Acids, Bases and Salts Practice Questions — JAMB UTME — Page 2

Question 21Question →

Which of the following methods is most suitable for preparing lead(II) sulfate ( $PbSO_4$ ), an insoluble salt, in the laboratory?

Neutralization of dilute tetraoxosulfate(VI) acid with aqueous lead(II) nitrate solution by titration

Direct combination of lead metal and concentrated tetraoxosulfate(VI) acid

Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate

Action of dilute tetraoxosulfate(VI) acid on insoluble lead(II) carbonate followed by evaporation to dryness

Show answer & explanation

Answer: Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate

Answer

Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate.

Insoluble salts such as lead(II) sulfate (

PbSO_4

) are prepared in the laboratory by double decomposition (precipitation), where two aqueous solutions containing soluble salts are mixed to precipitate the desired insoluble salt.

Step-by-Step Solution

1

Identify the solubility of the target salt

Lead(II) sulfate (

PbSO_4

) is insoluble in water.

The choice of preparation method depends primarily on whether the salt is soluble or insoluble.

2

Select the appropriate synthesis method for insoluble salts

Insoluble salts are best prepared by double decomposition (precipitation) using two soluble salts as starting materials.

Mixing solutions containing

Pb^{2+}

ions (e.g.,

Pb(NO_3)_2(aq)

) and

SO_4^{2-}

ions (e.g.,

Na_2SO_4(aq)

) results in the immediate precipitation of insoluble

PbSO_4(s)

.

3

Formulate the balanced chemical equation

Pb(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow PbSO_4(s) + 2NaNO_3(aq)

The insoluble lead(II) sulfate precipitates out and can easily be collected by filtration, washed with distilled water, and dried.

Key Concept

Preparation of insoluble salts by precipitation (double decomposition)

Question 22Question →

An aqueous solution of a weak monoacidic base, $\text{BOH}$ , has a concentration of $0.20\text{ mol dm}^{-3}$ and a base dissociation constant $K_b = 5.0 \times 10^{-6}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the pH of this solution.

Show answer & explanation

Answer: 11

Answer

The pH of the weak monoacidic base solution is 11.0.

For a weak base, partial equilibrium ionization gives

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{5.0 \times 10^{-6} \times 0.20} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The pOH is

-\log_{10}(1.0 \times 10^{-3}) = 3.0

. Using

\text{pH} = 14.0 - \text{pOH}

, the pH of the basic solution is

14.0 - 3.0 = 11.0

.

Step-by-Step Solution

1

Formulate the dissociation equilibrium equation for the weak base BOH.

The ionization is

\text{BOH}_{(aq)} \rightleftharpoons \text{B}^+_{(aq)} + \text{OH}^-_{(aq)}

, giving

K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}

.

Weak bases ionize incompletely in aqueous media.

2

Calculate the equilibrium hydroxide ion concentration [OH⁻].

[\text{OH}^-] = \sqrt{K_b \times C} = \sqrt{(5.0 \times 10^{-6})(0.20)} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Since the base is weak and degree of ionization is small,

[\text{BOH}] \approx 0.20\text{ mol dm}^{-3}

and

[\text{B}^+] = [\text{OH}^-]

.

3

Determine the pOH of the solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

pOH is defined as

-\log_{10}[\text{OH}^-]

.

4

Calculate the pH using the relationship between pH and pOH at 25°C.

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

.

At

25^\circ\text{C}

,

\text{pH} + \text{pOH} = 14.0

.

Key Concept

Weak base dissociation equilibrium and pH determination

Question 23Question →

An aqueous solution of ammonium trioxonitrate(V), $NH_4NO_3$ , turns blue litmus paper red. Which species and hydrolysis reaction account for this acidity?

Cation hydrolysis where

NH_4^+

reacts with water to produce

H^+

ions

Anion hydrolysis where

NO_3^-

reacts with water to produce

OH^-

ions

Neutralization between

NH_4^+

and

NO_3^-

ions in solution

Indicator interaction where

NO_3^-

directly alters litmus color without changing pH

Show answer & explanation

Answer: Cation hydrolysis where $NH_4^+$ reacts with water to produce $H^+$ ions

Answer

Cation hydrolysis where

NH_4^+

reacts with water to produce

H^+

ions

The ammonium salt

NH_4NO_3

is formed from a strong acid (

HNO_3

) and a weak base (

NH_3

). Upon dissolving in water, the cation

NH_4^+

reacts with water molecules (cation hydrolysis) according to the equilibrium

NH_4^+ + H_2O ⇌ NH_3 + H^+

. The generation of excess hydrogen ions makes the solution acidic and turns blue litmus red.

Step-by-Step Solution

1

Identify the parent acid and parent base of the salt

NH_4NO_3

is formed from a weak base (

NH_3

) and a strong acid (

HNO_3

).

Salts formed from weak bases and strong acids dissolve in water to produce acidic solutions.

2

Determine which ion undergoes hydrolysis

The ammonium cation (

NH_4^+

) hydrolyzes:

NH_4^+ (aq) + H_2O (l) ⇌ NH_3 (aq) + H^+ (aq)

.

