 50 g

ANS: D

Saturation is a point where the system is in equilibrium so that the rate of dissolution is equal to the rate

of crystallization. In the case of NaCl(s) and water this equilibrium is reached on dissolving 50 g of NaCl

in 100 g water at 90 oC. Any further addition of NaCl(s) does not dissolve in water.


LOC: 20–C1.10K TOP: Acids and Bases KEY: Solubility MSC: Knowledge

47. What amount of NaCl will make the above solution unstable at 90 oC?

A. 20 g

B. 30 g

C. 40 g

D. 50 g

ANS: D

10

20

30

40

50

0 10 20 30 40 50 60 70 80 90 100

Mas

s of

NaC

l (gr

ams)

Temperature (°C)

When 50 g of NaCl is added to 100 g of water, the solution becomes super saturated. Now in a super

saturated solution, the maximum number of solutes is present in the solution. This supersaturated solution

is very unstable and the excess amount of solute will precipitate.



MSC: Knowledge

48. The [H+(aq)] in a 0.0500 mol/L solution of ethanoic acid, CH3COOH(aq) is

A. 4.24 × 10–4 mol/L

B. 9.49 × 10–4 mol/L

C. 9.29 × 10–4 mol/L

D. 5.26 × 10–4 mol/L

ANS: B

The ethanoic acid dissociates as CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)

Ionization constant is given as

Ka = ]COOHCH[

]OH[]COOCH[

3

33+−

The concentration of hydronium ions is equal to the concentration of acetate ions.

Ka = ]COOHCH[

]OH[

3

23

or 1.8 × 10–5 = 5.0]OH[ 2

3

[H3O+]2 = 1.8 × 10–5 × 0.05

9.0 × 10–7 mol/L

[H3O+] = (90 × 10–8)1/2 = 9.49 × 10–4 mol L–1.



MSC: Knowledge

49. Which of the following expressions describing the concentration of a solution is not correct?

A. Molality = gramsinsolventofMass

soluteofMoles × 1000


B. Molarity = Lin solution ofVolume

soluteofMoles

C. Normality = mLinsolutionofVolume

soluteofsequivalentgramofNumber

D. Formality = Linsolution ofVolume

soluteofmassesformulaofNumber

ANS: C

The normality of a solution is defined as the number of gram equivalents of a solute dissolved per litre of

a solution.



MSC: Knowledge

50. Four compounds, carbon tetrachloride, trichlorofluoro methane, phosphorus pentachloride, and

ammonium nitrate, are dissolved in water and the energy changes during the process are noted.

Dissolution of which of the above substances will result in an endothermic process?

A. Carbon tetrachloride

B. Trichlorofluoromethane

C. Ammonium nitrate

D. Phosphorus pentachloride

ANS: C

Dissolving some salts (ionic compound) in water results in an endothermic process. Out of the given

substances, ammonium nitrate is a salt and the rest are covalent compounds. So, when ammonium nitrate

is dissolved in water, it takes in energy. Hence, it is an endothermic process.



MSC: Knowledge

51. Most of the cleaning detergents are in the form of liquids because

A. less quantity is required in the liquid form.

B. in the liquid form, the electrons are in the excited state.

C. in the liquid form, the surface area between the reacting substances is increased.


D. in the liquid form, there is a greater force of attraction between the reactants and the products.

ANS: C

The cleaning power of the detergent increases when the surface area between the reactive chemicals is

increased. Now in liquid form, the substances are broken down into molecules or ions and hence, surface

area between the reacting substances is increased. This results in increasing the cleaning power.



MSC: Knowledge

52. When iron nails are kept in a test tube containing water, they change their color after some time. This

is because a redox reaction takes place. Dissolving reactants in water is often a prerequisite for redox

chemical changes, because

A. atoms get excited when a substance is dissolved in water

B. catalyst is more effective when dissolved in water

C. ionization energy of reactants decreases when they are dissolved in water

D. the transference of electrons becomes possible when reactants are dissolved in water

ANS: D

Redox reaction involves the transference of electrons between the reactants. These reactions do not occur

by direct contact between the reactants, but by transfer of electron from reactants to water. Ionization of

reactants is possible only in water, such that one reactant is oxidized and the other is reduced. Hence,

chemical reaction takes place faster when reactants are dissolved in water.



MSC: Knowledge


Soda ash, chemically known as sodium carbonate (Na2CO3(s)), is used to soften water for laundering

clothes. When treated with hydrochloric acid, HCl(aq), the reaction produces a gas.

Which of the following gases is released in the above process?

A. Cl2(g)


B. H2(g)

C. CO(g)

D. CO2(g)

ANS: D

Carbonates and bicarbonates react with acids to give CO2 gas

Na2CO3(aq)+ 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)


LOC: 20–D1.1K TOP: Chemical Changes KEY: Chemical equations

MSC: Knowledge


Copper(II) nitrate, Cu(NO3)2(s) is a blue, crystalline solid in its anhydrous form. Nitric acid,

HNO3(aq), is formed when copper(II) nitrate is heated until it decomposes and the resulting gas

fumes are passed directly into water.

What are the products obtained after the decomposition as above?

