Acids, Bases and Salts Practice Questions — JAMB UTME — Page 4

Question 61Question →

Equal volumes of $0.10\text{ mol dm}^{-3}$ solutions of methanoic acid ( $\text{HCOOH}$ ) and hydrochloric acid ( $\text{HCl}$ ) are prepared at $25^\circ\text{C}$ . Which of the following statements correctly accounts for the lower electrical conductivity observed in the methanoic acid solution?

Methanoic acid has a lower molar concentration of dissolved acid molecules in solution than hydrochloric acid.

Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Methanoic acid requires an indicator to undergo complete ionization in an aqueous environment.

Methanoic acid molecules neutralize hydronium ions rapidly to reform un-ionized acid molecules, reducing ionic mobility.

Show answer & explanation

Answer: Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Answer

Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Electrical conductivity in aqueous solutions depends on the concentration of free, mobile ions. Hydrochloric acid is a strong monobasic acid that dissociates completely in water to yield a high concentration of

\text{H}_3\text{O}^+

and

\text{Cl}^-

ions. In contrast, methanoic acid is a weak monobasic acid that ionizes only partially in aqueous solution, leaving most acid molecules un-ionized. This lower ion concentration directly accounts for the reduced electrical conductivity.

Step-by-Step Solution

1

Analyze acid strength and ionization behavior of both acids

Hydrochloric acid (

\text{HCl}

) is a strong acid that ionizes completely:

\text{HCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{H}_3\text{O}^+_{(aq)} + \text{Cl}^-_{(aq)}

. Methanoic acid (

\text{HCOOH}

) is a weak acid that ionizes partially:

\text{HCOOH}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{HCOO}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)}

.

Electrical conductivity in aqueous solutions depends on the concentration of mobile ions present.

2

Compare ion concentrations at equal molar concentration

For

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

\text{HCl}

,

[\text{H}_3\text{O}^+] = 0.10\text{ mol dm}^{-3}

. For

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

\text{HCOOH}

,

[\text{H}_3\text{O}^+] \ll 0.10\text{ mol dm}^{-3}

due to partial ionization.

Fewer ions per unit volume results in lower electrical current transport through the methanoic acid solution.

Key Concept

Relative strength of acids depends on the extent of ionization in water, where weak acids establish an equilibrium yielding fewer mobile ions than strong acids of equal concentration.

Question 62Question →

A sample of $0.62\text{ g}$ of sodium oxide ( $\text{Na}_2\text{O}$ ) is reacted completely with distilled water to prepare $200\text{ cm}^3$ of stock solution. If $50\text{ cm}^3$ of this stock solution is diluted with distilled water to a final volume of $500\text{ cm}^3$ at $25^\circ\text{C}$ , what is the pH of the resulting diluted solution? $[\text{Na} = 23, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 12

Answer

The pH of the diluted solution is 12.0.

Dissolving

0.62\text{ g}

(

0.01\text{ mol}

) of

\text{Na}_2\text{O}

produces

0.02\text{ mol}

of

\text{OH}^-

ions in

200\text{ cm}^3

, giving a stock concentration of

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

. Diluting

50\text{ cm}^3

of this stock solution to

500\text{ cm}^3

reduces the

[\text{OH}^-]

tenfold to

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

. Taking the negative logarithm gives

\text{pOH} = 2.0

, which corresponds to a

\text{pH}

of

14.0 - 2.0 = 12.0

.

Step-by-Step Solution

1

Calculate the molar mass of sodium oxide (

\text{Na}_2\text{O}

) and determine the amount of moles dissolved.

\text{Molar mass of Na}_2\text{O} = (2 \times 23) + 16 = 62\text{ g mol}^{-1}

. Moles of

\text{Na}_2\text{O} = \frac{0.62\text{ g}}{62\text{ g mol}^{-1}} = 0.01\text{ mol}

.

Converting mass to moles is required to apply chemical stoichiometry.

2

Determine the moles of hydroxide ions (

\text{OH}^-

) formed upon complete reaction with water.