The conjugate acid of a weak base is strong enough to react with water, whereas the conjugate anion of a strong acid (

NO_3^-

) is a spectator ion.

3

Relate ion generation to solution acidity

Production of excess

H^+

ions turns blue litmus paper red.

Increased hydrogen ion concentration lowers pH below 7.

Key Concept

Hydrolysis of salts derived from strong acids and weak bases

Estimated Time:45s

Question 24Question →

A $10.0\text{ g}$ sample of hydrated magnesium tetraoxosulfate(VI), $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ , is heated to constant mass in a crucible, leaving a residue of $4.88\text{ g}$ of anhydrous salt. When the anhydrous residue is exposed to moist ambient air at room temperature, it absorbs water vapor until its mass increases back to $10.0\text{ g}$ without forming a liquid solution. What is the integer value of $x$ , and what atmospheric behavior does the anhydrous salt display during moisture absorption? $[\text{Relative atomic masses: } \text{Mg} = 24, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

x = 7

, and the salt displays hygroscopy

x = 7

, and the salt displays deliquescence

x = 5

, and the salt displays hygroscopy

x = 5

, and the salt displays efflorescence

Show answer & explanation

Answer: $x = 7$ , and the salt displays hygroscopy

Answer

x = 7

, and the salt displays hygroscopy

The mass of water lost from

10.0\text{ g}

of

\text{MgSO}_4 \cdot x\text{H}_2\text{O}

is

5.12\text{ g}

. Converting the mass of anhydrous

\text{MgSO}_4

(

4.88\text{ g}

) and water (

5.12\text{ g}

) to moles gives

0.04067\text{ mol}

and

0.2844\text{ mol}

respectively, yielding

x = 7

. Because the anhydrous salt absorbs atmospheric moisture without dissolving into a liquid solution, it demonstrates hygroscopy.

Step-by-Step Solution

1

Calculate the mass of water lost upon heating.

\text{Mass of } \text{H}_2\text{O} = 10.0\text{ g} - 4.88\text{ g} = 5.12\text{ g}

Heating to constant mass drives off all water of crystallization from the hydrated crystal structure.

2

Determine the molar masses of anhydrous

\text{MgSO}_4

and

\text{H}_2\text{O}

.

Molar mass of

\text{MgSO}_4 = 24 + 32 + (4 \times 16) = 120\text{ g/mol}

; Molar mass of

\text{H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}

.

Required to convert sample masses into chemical mole quantities.

3

Calculate the mole ratio of water of crystallization to anhydrous salt to find

x

.

\text{Moles of } \text{MgSO}_4 = \frac{4.88}{120} = 0.04067\text{ mol}

;

\text{Moles of } \text{H}_2\text{O} = \frac{5.12}{18} = 0.2844\text{ mol}

;

x = \frac{0.2844}{0.04067} = 7

.

The coefficient

x

represents the integer ratio of moles of water to moles of anhydrous salt.

4

Identify the atmospheric behavior of the anhydrous salt upon absorbing moisture without forming a solution.

The behavior is termed hygroscopy (or hygroscopic nature).

Hygroscopic substances absorb moisture from air without forming a liquid solution, whereas deliquescent substances absorb enough moisture to dissolve into a solution.

Key Concept

Stoichiometric determination of water of crystallization and conceptual differentiation between hygroscopy and deliquescence.

Question 25Question →

Consider the chemical reaction represented by the equation:

BF_3 + NH_3 \rightarrow F_3B:NH_3

Which of the following statements correctly describes the role of

BF_3

in this reaction according to acid-base theories?

It acts as a Lewis acid by accepting an electron pair.

It acts as a Brønsted-Lowry acid by donating a proton.

It acts as a Lewis base by donating an electron pair.

It acts as an Arrhenius acid by producing hydrogen ions in aqueous solution.

Show answer & explanation

Answer: It acts as a Lewis acid by accepting an electron pair.

Answer

Boron trifluoride (

BF_3

) acts as a Lewis acid because it accepts an electron pair from ammonia (

NH_3

).

Boron trifluoride (

BF_3

) has a central boron atom surrounded by six valence electrons, leaving an empty orbital. In the reaction, it accepts an unshared electron pair from nitrogen in ammonia (

NH_3

) to form a dative covalent bond. By definition, an electron-pair acceptor is a Lewis acid.

Step-by-Step Solution

1

Analyze the electronic structure of the reactants

Boron in

BF_3

has 6 valence electrons (an incomplete octet) and an empty p-orbital, whereas nitrogen in

NH_3

has a lone pair of non-bonding electrons.

Determining electron pair availability is necessary to apply the Lewis acid-base definition.

2

Identify the type of bond formed during the reaction

Nitrogen donates its lone pair into the empty orbital of boron to form a coordinate (dative) covalent bond (

F_3B:NH_3

).

According to the Lewis concept, an electron-pair acceptor is an acid and an electron-pair donor is a base.