A. NO(g), O2(g), Cu2O(s)

B. CuO(s), NO2(g), O2(g)

C. NO2(g), O2(g), Cu2O(s)

D. Cu(g), O2(g), NO2(g)

ANS: B

The decomposition reaction of Cu (NO3)2 is

2Cu(NO3)2(s) 2CuO(s) + 4NO2 (g) + O2 (g)

The brown colored gas is NO2(g), nitrogen dioxide.


LOC: 20–D1.1K TOP: Chemical Changes KEY: Chemical Reaction Equations

MSC: Knowledge

55. In a metallurgical process for the generation of tin metal, tin(IV) oxide is reacted with hydrogen gas to

produce tin metal and water vapour. When the chemical equation representing the above reaction is

balanced using lowest whole number coefficients, the coefficient for H2(g) is


A. One

B. Two

C. Three

D. Four

ANS: B

The skeleton equation is

SnO2(s) + H2(g) → Sn(s) + H2O(g)

Balancing the oxygen atom, we get

SnO2(s) + H2(g) → Sn(g) + 2H2O(g)

Now hydrogen atom becomes unbalanced. So, balancing hydrogen atom, we get

SnO2(s) + 2H2(g) → Sn(g) + 2H2O(g)

This is the balanced equation and the coefficient of H2 is two.


LOC: 20–D1.2K TOP: Chemical changes KEY: Chemical reaction equations

MSC: Knowledge

56. A chemist wants to change the oxidation state of iron from +3 to 0. For this purpose, he treats Fe2O3

with carbon. The coefficient of carbon in the balanced chemical equation for the above reaction will be

A. One

B. Two

C. Three

D. Four

ANS: C

(i) The skeleton equation for the above reaction

Fe2O3 + C Fe + CO

(ii) Now skeleton equation along with the oxidation number of each atom is

O2

232 COFe +

−+ 22OOCFe−+

+

(iii) The equation may be written as

O.N decreases by 2 per atom

McGraw H

o3

32 COFe +

+

2oCOFe+

+

O 3 × 2 per two atom

.N decreases by 3 per atom or

ill Ryerson © 2007 35

(iv) To balance the increase or decrease, multiply C by 3 and Fe2O3 by 1

Fe2O3 + 3C Fe + Co

(v) Balancing all other atoms we get

Fe2O3 + 3C 2Fe + 3CO

Coefficient of carbon is three.


LOC: 20–D1.2K TOP: Chemical changes KEY: Chemical reaction equations

MSC: Knowledge

57. In a metallurgical process, Pb3O4 undergoes a reaction with carbon as shown:

Pb3O4(s) + 2C(s) 3Pb(s) + 2CO2(g)

If 44.5 g of of Pb3O4(s) is, consumed, what mass of lead will be produced?

A. 39.8 g

B. 35.3 g

C. 43.1 g

D. 57.0 g

ANS: B

Molar mass of Pb3O4 = 310 g

According to the balanced chemical equation,

Pb3O4(s) + 2C(s) 3Pb(s) + 2CO2(g)

Amount of Pb produced by 310g of Pb3O4= 246 g

Amount of Pb produced by 1g of Pb3O4= 310246 g

Amount of Pb produced by 44.5g of Pb3O4 = 310246 x 44.5 = 35.3 g


LOC: 20–D1.3K TOP: Chemical Changes KEY: Stoichiometry

MSC: Knowledge


Potassium dichromate is used in industries, in calico printing, in tanning of leather. It is used in

photography for hardening of gelatin film.


As an oxidizing agent, acidified potassium dichromate oxidizes tin(II) salts to tin(IV) salts. What is the

balanced ionic equation for this reaction?

A. Cr2O72– + 14H+ + 3Sn2+ 2Cr3+ + 3Sn4+ + 7H2O

B. Cr2O72– + 7H+ + Sn2+ Cr3+ + Sn4+ + 7H2O

C. Cr2O72– + Sn2+ + 7H2 2Cr3+ + Sn4+ + 7H2O

D. Cr2O72– + 14H+ + 6e– + Sn2+ + 2Cr3+ + 3Sn4+ + 7H2O

ANS: A

Tin(II) salt (Sn2+) is changed to tin(IV) salt (Sn4+) in the following way

Cr2O72– + 14H+ + 3Sn2+ 2Cr3+ + 3Sn4+ + 7H2O


LOC: 20–D1.4K TOP: Chemical Changes KEY: Ionic Equation

MSC: Knowledge


In volumetric analysis of redox reaction, potassium permanganate (KMnO4) is used as an oxidizing

agent. Estimation of ferrous salts, oxalates etc is done in such a reaction between KMnO4 and Mohr’s

salt [FeSO4 (NH4)2 SO4 • 6H2O]. KMnO4 acts as a self indicator.

What is the net ionic equation in this redox reaction?

A. MnO4– + Fe2+ + H2 Mn2+ + Fe3+ + 2H+ + 2O2

B. MnO4– + Fe2+ + 2OH– Mn2+ + Fe3+ + H2O + 3O2

C. MnO4– + Fe2+ + 4H2 Mn2+ + Fe3+ + 4H2O

D. MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

ANS: D

In acidic medium, KMnO4 acts as a strong oxidizing agent.

MnO4– + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O


LOC: 20–D1.4K TOP: Chemical Changes KEY: Ionic Equation

MSC: Knowledge


60. Use the following information (along with the graphs) to answer the next two questions.


In an experiment of acid - base titration HCl, NaOH, NH3, and CH3COOH are titrated against a strong acid

or a strong base. The data obtained are represented in the following curves.