The balanced equation is

\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \rightarrow 2\text{Na}^+ + 2\text{OH}^-

. Therefore,

0.01\text{ mol}

of

\text{Na}_2\text{O}

yields

0.02\text{ mol}

of

\text{OH}^-

.

Sodium oxide is a basic oxide that reacts with water in a 1:2 mole ratio to yield hydroxide ions.

3

Calculate the hydroxide ion concentration in the

200\text{ cm}^3

(

0.2\text{ dm}^3

) stock solution.

[\text{OH}^-]_{\text{stock}} = \frac{0.02\text{ mol}}{0.2\text{ dm}^3} = 0.1\text{ mol dm}^{-3}

.

Molarity is defined as moles of solute per cubic decimeter of solution.

4

Apply the dilution formula

C_1 V_1 = C_2 V_2

to find the hydroxide ion concentration after dilution.

[\text{OH}^-]_{\text{diluted}} = \frac{0.1\text{ mol dm}^{-3} \times 50\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Diluting

50\text{ cm}^3

to

500\text{ cm}^3

decreases the concentration by a factor of 10.

5

Calculate the pOH and subsequently the pH of the diluted solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-2}) = 2.0

. Using

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

,

\text{pH} = 14.0 - 2.0 = 12.0

.

The logarithmic scale defines

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

and at

25^\circ\text{C}

,

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

.

Key Concept

Stoichiometric reaction of basic oxides with water combined with dilution calculations to determine solution pH and pOH.

Question 63Question →

An aqueous solution of sodium chloride ( $NaCl$ ) undergoes salt hydrolysis, causing it to turn blue litmus paper red.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Sodium chloride (

NaCl

) is a neutral salt of a strong acid (

HCl

) and a strong base (

NaOH

); its ions do not undergo hydrolysis in aqueous solution.

The statement is false because sodium chloride (

NaCl

) originates from a strong acid (

HCl

) and a strong base (

NaOH

). Since neither cation (

Na^+

) nor anion (

Cl^-

) reacts hydrolytically with water, the aqueous solution maintains a neutral

\text{pH}

of 7 and has no effect on blue litmus paper.

Step-by-Step Solution

1

Identify the parent acid and base that form sodium chloride (

NaCl

).

The parent acid is hydrochloric acid (

HCl

, a strong acid) and the parent base is sodium hydroxide (

NaOH

, a strong base).

Knowing the relative strengths of the parent acid and base determines whether hydrolysis occurs.

2

Determine the hydrolysis tendency of the constituent ions (

Na^+

and

Cl^-

).

Neither

Na^+

nor

Cl^-

undergoes hydrolysis in aqueous solution.

Conjugate ions of strong acids and strong bases are extremely weak and do not react with water molecules.

3

Deduce the solution acidity/alkalinity and its effect on litmus paper.

The concentrations of

H^+

and

OH^-

remain equal, giving a neutral solution (

\text{pH} = 7

) that does not turn blue litmus paper red.

Without hydrolysis generating excess

H^+

ions, the solution cannot display acidic behavior.

Key Concept

Salts derived from strong acids and strong bases yield neutral aqueous solutions because their ions do not hydrolyze.

Estimated Time:45s

Question 64Question →

When transparent crystals of washing soda, $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ , are left exposed to dry laboratory air, they gradually lose water of crystallization to become a white powdery monohydrate, $\text{Na}_2\text{CO}_3 \cdot \text{H}_2\text{O}$ . Which phenomenon is demonstrated by this observation?

Deliquescence

Efflorescence

Hygroscopy

Hydrolysis

Show answer & explanation

Answer: Efflorescence

Answer

Efflorescence is demonstrated when washing soda crystals lose water of crystallization upon exposure to dry air.

Efflorescence is the phenomenon where a hydrated compound loses its water of crystallization spontaneously when exposed to air because its hydration vapor pressure exceeds atmospheric moisture vapor pressure.

Step-by-Step Solution

1

Analyze the observed physical change.

Hydrated washing soda crystals (decahydrate) lose 9 molecules of water of crystallization to form a monohydrate powder in dry air.