3

Evaluate the applicable acid-base theory

Because no proton (

H^+

) transfer occurs and no aqueous

H^+

ions are produced, neither Brønsted-Lowry nor Arrhenius definitions apply, making it strictly a Lewis acid-base reaction.

Lewis theory extends acid-base chemistry to reactions occurring without hydrogen or solvent participation.

Key Concept

Lewis Theory of Acids and Bases

Estimated Time:1m 0s

Question 26Question →

Complete the following statement regarding acid-base behavior according to the Brønsted-Lowry theory.

Fill in the blanks below

In the reversible reaction

HSO_4^- + H_2O \rightleftharpoons SO_4^{2-} + H_3O^+

, the species that acts as the conjugate base of hydrogen sulfate ion (

HSO_4^-

) is the ion.

Show answer & explanation

Answer

sulfate (or tetraoxosulfate(VI),

SO_4^{2-}

)

Under the Brønsted-Lowry definition, an acid is a proton donor and its conjugate base is the species remaining after the loss of that proton. When the hydrogen sulfate ion (

HSO_4^-

) donates a proton (

H^+

), it forms the sulfate ion (

SO_4^{2-}

), making sulfate the conjugate base.

Step-by-Step Solution

1

Identify the Brønsted-Lowry acid in the forward reaction.

The hydrogen sulfate ion (

HSO_4^-

) acts as an acid by donating a proton (

H^+

).

By definition, a Brønsted-Lowry acid is a proton donor.

2

Determine the remaining species after proton donation.

Removing

H^+

from

HSO_4^-

yields

SO_4^{2-}

.

When an acid loses a single proton, the remaining chemical species formed is its conjugate base.

Key Concept

Brønsted-Lowry Conjugate Acid-Base Pairs

Question 27Question →

Match each sparingly soluble salt with its correct solubility product ( $K_{sp}$ ) expression, where $s$ represents the molar solubility of the salt in $\text{mol dm}^{-3}$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

Calcium sulfate matches

K_{sp} = s^2

; Silver chromate matches

K_{sp} = 4s^3

; Aluminium hydroxide matches

K_{sp} = 27s^4

; Calcium phosphate matches

K_{sp} = 108s^5

.

Each salt dissociates according to its stoichiometry

A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}

, giving

K_{sp} = x^x y^y s^{x+y}

. Calcium sulfate (a 1:1 salt) gives

s^2

; silver chromate (a 2:1 salt) gives

4s^3

; aluminium hydroxide (a 1:3 salt) gives

27s^4

; and calcium phosphate (a 3:2 salt) gives

108s^5

.

Step-by-Step Solution

1

Write the balanced dissolution equilibrium for a generic sparingly soluble salt

A_x B_y

.

A_x B_y(s) \rightleftharpoons x A^{y+}(aq) + y B^{x-}(aq)

This establishes the stoichiometry of the dissolved ions in solution.

2

Express the concentration of each constituent ion in terms of molar solubility

s

.

[A^{y+}] = xs

and

[B^{x-}] = ys

Molar solubility

s

represents the moles of salt dissolved per

\text{dm}^3

of saturated solution.

3

Substitute the ionic concentrations into the solubility product expression

K_{sp} = [A^{y+}]^x [B^{x-}]^y

.

K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)}

This yields the general mathematical relationship between

K_{sp}

and

s

for any ionic solid.

4

Apply the general formula to each specific salt based on its stoichiometric coefficients

x

and

y

.

For

\text{CaSO}_4

(

1:1

),

K_{sp} = 1^1 \cdot 1^1 \cdot s^{1+1} = s^2

. For

\text{Ag}_2\text{CrO}_4

(

2:1

),

K_{sp} = 2^2 \cdot 1^1 \cdot s^{2+1} = 4s^3

. For

\text{Al(OH)}_3

(

1:3

),

K_{sp} = 1^1 \cdot 3^3 \cdot s^{1+3} = 27s^4

. For

\text{Ca}_3(\text{PO}_4)_2

(

3:2

),

K_{sp} = 3^3 \cdot 2^2 \cdot s^{3+2} = 108s^5

.

Matching each salt with its stoichiometric coefficients produces the correct mathematical relationship.

Key Concept

Derivation of Solubility Product (

K_{sp}

) expressions from stoichiometric dissolution equilibria.

Question 28Question →

Concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ) is added to hydrated copper(II) tetraoxosulfate(VI) crystals ( $CuSO_4 \cdot 5H_2O$ ), turning their blue color to white. In a separate vessel, dilute tetraoxosulfate(VI) acid reacts with zinc granules to evolve hydrogen gas ( $H_2$ ). Which chemical properties of tetraoxosulfate(VI) acid are demonstrated in the first and second reactions, respectively?

Dehydrating property and acidic property

Oxidizing property and acidic property

Dehydrating property and oxidizing property

Acidic property and dehydrating property

Show answer & explanation

Answer: Dehydrating property and acidic property

Answer

Dehydrating property and acidic property

The removal of water of crystallization from blue

CuSO_4 \cdot 5H_2O

to leave white anhydrous

CuSO_4

demonstrates the dehydrating property of concentrated

H_2SO_4

. The reaction of dilute

H_2SO_4

with zinc to yield hydrogen gas demonstrates a fundamental chemical property of acids.