7

14

Graph 2

p

H V

alue

Volume of base (in mL)

7

14

Graph 1

Volume of acid (in mL)

7

14

Graph 3

p

H V

alue

Volume of acid added (in mL)

7

14

Graph 4

pH V

alue

Volume of base added (in mL)

p

H V

alue

Match each of the graphs, as numbered above, with the corresponding titration species listed below. HCl – strong base _______

NaOH – strong acid _______

NH3 – strong acid _______

CH3COOH – strong base _______

ANS: 2, 1, 3, 4

HCl – strong base 2

NaOH – strong acid 1


NH3 – strong acid 3

CH3COOH – strong base 4


LOC: 20–D2.5K TOP: Chemical Changes KEY: Titration

MSC: Knowledge

61. Out of the graphs plotted above, which titration curve corresponds to a strong monoprotic base and a

strong monoprotic acid?

A. Graph 1

B. Graph 2

C. Graph 3

D. Graph 4

ANS: A

In the beginning, strong monoprotic base is present, there the pH is high. When a strong monoprotic acid

is run into it, there is a rapid fall of pH in the beginning, but the rate of fall slows down. The equivalence

point in a strong acid/strong base titration is at pH 7.



MSC: Knowledge

62. A sample mixture of acids, amines and salts is given to students for analysis. Determination of which

of the following characteristics in the above sample will come under the category of qualitative analysis?

A. Number of moles of acids present.

B. Concentration of amines expressed in percentage by mass.

C. Parts per million of salt present.

D. The presence of –COOH group in the sample with the help of spectroscopy.

ANS: D

The purpose of qualitative analysis is to confirm the presence of an element, compound, phase, functional

group, organic compound or liquid in a sample. Qualitative analysis does not involve amounts.


LOC: 20–D1.3K TOP: Chemical changes KEY: Reaction stoichiometry

MSC: Knowledge


Iron is available in three forms: cast iron, wrought iron and steel. These forms differ from each other

in the carbon content and, of course, traces of certain other elements.

An iron sample of cast iron is 1.5% impure. The following reaction shows the formation of pentacarbonyl

iron:

Fe + 5CO essurePr⎯→⎯∆

ironnylPentacarbo5)CO(Fe[

If 4.2 g of this sample is used in the formation of pentacarbonyl iron, what is the actual yield of

[Fe(CO)5]?

A. 25.2 g

B. 12.9 g

C. 14.5 g

D. 27.1 g

ANS: C

56 g of Fe can give 196 g of [Fe(CO)5].

Since the iron sample is 1.5% impure, mass of impure iron = 4.2 × 100

5.1 = 0.063 g

Mass of pure iron = 4.2 – 0.063 = 4.137 g

Amount of [Fe(CO)5] produced by 4.136 g of pure iron = 56

196 × 4.137 = 14.5 g


LOC: 20–D2.3K TOP: Chemical Changes KEY: Actual, theoretical and percent yield

MSC: Knowledge


Carbon monoxide gas reacts with solid iron(III) carbonate to produce iron metal and carbon dioxide

gas. The reaction that takes place is

Fe2CO3(s) + CO(g) 2Fe(s) + 2CO2(g)


Assuming stoichiometric quantities of all reactants and products, what mass of Fe2CO3(s) is consumed

when it is reacted with 5.6 g CO(g), producing 22.4 g of Fe(s) and 17.6 g of CO2(g)?

A. 17.4 g

B. 34.4 g

C. 37.2 g

D. 41.6 g

ANS: B

+ → + ?

32COFeg6.5

COg4.22

Fe2g6.172CO2

According to the law of conservation of mass:

Mass of reactant = mass of products.

Mass of Fe2CO3 + mass of CO = mass of iron + mass of CO2

Mass of Fe2CO3 + 5.6 g = 22.4 g + 17.6 g

Mass of FeCO3 = 40 – 5.6 = 34.4 g


LOC: 20–D2.1K TOP: Chemical Changes KEY: Reaction Stoichiometry

MSC: Knowledge

65. In a fertilizer factory, 410 kg of ammonium hydroxide, NH4OH(aq), reacts with 738 kg of nitric acid,

HNO3(aq), to produce 937.14 kg of the fertilizer ammonium nitrate, NH4NO3(s). Assuming all reactants

and products are in stoichiometric quantities, what is the mass of all other products is formed in this

reaction?

A. 412 kg

B. 383.65 kg

C. 327.1 kg

D. 210.86 kg

ANS: C

The chemical equation is

+ → + kg410

4OHNHkg738

3HNOkg14.937

34NONH?2OH

According to the law of conservation of mass:

Mass of reactants = mass of products

410 kg+ 738 kg = 937.14 kg + mass of H2O

Mass of H2O = 1148 – 937.14 = 210.86 kg.



LOC: 20–D2.1K TOP: Chemical Changes KEY: Reaction Stoichiometry

MSC: Knowledge


In a precipitation reaction, 8.5 g of Na2SO4(s) is added to an aqueous solution containing of 15 g of

BaCl2(aq). A white precipitate, insoluble BaSO4(s), is formed.

Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)

Which of the following compounds acts as a limiting reagent in this chemical change?