The vapor pressure of water in the hydrated crystal is greater than the partial pressure of water vapor in the surrounding air.

2

Match the observed behavior with the appropriate chemical term.

The spontaneous loss of water of crystallization to the atmosphere is defined as efflorescence.

Efflorescent salts lose moisture to dry air until an equilibrium is reached.

Key Concept

Efflorescence in hydrated salts

Question 65Question →

Complete the following statement regarding salt classification and laboratory preparation methods by filling in the missing terms.

Fill in the blanks below

Potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, is classified as a salt because it completely dissociates in water into all its constituent simple ions, whereas an insoluble salt like lead(II) sulfate (

PbSO_4

) is prepared in the laboratory using .

Show answer & explanation

Answer

The first blank is 'double' (or 'double salt') and the second blank is 'double decomposition' (or 'precipitation').

Potash alum dissociates into

K^+

,

Al^{3+}

, and

SO_4^{2-}

ions in solution, making it a double salt. Lead(II) sulfate (

PbSO_4

) is insoluble in water and must be prepared by double decomposition (precipitation) by combining two aqueous solutions containing the requisite ions.

Step-by-Step Solution

1

Identify the classification of potash alum

Potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, contains two different cations (

K^+

and

Al^{3+}

) and ionizes completely into simple ions in aqueous solution, which defines a double salt.

Double salts retain the chemical properties of their constituent individual salts when dissolved in water, unlike complex salts which form complex ions.

2

Determine the appropriate laboratory preparation method for lead(II) sulfate (

PbSO_4

)

Since

PbSO_4

is an insoluble salt, it is prepared by mixing solutions of two soluble salts containing

Pb^{2+}

and

SO_4^{2-}

ions respectively.

Insoluble salts are synthesized via precipitation (double decomposition) reactions.

Key Concept

Classification of salts (double salts vs complex salts) and preparation of insoluble salts via precipitation/double decomposition.

Estimated Time:1m 0s

Question 66Question →

An aqueous solution of aluminium chloride ( $AlCl_3$ ) is acidic because the chloride ion ( $Cl^-$ ) undergoes anion hydrolysis in water to produce hydrogen ions ( $H^+$ ).

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Chloride ions do not undergo salt hydrolysis; the acidity of an aluminium chloride solution arises from cation hydrolysis of the hydrated aluminium ion.

The statement is false because chloride ions are spectator ions in aqueous solution and do not hydrolyze. The acidic nature of an aluminium chloride solution is instead driven by cation hydrolysis, in which the high charge density of the hydrated aluminium ion causes it to act as a Brønsted-Lowry acid by donating protons to water.

Step-by-Step Solution

1

Analyze the dissolution of

AlCl_3

in aqueous medium.

AlCl_3(s) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq) + 3Cl^-(aq)

Aluminium chloride dissociates fully in water to yield hydrated aluminium cations and chloride anions.

2

Evaluate the hydrolysis potential of the chloride anion (

Cl^-

).

Cl^-(aq) + H_2O(l) \rightarrow \text{No Reaction}

Chloride is the conjugate base of the strong acid

HCl

; hence, it is a spectator ion with negligible basicity and cannot undergo hydrolysis.

3

Evaluate the hydrolysis reaction of the hexaaquaaluminium(III) cation (

[Al(H_2O)_6]^{3+}

).

[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)

The high charge density of

Al^{3+}

weakens O-H bonds in coordinated water molecules, enabling proton transfer to free water molecules and increasing

[H_3O^+]

.

Key Concept

Cation Hydrolysis of Polyvalent Metal Ions

Question 67Question →

When solid ammonium chloride ( $NH_4Cl$ ) is dissolved in water, the resulting solution exhibits an acidic pH. Which species undergoes hydrolysis to cause this acidity?

Ammonium ion (

NH_4^+

)

Chloride ion (

Cl^-

)

Both ammonium (

NH_4^+

) and chloride (

Cl^-

) ions

Un-ionized ammonium chloride molecule (

NH_4Cl

)

Show answer & explanation

Answer: Ammonium ion ( $NH_4^+$ )

Answer

The ammonium ion (

NH_4^+

) undergoes cation hydrolysis to form hydronium ions (

H_3O^+

), making the solution acidic.