Step-by-Step Solution

1

Analyze the first reaction involving concentrated

H_2SO_4

and hydrated copper(II) tetraoxosulfate(VI)

Concentrated

H_2SO_4

acts as a dehydrating agent by removing the five molecules of water of crystallization from blue

CuSO_4 \cdot 5H_2O

to form white anhydrous

CuSO_4

.

Dehydration is the chemical removal of chemically combined water molecules from a compound.

2

Analyze the second reaction involving dilute

H_2SO_4

and zinc granules

Dilute

H_2SO_4

reacts with zinc (

Zn

) according to

Zn_{(s)} + H_2SO_{4(aq)} \rightarrow ZnSO_{4(aq)} + H_{2(g)}

, liberating hydrogen gas.

A characteristic property of acids is reacting with metals above hydrogen in the reactivity series to form a salt and hydrogen gas.

3

Combine the identified properties in sequential order

The first reaction exhibits a dehydrating property, and the second reaction exhibits an acidic property.

The question asks for the properties demonstrated in the first and second experiments respectively.

Key Concept

Distinct behavior of concentrated vs dilute tetraoxosulfate(VI) acid

Question 29Question →

Match each chemical reaction involving acids or bases listed on the left with its correct characteristic observation or product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Reaction of dilute

HCl

with

Na_2CO_3

pairs with effervescence of carbon(IV) oxide gas; heating

NH_4Cl

with aqueous

NaOH

pairs with evolution of pungent ammonia gas; reaction of dilute

H_2SO_4

with zinc metal pairs with liberation of hydrogen gas; and reaction of aqueous

NaOH

with dilute

HNO_3

pairs with neutralization forming a salt and water without gas evolution.

Each acid and base chemical property produces a distinct, diagnostic observation: acid + carbonate yields carbon(IV) oxide (

CO_2

), base + ammonium salt yields ammonia (

NH_3

), acid + reactive metal yields hydrogen (

H_2

), and acid + base (hydroxide) results in neutralisation forming salt and water.

Step-by-Step Solution

1

Analyze acid-trioxocarbonate(IV) reaction property.

2HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + CO_{2(g)}

. The evolved

CO_2

gas causes effervescence and turns lime water milky.

Acids liberate carbon(IV) oxide gas from trioxocarbonates(IV).

2

Analyze base-ammonium salt reaction property.

NH_4Cl_{(s)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)} + NH_{3(g)}

. Ammonia gas is pungent and alkaline, turning moist red litmus paper blue.

Soluble bases displace ammonia from ammonium salts when heated.

3

Analyze acid-metal displacement reaction property.

Zn_{(s)} + H_2SO_{4(aq)} \rightarrow ZnSO_{4(aq)} + H_{2(g)}

. Hydrogen gas burns with a pop sound.

Dilute acids react with metals above hydrogen in the electrochemical series to yield hydrogen gas.

4

Analyze acid-base neutralization property.

NaOH_{(aq)} + HNO_{3(aq)} \rightarrow NaNO_{3(aq)} + H_2O_{(l)}

. This is a neutralisation reaction yielding salt and water.

Soluble metal hydroxides react with acids to form salt and water only.

Key Concept

Chemical properties of acids and bases including gas evolution tests and neutralization

Question 30Question →

An aqueous solution of barium hydroxide, $\text{Ba(OH)}_2$ , has a molar concentration of $0.0005\text{ mol dm}^{-3}$ . Assuming complete dissociation of the base at $25^\circ\text{C}$ , what is the pH of this solution?

11.0

3.0

10.7

3.3

Show answer & explanation

Answer: $11.0$

Answer

The pH of the solution is

11.0

.

Barium hydroxide (

\text{Ba(OH)}_2

) produces two hydroxide ions (

\text{OH}^-

) per formula unit upon complete dissociation. Thus, a

0.0005\text{ mol dm}^{-3}

solution produces

[\text{OH}^-] = 2 \times 0.0005 = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The pOH is

-\log_{10}(1.0 \times 10^{-3}) = 3.0

. Subtracting pOH from 14.0 yields a pH of

11.0

.

Step-by-Step Solution

1

Determine the concentration of hydroxide ions

[\text{OH}^-]

from the mole ratio of

\text{Ba(OH)}_2

.

Since each mole of

\text{Ba(OH)}_2

dissociates into two moles of

\text{OH}^-

,

[\text{OH}^-] = 2 \times 0.0005\text{ mol dm}^{-3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Barium hydroxide is a strong dibasic base that fully dissociates in aqueous solution.

2

Calculate the pOH of the solution.

\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

The pOH scale is defined as the negative logarithm of the hydroxide ion concentration.

3

Calculate the pH using the relationship

\text{pH} + \text{pOH} = 14.0

.

\text{pH} = 14.0 - 3.0 = 11.0

.

At

25^\circ\text{C}

, the sum of pH and pOH for any dilute aqueous solution equals 14.0.