A. Na2SO4

B. BaCl2

C. BaSO4

D. NaCl

ANS: A

142 g Na2SO4 reacts with BaCl2 = 208 g.

Now 15 g BaCl2 will react with NaSO4 = 208142 . × 15 = 10.2 g

Mass of Na2SO4 present = 8.5 g.

Mass of Na2SO4 required = 10.2 g.

Hence, Na2SO4 is the limiting reagent.

PTS: 1 DIF: Average-Difficult OBJ: Section 1.4

LOC: 20–D2.2K TOP: Chemical Changes KEY: Limiting and excess species

MSC: Knowledge

67. Iron can be extracted from iron (III) oxide in the reaction Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g). If

0.50 moles of Fe2O3(s) is present, carbon will be limiting reagent if its number of moles are

A. 0.40 mol or less

B. 1.5 mol or less

C. 1.6 mol or less

D. 2.5 mol or less

ANS: A McGraw Hill Ryerson © 2007 42

We have the following chemical equation

Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g)

Moles of Fe2O3(s) = 0.5 mol

The mole ratio of reactants in the chemical reaction is given as

Fe2O3: C or C: Fe2O3

1 : 3 1 : 31

If 0.50 mol Fe2O3(s)is consumed in the reaction, it would require 0.50 mol Fe2O3(s)× 3 mol C(s)/1 mol

Fe2O3(s)= 1.5 moles of C(s). Now if there is only 0.40 mol of C(s) present, this is less than the required

1.5 mol. So if only 0.4 moles of carbon is provided, carbon, C(s), will be the limiting reagent.


LOC: 20–D2.2K TOP: Chemical changes KEY: Limiting and excess species

MSC: Knowledge

68. Which indicator is used for the titration between hydrochloric acid and ammonia?

A. Phenolphthalein

B. Methyl orange

C. Thymol blue

C. Cresol red

ANS: B

In this kind of titration, the equivalence point is below 7 because the salt (NH4Cl) formed at the

neutralization reacts with water to give H+ ions. The equivalence point lies at about pH 5.3. It is,

therefore, necessary to use an indicator with pH range slightly on the acidic side. Therefore, methyl

orange can be used.


LOC: 20–D2.6K TOP: Chemical changes KEY: Titration

MSC: Knowledge

69. While performing an acid–base titration between HCl(aq) and Na2CO3(aq) solutions, a student forgot

to use the indicator. What would be an appropriate indicator to use?

A. Phenolphthalein

B. Methyl red

C. Methyl orange


D. Bromthymol blue

ANS: C

The titration is between a strong acid (HCl(aq)) and a weak base Na2CO3(aq). The pH at the final

equivalence point will be below 7. The most appropriate indicator will have a Ka halfway between the

hydrogen carbonate ion, HCO3–(aq), and water, which would be methyl orange.



MSC: Knowledge

70. Solid sulfur, S8(s), and fluorine gas, F2(g), react to produce SF6(s). If 384 grams of S6(s) is reacted

with a limited amount of F2(s) producing 1420 g of SF6(s), what is the percent yield?

A. 44.3%

B. 59.7%

C. 81.1%

D. 93.2%

ANS: C

+ → )s(6S

)g(2F18

)g(6SF6

192 g S6 produces SF6 = 876 g.

384 g S6 produces SF6 = 192876 × 384 = 1752 g.

Percent yield = yieldPredicted

yieldActual × 100 =

17521420 × 100 = 81.1%.



MSC: Knowledge



Volume of HCl(aq) added (mL)

HC

l(aq)

7

14 4

1 2

3

The pH curve for the titration of HCl(aq) and NaOH(aq).

The equivalence point in the titration curve between HCl(aq) and NaOH(aq), shown above, is numbered

A. 1

B. 2

C. 3

D. 4

ANS: B

The equivalence point for a strong acid–strong base occurs when pH is equal to 7, labeled #2 on the

graph.


LOC: 20–D2.7K TOP: Chemical Changes KEY: Equivalence point

MSC: Knowledge


NUMERIC RESPONSE

1. The electron dot diagrams of four elements, A, B, C, and D, which are in the same period of the

periodic table are shown below:

A

1.

B

2.

C

3.

D

4.

The elements in increasing order of electronegativity are _____, _____, _____, and _____, respectively.

ANS: 4, 3, 2, 1

The number of valence electrons of elements increases as you move across a period from left to right. The

electron dot diagram of an atom shows its number of valence electrons. The atom having the fewest

number of valence electrons will be the least electronegative and the element having the highest number

of valence electrons will be the most electronegative. Hence, the elements in the increasing order of

electronegativity are D < C < B < A.



MSC: Knowledge


2. Use the following information to answer the next question

In a leading chemical factory, fractional distillation is used to separate ethanol from water. In this

process, ethanol separates from water when the temperature reaches 352 K.

The boiling point of a substance is 22 oC less than the boiling point of ethanol. The boiling point of that

substance is _____K.

(Record your answer in the numerical-response section on the answer sheet.)

ANS: 330 K

The boiling point of the substance is 22 oC less than the boiling point of ethanol on the Celsius scale. So,

the boiling point of the substance on the Celsius scale will be 79 oC – 22 oC = 57 oC, and its boiling point

on the Kelvin scale will be = 57 oC + 273 = 330 K.


LOC: 20–B1.2K TOP: Forms of matter–gases KEY: Celsius and Kelvin scales

MSC: Knowledge

3. A 2.00 L flask is filled with 3.20 g H2(g). If the temperature inside of the tube is 27.0 oC, the gas will

exert a pressure of _____.


ANS: 1 × 10–5 atm

The number of hydrogen molecules left in the tube = 4.879 × 1017

The number of moles of hydrogen molecules left = 23

17

10022.610879.4

×

× = 8.1 × 10–7moles

Now we know PV = nRT

P = x

R = 0.082 litre atm K–1 mol–1

V = 2 L

T = 27 + 273 = 300 K

n = 8.1 × 10–7 mol

Substituting the values

8.1 × 10–7 mol = 300082.0

2x×

×

x = 1 × 10–5 atm



LOC: 20–B1.4K TOP: Forms of matter–gases KEY: Ideal Gas Law

MSC: Knowledge

4. You want to prepare 1.00 L of a 0.0200 mol/L H2SO4(aq) solution from a 90% (M/M) solution of

H2SO4(aq), with a density 1.80 g/mL. The volume of the 90% H2SO4 solution required to prepare the

1.00 L of 0.0200 mol/L H2SO4(aq) is _____ mL.


ANS: 1.21 mL

Number of moles of H2SO4 = 9890 .

In 100 mL of solution:

nH2SO4(aq) = Mm

=

molSOH g 98

SOH g 9042

42

= 0.918 mol H2SO4(aq)

Volume of solution = density

solutionofmass

Mass of solution = 100 g

Density = 1.80 g mL–1

Volume of solution = g8.1

100 = 55.6 mL

Therefore the molar concentration of the original solution is:

CH2SO4(aq) = Vn

= L 0.0556

(aq)SOH mol 0.918 42

= 16.5 mol/L H2SO4(aq)

To calculate the quantity of the solution required, use the dilution formula,

C1V1 = C2V2

16.5 mol/L × V1 = 0.0200 mol/L × 1.00 L


V1 = 5.16000102.0 × = 1.21 mL.



MSC: Knowledge


Ammonium chloride, NH4Cl(s), has extensive application in completion of well and prevention of

damage that would result from interaction of clay and water. It is easily soluble in fresh water. The pH

of ammonium chloride brines lies in the range of 5 – 7.

The number of significant digits reported for the concentration of H3O+(aq) ions in a solution of

ammonium chloride with a pH of 5.1 is _____.

(Record your answer on the numerical-response sheet provided).

ANS: 2

pH of the solution is 5.1

pH = –log [H+]

log [H+] = –pH = –5.1

[H+] = antilog [–5.1]

= 7.46 × 10–6

Since pH value has only 1 significant digit, the concentration of [H+] can be represented as

7 × 10–6 mol/L, which has one significant digit.



MSC: Knowledge


The following steps are used to calculate of volume of dilute solutions from concentrated solutions:

(1) Apply molarity equation.

(2) Determine volume of solution.

(3) Determine the density of acid in solution.

(4) Calculate the molar mass and number of moles of acid.


The correct order of the above steps for calculating the volume of 70% sulfuric acid, by weight (density =

1.8 g mL–1), required to prepare 500 mL of 0.02 M H2SO4, is _____; _____; _____; and _____.

ANS: 4, 3, 2, 1

70% H2SO4 solution means that 70g of sulfuric acid is present in 100g of solution in water.

Now molecular mass of H2SO4 = 2 + 32 + 64 = 98

Weight of H2SO4 = 70 g.

Moles of H2SO4 present = 9870 .

Mass of the acidic solution = 100 g.

Density of the acid in solution = 1.8 g mL–1.

Volume of the solution = DensityMass =

8.1100 = 55.5 mL

Molarity of solution = 5.5598

000170×× = 12.9 M

Applying molarity equation,

(concentrated) M1V1 = M2V2 (dilute)

12.9 × V1 = .02 × 500

V1 = 9.1250002.0 × = 0.77 mL

So correct order of steps is 4, 3, 2, 1.



MSC: Knowledge


The following compounds were provided to students for carrying out various syntheses in a practical

exam. However, when dissolved in water, all the chemicals do not act the same way.

1. Sodium hydroxide

2. Ammonium nitrate

3. Table sugar

The respective order in which the above chemicals will release energy; absorb energy; and neither release

nor absorb energy, is _____, _____, and _____ respectively.


ANS: 1, 2, 3

Sodium hydroxide, when dissolved in water, releases energy to give an exothermic reaction. Ammonium

nitrate absorbs energy, when it is dissolved in water, to give an endothermic reaction. Table salt neither

absorbs energy nor releases energy, when it is dissolved in water. So the correct order is 1, 2, and 3.


LOC: 20–C1.3K TOP: Acids and Bases KEY: Solubility MSC: Knowledge

8. A solution with a concentration of 0.387 mo/L is prepared by dissolving 1.55 g sodium hydroxide,

NaOH(aq) The volume of solution obtained is a.bc × 10–d L. The values of a, b, c, and d are _____,

_____, _____, and _____.


ANS: 1001

The concentation of solution is 0.387 mol/L

mass of NaOH(s) = 1.55 g

nNaOH = Mm

=

NaOH molNaOH g 40.00

NaOH g 1.55

= 0.0388 mol NaOH

CNaOH = Vn

V = Cn

LNaOH mol 0.387NaOH mol 0.0388

= 0.100 L

When converted to scientific notation, 0.100 L is reported as 1.00 x 10–1 mol/L. The correct answer is

1001.