Key Concept

pH and pOH calculations for strong basic solutions considering stoichiometry

Estimated Time:1m 0s

Question 31Question →

Complete the statement below by calculating the missing pOH and pH values for the given alkaline solution.

Fill in the blanks below

A solution of potassium hydroxide (

\text{KOH}

) has a hydroxide ion concentration of

0.01\text{ mol dm}^{-3}

at

25^\circ\text{C}

. The pOH of this solution is and its pH is .

Show answer & explanation

Answer

The pOH of the potassium hydroxide solution is 2 and its pH is 12.

Potassium hydroxide (

\text{KOH}

) dissociates completely in water to yield a hydroxide ion concentration of

[\text{OH}^-] = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

. The pOH is calculated as

-\log_{10}(10^{-2}) = 2

. Since

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

, subtracting 2 from 14 gives a pH of 12.

Step-by-Step Solution

1

Calculate the pOH from the given hydroxide ion concentration

[\text{OH}^-]

.

\text{pOH} = -\log_{10}(0.01) = -\log_{10}(10^{-2}) = 2

\text{KOH}

is a strong monobasic base that fully dissociates in water, giving

[\text{OH}^-] = 0.01\text{ mol dm}^{-3}

.

2

Calculate the pH using the relationship

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

.

\text{pH} = 14 - \text{pOH} = 14 - 2 = 12

The sum of pH and pOH for any aqueous solution at standard temperature (

25^\circ\text{C}

) equals 14.

Key Concept

Determination of pOH and pH for strong base solutions using the ion product constant of water.

Estimated Time:1m 0s

Question 32Question →

A solution is prepared by dissolving $0.04\text{ g}$ of sodium hydroxide ( $\text{NaOH}$ ) in distilled water to make $1.0\text{ dm}^3$ of solution at $25^\circ\text{C}$ . What is the $\text{pH}$ of the resulting solution? [Molar mass of $\text{NaOH} = 40\text{ g mol}^{-1}$ ]

3.0

11.0

1.4

12.6

Show answer & explanation

Answer: $11.0$

Answer

The pH of the resulting solution is

11.0

.

The solution contains

0.04\text{ g}

of

\text{NaOH}

in

1.0\text{ dm}^3

, which corresponds to

0.001\text{ mol dm}^{-3}

or

1.0 \times 10^{-3}\text{ mol dm}^{-3}

. Since

\text{NaOH}

is a strong base,

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

, giving a

\text{pOH}

of

3.0

. Using the relationship

\text{pH} + \text{pOH} = 14

, the

\text{pH}

is

14.0 - 3.0 = 11.0

.

Step-by-Step Solution

1

Calculate the molar concentration of sodium hydroxide (

\text{NaOH}

).

\text{Molar concentration} = \frac{0.04\text{ g dm}^{-3}}{40\text{ g mol}^{-1}} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Concentration must be in

\text{mol dm}^{-3}

before calculating ion concentration and

\text{pOH}

.

2

Determine the hydroxide ion concentration

[\text{OH}^-]

.

[\text{OH}^-] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

\text{NaOH}

is a strong monobasic base that completely dissociates into

\text{Na}^+

and

\text{OH}^-

ions.

3

Calculate the

\text{pOH}

of the solution.

\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

\text{pOH}

is defined as the negative base-10 logarithm of the hydroxide ion concentration.

4

Convert

\text{pOH}

to

\text{pH}

.

\text{pH} = 14.0 - \text{pOH} = 14.0 - 3.0 = 11.0

.

At

25^\circ\text{C}

,

\text{pH} + \text{pOH} = 14.0

.

Key Concept

Calculating pH from the mass concentration of a strong base via pOH conversion

Estimated Time:1m 30s

Question 33Question →

A chemist needs to prepare a pure, hydrated sample of copper(II) tetraoxosulfate(VI) ( $CuSO_4 \cdot 5H_2O$ ) starting from insoluble copper(II) oxide ( $CuO$ ). Which of the following experimental procedures is correct for this preparation?

Add excess copper(II) oxide to dilute tetraoxosulfate(VI) acid, filter the mixture, concentrate the filtrate, and allow it to cool for crystallization.

Mix aqueous copper(II) nitrate with aqueous sodium tetraoxosulfate(VI), filter off the precipitate formed, and dry it in an oven.

Titrate copper(II) oxide solution against dilute tetraoxosulfate(VI) acid using phenolphthalein, then evaporate the liquid to complete dryness.

Dissolve copper(II) oxide in hot distilled water, add sodium chloride solution, and collect the resulting salt crystals.

Show answer & explanation

Answer: Add excess copper(II) oxide to dilute tetraoxosulfate(VI) acid, filter the mixture, concentrate the filtrate, and allow it to cool for crystallization.

Answer

The correct procedure is to react excess insoluble copper(II) oxide with dilute tetraoxosulfate(VI) acid, filter out the unreacted solid, concentrate the filtrate by gentle heating, and cool to form crystals.

The standard method for preparing a soluble salt from an insoluble oxide involves adding excess oxide to hot dilute acid to ensure complete acid neutralization, filtering out the unreacted solid, and partially evaporating the filtrate so that hydrated crystals form upon cooling.