MSC: Knowledge



In case of a very dilute acidic solution, H+ ion concentrations from acid and water are comparable. In

this case the concentration of H+ ions from water can not be neglected. Therefore,

[H+]total = [H+]acid +[ H+]water

Total concentration of H+ ions in a 10–8 M HCl solution, in scientific notation is abc × 10–d. The values of

a, b, c, and d are _____, _____, _____, and _____, respectively.

ANS: 1, 0, 5, 7

Since HCl is a strong acid and is completely ionized,

[H+] = 1.0 × 10–8

The concentration of H+ ions in water is equal to concentration of OH– ions in water.

= = x OH2]H[ +

OH2]OH[ −

= 1.0 × 10Total]H[ + –8 + x

But [H+] [OH–] = 1.0 × 10–14

(1.0 × 10–8 + x) (x) = 1.0 × 10–14

x2 + 10–8x –10–14 = 0

x = 1.2

)10.(1.4)10(10 14288 −−− −−±−

Solving for x we get

x = 9.5 × 10–8

[H+]total = 1.0 × 10–8 + 9.5 × 10–8

= 10.5 × 10–8

= 1.05 × 10–7

So, the values of a, b, c, and d are 1, 0, 5, and 7, respectively.



MSC: Knowledge

10. Use the following information to the answer the next question.


Sodium acetate (NaCH3COO(aq)) is a strong electrolyte and it dissociates completely when dissolved

in water. Ethanoic acid (CH3COOH(aq)) is a weak electrolyte which ionizes only partially when

dissolved in water.

The concentration of hydronium ion in a mixture of 0.01 M CH3COOH and 0.2 M CH3COONa will be

_____. (Ka for ethanoic acid = 1.8 × 10–5)


ANS: 0.09 × 10–5 mol L–1

From the above information, the ionization of ethanoic acid and sodium acetate may be represented as

CH3COOH + H2O ⇌ CH3COO– + H3O+

CH3COONa CH3COO– + Na+

Now let us assume that the concentration of acetate ions obtained from ethanoic acid is A. Since sodium

acetate is completely ionized, the concentration of acetate ions from sodium acetate is = 0.2 M

Total concentration of acetate ion is = A + 0.2

Concentration of unionized ethanoic acid = 0.01 – A

Now, A is very small as compared to 0.2. So, we can write

[CH3COO–]= 0.2 + A = 0.2 mol L–1

[CH3COOH] = 0.01 – A = 0.01 mol L–1

Now the value of dissociation constant is given as

K = ]COOHCH[

]COOCH[]OH[

3

33−+

1.8 × 10–5 = 01.0

2.0]OH[ 3 ×+

[H3O+] = 2.0

01.0108.1 5 ×× −

[H3O+] = 0.09 × 10–5 mol L–1



MSC: Knowledge


A solution is made by mixing two substances A of molecular mass 10 and B of molecular mass 15.


If the mole fraction of A in a solution with B is 0.30, the mass percentage of A in the solution will be

_____.


ANS: 23.23%

Let the mass percentage of A in solution be x. So 100 g of solution contains x g of A and (100 – x) g of B

Moles of A = Aofmass.mol

Aofmass = 10x

Moles of B = 15

x100 −

Now mole fraction of A = 0.3

0.3 = BofmolesAofmoles

Aofmoles+

0.3 =

15x100

10x

10x

−+

0.3 =

150)x100(10x15

10x

−+

0.3 = 10x ×

)x100(10x15150

−+

0.3 = )x100(10x15

15x−+

×

0.3 (15x + 1 000 – 10x) = 15x

0.3 (5x + 1 000) = 15x`

5x + 1 000 = 3.0x15 = 50x

1 000 = 50x – 5x

45x = 1 000

x = 450001 = 22.23

So, mass percentage of A = 23.23%.


LOC: 20–C1.5K TOP: Acids and Bases KEY: Percent by mass

MSC: Knowledge



A scientist was working on the ionization of water. He calculated ionic product of water at different

temperatures. At a specific temperature, ionic product of water comes out as 8.25 × 10–14.

The pH of the water at the same temperature will be _____.


ANS: 6.54

Ionic product of water = 8.25 × 10–14 or [H3O+] [OH–] = 8.25 × 10–14

For water, concentration of hydronium ions is equal to concentration of hydroxide ions. So we can write

[H3O+] [H3O+] = 8.25 × 10–14

[H3O+] = 141025.8 −×

= 25.8 × 10–7

= 2.88 × 10–7

Now pH = –log [H3O+]

= –log (2.88 × 10–7)

= –[log 2.88 – 7]

= –[.46 – 7]

= –[–6.54]

pH = 6.54


LOC: 20–C2.4K TOP: Acids and Bases KEY: Hydronium ion/pH

MSC: Knowledge

13. The reaction representing the decomposition of nitramide is is shown below.

NH2NO2 (aq) N2O(g) + H2O (l)

If 6.14 g of NH2NO2(aq) decomposes completely, the volume of N2O(g) produced at STP is

____________ L.


ANS: 2.18 L

Molar mass of NH2NO2 = 63 g


Stoichiometrically,

63 g of NH2NO2 produced N2O at STP is 22.4 L

6.14 g of NH4 NO2 will produce N2O at STP = 63

4.22 × 6.14 = 2.18 L



MSC: Knowledge


Eudiometry is a branch of science that deals with the analysis of a gaseous mixture, including the

determination of molecular formula of a gaseous hydrocarbon. The general chemical equation for the

combustion of a hydrocarbon is given as under:

CxHy + ⎟⎠

⎞⎜⎝

⎛ +4yx O2 xCO2 + 2

y H2O

Butane, a gaseous hydrocarbon is combusted in the presence of oxygen. CO2 gas and 90 g of water are

recovered. How many moles of CO2 gas are recovered?

ANS: 4

The formula of the hydrocarbon = C4H10

Number of moles of H2O = 1890 = 5 or x = 4,

2y = 5

x = 4, y = 10

Hence, CxHy = C4H10

Number of moles of CO2 = 4



MSC: Knowledge

15. 2.66 g of phosphorus, P4(s), is heated in the presence of 2.43 of oxygen gas, O2(g) The theoretical

yield of phosphorus pentoxide (P2O5(s)) is _____ g.


ANS: 4.31 g


+ → g66.2

4Pg43.22O5

g31.452OP2

Since Amount of P2O5 produced by 124 g of phosphorus = 2 × 142 g

Amount of P2O5 produced by 2.66 g of phosphorus= 124

1422 × × 2.66 = 6.092 g

Theoretical yield of P2O5 = 6.092 g

(s)OPg6.10(s)OPmol1

(s)OPg141.94(s)Pmol1

(s)OPmol2(s)Pg123.88

(s)Pmol1(s)Pg2.66(s)OPgx

limiting)is(s)P(ifproduced(s)OPmass

52

52

52

4

52

4

4452

452

=

×××=

(s)OPg4.31(s)OPmol1

(s)OPg141.94(g)Omol5(s)OPmol2

(g)Og32.00(g)Omol1(g)Og2.43(s)OPgx

limiting)is(g)O(ifproduced(s)OPmass

52

52

52

2

52

2

2252

252

=

×××=

Since the theoretical mass of P2O5(s) produced with 2.43 g of oxygen gas, O2(g) is less than with

phosphorus, P4(s), oxygen gas is the limiting reagent, and the theoretical mass of P2O5(s) produced is 4.31

g, the theoretical mass produced when all of the limiting reagent is consumed.



MSC: Knowledge


In a lab experiment, a solution of an unknown ionic compound AB(s) (molar mass

80 g/mol) is prepared by dissolving 30 g of AB(s) in a 250 mL volumetric flask,

which is then filled with water up to the mark on the neck.

The molar concentration of the solution formed in the above experiment will be

A. 0.015 m

B. 0.025 m

C. 0.030 m

D. 0.035 m

ANS: A

Mass of the compound AB dissolved = 0.30g

Molar mass of AB = 80 g/mol


Moles of AB dissolved = 8030.

Volume of solution = 250 mL

So, concentration of the solution will be = 250

80/30.0 × 1 000

= 25080

30.0×

× 1 000 = 0.015 M.


LOC: 20–C1.5K TOP: Solutions KEY: Calculating Concentration

MSC: Knowledge


WRITTEN RESPONSE


Element Number of Valence Electrons Electronegativity

A 2 2.5

B 7 1.9

C 7 1.5

D 4 1.6

A and B combine to form an ionic compound whereas C and D do not, why? Which will form ionic

bonds, covalent bonds, and polar covalent bonds?

ANS: If difference in electro negativity between two atoms is high, the bonding between them takes place

with complete transfer of electrons, resulting in an ionic bond. In this type of bonding, one of the atoms

loses electrons while the other gains and acquires the nearest noble gas configuration. The bonding

between atoms A and B can be represented by Lewis dot structure as shown below:


A2+ , B– [A]2+ 2[B]– or AB2 B A

During the formation of this bond, A will lose its two electrons. Now, B requires only one electron to

complete its octet. So, for the formation of a neutral ionic compound, A will combine with two atoms of

B to form AB2. No such bonding is possible in the case of atoms C and D, which have almost the same

electronegativity. C and D will combine by sharing electrons to form a covalent bond.


LOC: 20–A1.4K TOP: Diversity of matter KEY: Valence electron and Lewis Structure

MSC: Knowledge


Ink and air both are form of mixture. Ink is a liquid, which is composed of various pigments

while air is a mixture of various gases in no definite proportion.

Why ink has visible boundary of separation while air does not?

ANS: Ink is an example of heterogeneous mixture in which one component of mixture is suspended into

the other component while air is an example of non-homogeneous mixture which has same composition

through out. Heterogeneous mixture does contain visible boundary of separation while homogeneous

mixtures do not contain visible boundary of separation. So, ink has visible boundary of separation while

air does not.


LOC: 20–C1.1K TOP: Acids and Bases KEY: Homogenous mixture

MSC: Knowledge


Most of the common substances used in our home are acidic; slightly acidic; basic; or slightly basic.

The following substances were tested with a pH meter.