Step-by-Step Solution

1

Identify the solubility of the product salt and reactants

Copper(II) tetraoxosulfate(VI) (

CuSO_4

) is a soluble salt, while copper(II) oxide (

CuO

) is an insoluble base.

Soluble salts prepared from insoluble bases require the excess insoluble reactant method.

2

Neutralize dilute acid with excess insoluble base

Warm

H_2SO_4(aq)

reacts with excess

CuO(s)

according to:

CuO(s) + H_2SO_4(aq) \rightarrow CuSO_4(aq) + H_2O(l)

.

Using excess base ensures all acid is fully consumed.

3

Separate unreacted solid and crystallize

Filter off unreacted

CuO

, heat the filtrate to crystallizing point, and cool slowy to obtain hydrated crystals (

CuSO_4 \cdot 5H_2O

).

Filtration removes insoluble excess, and gentle cooling preserves water of crystallization.

Key Concept

Preparation of soluble salts using dilute acids and insoluble bases

Estimated Time:1m 30s

Question 34Question →

Complete the following statement regarding laboratory methods of salt preparation and salt classification.

Fill in the blanks below

The preparation of anhydrous iron(III) chloride (

FeCl_3

) by reacting red-hot iron filings directly with dry chlorine gas is an example of , whereas potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, is classified as a salt.

Show answer & explanation

Answer

The preparation method is direct combination (or synthesis), and potash alum is classified as a double salt.

Anhydrous binary salts like iron(III) chloride must be synthesized by direct combination of their elements in dry conditions to avoid hydrolysis. Potash alum is formed by equimolar crystallization of potassium sulfate and aluminium sulfate, yielding a double salt.

Step-by-Step Solution

1

Determine the preparation method for anhydrous iron(III) chloride.

Direct combination of elements (

2Fe_{(s)} + 3Cl_{2(g)} \rightarrow 2FeCl_{3(s)}

) is used because aqueous crystallization causes salt hydrolysis.

Reacting two elements directly to form a compound without water is direct combination/synthesis.

2

Classify potash alum based on its chemical formula and ionic behavior.

Potash alum contains two distinct metallic cations (

K^+

and

Al^{3+}

) combined with sulfate anions (

SO_4^{2-}

), making it a double salt.

Salts composed of two simple salts crystallized together in equimolar proportions that dissociate completely into constituent ions are double salts.

Key Concept

Salt Preparation Methods and Classifications

Question 35Question →

The solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at $40^\circ\text{C}$ is $6.0\text{ mol dm}^{-3}$ . What mass of $\text{KNO}_3$ , in grams, is required to prepare a saturated solution in $250\text{ cm}^3$ of water at this temperature? (Molar mass of $\text{KNO}_3 = 101\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 151.5

Answer

The mass of

\text{KNO}_3

required to prepare a saturated solution in

250\text{ cm}^3

of water at

40^\circ\text{C}

is

151.5\text{ g}

.

To find the mass of solute required for saturation, convert the given volume of water to cubic decimeters (

250\text{ cm}^3 = 0.25\text{ dm}^3

). Multiply the volume by the molar solubility (

6.0\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 1.5\text{ mol}

) to obtain the number of moles, then multiply by the molar mass (

1.5\text{ mol} \times 101\text{ g mol}^{-1} = 151.5\text{ g}

).

Step-by-Step Solution

1

Convert volume from

\text{cm}^3

to

\text{dm}^3

0.25\text{ dm}^3

Molar solubility is expressed per

\text{dm}^3

, so the volume of solvent must be in

\text{dm}^3

.

2

Determine moles of

\text{KNO}_3

needed for saturation

1.5\text{ mol}

Multiply molar solubility by the volume in

\text{dm}^3

.

3

Convert moles to mass in grams

151.5\text{ g}

Multiply moles by the molar mass of

\text{KNO}_3

(

101\text{ g mol}^{-1}

).

Key Concept

Calculating solute mass for saturation using molar solubility and volume

Question 36Question →

In the laboratory preparation of zinc tetraoxosulfate(VI) crystals ( $ZnSO_4 \cdot 7H_2O$ ) from dilute tetraoxosulfate(VI) acid ( $H_2SO_4$ ) and zinc metal, why is excess zinc metal added to the acid?

To ensure all the acid is completely reacted so that the salt solution is not contaminated with acid

To prevent the zinc tetraoxosulfate(VI) salt from undergoing hydrolysis in the aqueous solution

To act as a heterogeneous catalyst to increase the rate of crystallization

To convert the resulting normal salt into a basic salt by excess metal accumulation

Show answer & explanation

Answer: To ensure all the acid is completely reacted so that the salt solution is not contaminated with acid

Answer

Excess zinc metal is added to ensure all the acid is completely reacted so that the salt solution is not contaminated with acid.

When preparing a soluble salt from an acid and an insoluble solid (such as a metal, oxide, or carbonate), the solid is added in excess to ensure that every molecule of acid reacts completely. Because both the acid and the formed salt are soluble in water, any unreacted acid would remain mixed with the salt during crystallization and contaminate the product. The excess solid metal is easily removed by filtration before evaporating the filtrate.