Compound Color pH value

Soda pop Red 3.0

Lemon juice Dark red 2.2

Coffee Yellow 5.0

Milk Green 6.5

Sea water Blue 7.8

(A) Determine the concentration of H+ ions in a sample of milk

(B) Which of the above substances has the highest hydronium ion concentration?

ANS: (A) In a sample of milk, pH paper shows light green color which corresponds to a pH value of 6.5.

Now pH = 6.5 or –log [H3O+] = 6.5

log [H3O+] = –6.5

[H3O+] = antilog (–6.5)

[H3O+] = 3.16 × 10–7

(B) The pH paper turns from blue to red as the acidity of the solution increases. Out of the above

substances lemon juice gives the darkest red color which corresponds to the lowest pH value and the

highest concentration of hydronium ions. So among these substances, lemon juice is the most acidic.



MSC: Knowledge


4. (A) On a graph paper, draw a graph that represents the titration curve of 0.3 M hydrochloric acid and

0.3 M sodium hydroxide solution mark the equivalence point of.

10 20

Equivalence point

4030

10

8


A. pH 6

4 2 0

50 Volume of NaOH

(B) In an experiment, titration is being performed between 25 ml of a solution of HCl and a standard

NaOH solution (concentration of both is 0.1M). What will be the value of pH at the equivalence point?

ANS: (A) When we add a solution of NaOH to the solution of HCl, pH progressively increases. This is

because OH– ions from the base will react with the H+ ions from the acid to form water. This decreases

the concentration of H+ ions and pH increases. Near the stoichiometric point there is a sudden jump in pH.

It has been observed that at the equivalence point pH is 7 and the pH increases sharply afterwards.

(B) Initially the pH of HCl is given as

pH = –log [0.1] = 1.0

Now we add 10 ml of 0.1 m NaOH to HCl. The total volume is then = 25 + 10 = 35 ml.

Concentration of H+ ions initially present = 0001

251.0 × = 2.5 × 10–3M

Concentration of OH– ions in the added NaOH solution = 0001

101.0 × = 1.0 × 10–3M

After the neutralization of the OH– ions of the added solution, the concentration of H+ ions left

= 2.5 × 10–3 - 1.0 × 10–3

= 1.5 × 10–3 M

Concentration of H+ ions = 35

0001105.1 3 ×× −= 0.043 M

pH = –log (0.043) = 1.37

Processing as above, we can calculate the pH of the solution on the addition of 20 ml, 24 ml, 24.9 ml and

24.95 ml of NaOH solution. The pH values come to 1.95, 2.69, 3.70 and 4.00 respectively. Now, on

addition of 25 ml of NaOH, the acid is completely neutralized giving NaCl. The pH of the resulting

solution is 7.


LOC: 20–D2.7K TOP: Chemical changes KEY: Titration graphs, acids, bases

MSC: Knowledge


In qualitative analysis, group 2 and group 4 cations are precipitated as sulfides by passing H2S(g)

through the aqueous mixture. But in case of group 2, the solution must be acidified with dilute

HCl, before passing the H2S(g), so that the cations of group 4 are not precipitated, if present in

the mixture as sulfide.

Group 2 cations – Hg2+, Pb2+, Bi3+, Cu2+, Cd2+, As3+, Sb3+, Sn2+

Group 4 cations – CO2+, Ni2+, Mn2+, Zn2+

(A) Explain the function of acid in the aqueous mixture solution, along with the relevant chemical

equation(s).

(B) What is the minimum required concentration of S2– ions, to cause precipitation of ZnS from a

1 × 10–3 M Zn (NO3)2 solution? (Ksp for ZnS = 7.0 × 10–16)

ANS: (A) When H2S is passed, the acid is added to suppress the S2– ion (sulfide ion) concentrate in the

solution, due to the common ion effect.

H2s(aq) → 2H+ (aq) + S2–(aq)

HCl(aq) → H+(aq) + Cl–(aq)

common ion

In this way S2– ion concentration in the solution is controlled and any cation of group 4 is not precipitated

along with the cation of group 2. Also the Ksp (solubility product values) of the sulfides of second group

cations are low and the concentration of S2– ions released by H2S is sufficient to precipitate these sulfides

because the ionic product exceeds the solubility product. As the Ksp values of the sulfides of cations of

fourth group are higher, the ionic products in their case will remain lesser than the solubility products.

Hence, these cations will not be precipitated in the second group. If the medium is not made acidic, the

cations of fourth group may also be precipitated in the second group.

(B) To get precipitates of ZnS, the product [Zn2+] [S2–] should exceed the Ksp for ZnS i.e. 7.0 × 10–16

or [Zn2+] [S2–] > 7.0 × 10–16

As [Zn2+] = 1 × 10–3 mole L–1 in Zn (NO3)2 solution

[Zn2+] [S–2] = 7.0 × 10–16

[1 × 10–3] [S2–] = 7.0 × 10–16


[S2–] = 3

16

101100.7

−

−

×

× = 7.0 × 10–3 mole L–1

Thus, minimum concentration of S2– ions required for the precipitation of ZnS is 7.0 × 10–13 mole L–1


LOC: 20–D1.3K TOP: Chemical Changes KEY: Chemical equations, Stoichiometry

MSC: Knowledge


Documents

AB Chemistry CAB Sample