Step-by-Step Solution

1

Identify the type of reaction and salt being prepared

Zinc tetraoxosulfate(VI) (

ZnSO_4

) is a soluble salt prepared by reacting a dilute acid (

H_2SO_4

) with a moderately reactive insoluble metal (

Zn

).

Soluble salts of reactive metals are prepared by acid-metal reactions.

2

Analyze the separation and purification step

If acid remains in excess, it cannot be separated from the soluble salt solution by filtration or simple evaporation.

Both the salt and acid are soluble in water, so unreacted acid would contaminate the final crystalline product.

3

Determine the role of adding excess insoluble solid reactant

Adding excess zinc metal guarantees 100% consumption of the acid. The excess insoluble zinc can easily be filtered off, leaving a pure aqueous salt solution.

Insoluble solids are easy to separate from solutions via filtration.

Key Concept

Method of preparing soluble salts by reacting dilute acid with an excess of an insoluble metal, base, or trioxocarbonate(IV).

Estimated Time:1m 0s

Question 37Question →

Match each aqueous salt solution ( $0.1\text{ mol dm}^{-3}$ ) to its corresponding pH nature resulting from salt hydrolysis.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium ethanoate matches with Alkaline solution (pH > 7) due to anion hydrolysis; Ammonium chloride matches with Acidic solution (pH < 7) due to cation hydrolysis; Sodium chloride matches with Neutral solution (pH = 7) with negligible hydrolysis; Ammonium ethanoate matches with Neutral solution (pH ≈ 7) due to equal mutual hydrolysis of both ions.

Potassium ethanoate (

CH_3COOK

) undergoes anion hydrolysis to yield hydroxide ions, producing an alkaline solution (

pH > 7

). Ammonium chloride (

NH_4Cl

) undergoes cation hydrolysis to yield hydroxonium ions, producing an acidic solution (

pH < 7

). Sodium chloride (

NaCl

) contains spectator ions from a strong acid and strong base, leading to negligible hydrolysis and a neutral solution (

pH = 7

). Ammonium ethanoate (

CH_3COONH_4

) undergoes mutual hydrolysis of both cation and anion to equal extents, resulting in a neutral solution (

pH \approx 7

).

Step-by-Step Solution

1

Classify each salt based on the strengths of its parent acid and parent base.

Potassium ethanoate comes from a weak acid (

CH_3COOH

) and strong base (

KOH

). Ammonium chloride comes from a weak base (

NH_3

) and strong acid (

HCl

). Sodium chloride comes from a strong acid (

HCl

) and strong base (

NaOH

). Ammonium ethanoate comes from a weak acid (

CH_3COOH

) and weak base (

NH_3

).

Only ions originating from weak electrolytes undergo significant reaction with water (hydrolysis).

2

Determine the hydrolyzing species and the resulting ion produced in water.

For

CH_3COOK

,

CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-

(alkaline). For

NH_4Cl

,

NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+

(acidic). For

NaCl

, no hydrolysis occurs. For

CH_3COONH_4

, both ions hydrolyze equally.

Anion hydrolysis increases

OH^-

concentration, whereas cation hydrolysis increases

H_3O^+

concentration.

3

Correlate each salt to its solution pH characteristics.

CH_3COOK \rightarrow pH > 7

,

NH_4Cl \rightarrow pH < 7

,

NaCl \rightarrow pH = 7

, and

CH_3COONH_4 \rightarrow pH \approx 7

.

The nature of the solution depends on which ion hydrolyzes or whether both hydrolyze to equal extents.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Question 38Question →

Match each hydration phenomenon or property on the left with its corresponding vapor pressure behavior or physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Deliquescence matches with the condition where the saturated solution's vapor pressure is lower than ambient water vapor pressure, causing complete dissolution; Efflorescence matches with the condition where the crystal's vapor pressure exceeds ambient water vapor pressure, causing loss of hydration water; Hygroscopy matches with the absorption of moisture without dissolving; Water of Crystallization matches with the definite stoichiometric water molecules bound in the crystal lattice.

Each phenomenon is correctly paired based on its vapor pressure relationship and physical characteristics: Deliquescence requires the vapor pressure of the saturated solution to be lower than ambient water vapor pressure, causing liquefaction. Efflorescence requires the crystal's vapor pressure to exceed ambient water vapor pressure, leading to water loss. Hygroscopy involves moisture absorption without dissolving, and water of crystallization represents the fixed stoichiometric water bound in the crystal lattice.

Step-by-Step Solution

1

Analyze the thermodynamic driving force for deliquescence.

A substance deliquesces when it absorbs water from the air until it dissolves. This occurs because the saturated solution formed has a vapor pressure lower than the partial pressure of water vapor in the atmosphere.

Water spontaneously condenses into a system with lower vapor pressure.

2

Analyze the thermodynamic driving force for efflorescence.

A hydrated salt effloresces when its internal hydration vapor pressure is greater than the atmospheric water vapor pressure, causing water molecules to escape into the atmosphere.

Water leaves the crystal lattice when the internal vapor pressure is higher than ambient humidity.

3

Distinguish hygroscopy from deliquescence.

Hygroscopic substances absorb moisture from the atmosphere, but unlike deliquescent substances, they do not dissolve to form liquid solutions.

Hygroscopic materials absorb moisture into their bulk or surface without forming a solution.

4

Define water of crystallization.

Water of crystallization refers to water molecules chemically trapped in a fixed mole ratio inside the crystal lattice of salts.

It determines the specific hydrated chemical formula of the salt.

Key Concept

Atmospheric behavior of hydrated and anhydrous salts based on vapor pressure relationships.

Question 39Question →

The solubility of a sparingly soluble salt $MX_2$ (molar mass $= 200\text{ g mol}^{-1}$ ) in water at $25^\circ\text{C}$ is $0.20\text{ g dm}^{-3}$ . What is the solubility product ( $K_{sp}$ ) of $MX_2$ at this temperature?

1.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

2.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

3.2 \times 10^{-2}\text{ mol}^3\text{ dm}^{-9}

Show answer & explanation

Answer: $4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}$

Answer

The solubility product (

K_{sp}

) of

MX_2

at

25^\circ\text{C}

is

4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

.

To find the solubility product (

K_{sp}

), first convert the solubility from

\text{g dm}^{-3}

to molar solubility (

s

) in

\text{mol dm}^{-3}

by dividing by the molar mass:

s = \frac{0.20}{200} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. The dissociation equation

MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)

yields

[M^{2+}] = s

and

[X^-] = 2s

. Substituting these into the solubility product expression gives

K_{sp} = [M^{2+}][X^-]^2 = 4s^3 = 4 \times (1.0 \times 10^{-3})^3 = 4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

.

Step-by-Step Solution

1

Convert solubility from

\text{g dm}^{-3}

to molar solubility (

s

) in

\text{mol dm}^{-3}

s = \frac{\text{Solubility in g dm}^{-3}}{\text{Molar Mass}} = \frac{0.20\text{ g dm}^{-3}}{200\text{ g mol}^{-1}} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

Solubility product calculations require concentration units in

\text{mol dm}^{-3}

.

2

Write the ionic dissociation equation and express equilibrium concentrations

MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^-(aq)

[M^{2+}] = s = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

[X^-] = 2s = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

Each mole of

MX_2

yields 1 mole of

M^{2+}

and 2 moles of

X^-

upon dissolution.

3

Write the

K_{sp}

expression and calculate the numerical value

K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3 = 4 \times (1.0 \times 10^{-3})^3 = 4.0 \times 10^{-9}\text{ mol}^3\text{ dm}^{-9}

Substitute the molar equilibrium concentrations into the equilibrium constant expression.

Key Concept

Relationship between Molar Solubility and Solubility Product (

K_{sp}

)

Question 40Question →

In an acid-base titration, $25.0\text{ cm}^3$ of a $0.10\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution required $12.5\text{ cm}^3$ of a tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution for complete neutralization. What is the concentration of the acid solution in $\text{g dm}^{-3}$ ? $[\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]$

9.80\text{ g dm}^{-3}

19.60\text{ g dm}^{-3}

4.90\text{ g dm}^{-3}

0.10\text{ g dm}^{-3}

Show answer & explanation

Answer: $9.80\text{ g dm}^{-3}$

Answer

The concentration of the tetraoxosulfate(VI) acid solution is

9.80\text{ g dm}^{-3}

.

The balanced chemical equation

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

establishes a mole ratio (

n_a : n_b

) of

1 : 2

. Substituting the given values into

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

gives

C_a = 0.10\text{ mol dm}^{-3}

. Multiplying this by the molar mass of

\text{H}_2\text{SO}_4

(

98.0\text{ g mol}^{-1}

) yields

9.80\text{ g dm}^{-3}

.

Step-by-Step Solution

1

Write the balanced chemical equation to determine the mole ratio.

\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

, giving

n_a = 1

and

n_b = 2

.

Stoichiometric coefficients define the mole ratio required for complete neutralization.

2

Calculate the molar concentration of tetraoxosulfate(VI) acid (

C_a

) using the titration formula

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

.

\frac{C_a \times 12.5}{0.10 \times 25.0} = \frac{1}{2} \implies C_a \times 12.5 = 1.25 \implies C_a = 0.10\text{ mol dm}^{-3}

.

Relates the volumes and concentrations of acid and base according to their stoichiometry.

3

Convert the molar concentration to mass concentration in

\text{g dm}^{-3}

.

Molar mass of

\text{H}_2\text{SO}_4 = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-1}

. Mass concentration

= 0.10\text{ mol dm}^{-3} \times 98.0\text{ g mol}^{-1} = 9.80\text{ g dm}^{-3}

.

Mass concentration equals molar concentration multiplied by relative molar mass.

Key Concept

Volumetric calculations involving diprotic acid-monoprotic base titrations and conversion between molarity and mass concentration.

Acids, Bases and Salts

Answer

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