Ammonium chloride (

NH_4Cl

) completely dissociates in water into

NH_4^+

and

Cl^-

. Since

NH_4^+

is the conjugate acid of the weak base

NH_3

, it reacts with water molecules (cation hydrolysis) to donate a proton, producing excess

H_3O^+

ions which lower the pH below 7.

Step-by-Step Solution

1

Identify the parent acid and base of the salt.

Ammonium chloride (

NH_4Cl

) is formed from the weak base ammonia (

NH_3

) and the strong acid hydrochloric acid (

HCl

).

Salts composed of a weak base and strong acid form acidic aqueous solutions due to cation hydrolysis.

2

Determine which ion hydrolyzes in aqueous solution.

The ammonium ion (

NH_4^+

) reacts with water:

NH_4^+_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + H_3O^+_{(aq)}

.

Conjugate acids of weak bases are strong enough to donate protons to water molecules, generating hydronium ions.

Key Concept

Salt Hydrolysis of Weak Base and Strong Acid Salts

Question 68Question →

A $14.3\text{ g}$ sample of hydrated sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , was heated strongly in a crucible to constant mass. After complete heating, the mass of the remaining anhydrous residue was found to be $5.3\text{ g}$ . Given the relative atomic masses $(\text{Na} = 23, \text{C} = 12, \text{O} = 16, \text{H} = 1)$ , what is the value of $x$ in the formula of the hydrated salt?

10

7

5

1

Show answer & explanation

Answer: $10$

Answer

The value of

x

is

10

, giving the formula

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Heating to constant mass completely removes the water of crystallization (

9.0\text{ g}

of

\text{H}_2\text{O}

, equal to

0.50\text{ mol}

). The remaining

5.3\text{ g}

of anhydrous

\text{Na}_2\text{CO}_3

equals

0.05\text{ mol}

. Dividing

0.50\text{ mol}

by

0.05\text{ mol}

yields a mole ratio of

10

, which means

x = 10

.

Step-by-Step Solution

1

Calculate the mass of water lost during heating

\text{Mass of H}_2\text{O} = 14.3\text{ g} - 5.3\text{ g} = 9.0\text{ g}

The loss in mass upon heating to constant mass corresponds directly to the driven-off water of crystallization.

2

Determine the molar masses of anhydrous

\text{Na}_2\text{CO}_3

and

\text{H}_2\text{O}

\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}

;

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

Molar masses are required to convert the given masses into mole quantities.

3

Calculate the number of moles of anhydrous salt and water

\text{Moles of Na}_2\text{CO}_3 = \frac{5.3\text{ g}}{106\text{ g/mol}} = 0.05\text{ mol}

;

\text{Moles of H}_2\text{O} = \frac{9.0\text{ g}}{18\text{ g/mol}} = 0.50\text{ mol}

Moles equal mass divided by molar mass.

4

Calculate the mole ratio to find

x

x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.50}{0.05} = 10

The coefficient

x

represents the stoichiometric ratio of moles of water of crystallization per mole of anhydrous salt.

Key Concept

Determination of Water of Crystallization in Hydrated Salts

Question 69Question →

An aqueous solution of ammonium ethanoate ( $CH_3COONH_4$ ) is approximately neutral with a pH close to 7 because neither the ammonium cation ( $NH_4^+$ ) nor the ethanoate anion ( $CH_3COO^-$ ) undergoes hydrolysis in water.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Both the cation (

NH_4^+

) and anion (

CH_3COO^-

) undergo hydrolysis in water, but because the acid dissociation constant of the parent weak acid (

K_a

) equals the base dissociation constant of the parent weak base (

K_b

), the concentrations of generated hydronium and hydroxide ions are equal, producing a neutral solution.

The statement incorrectly attributes the neutral pH of aqueous ammonium ethanoate to an absence of hydrolysis. In reality, both the conjugate acid (

NH_4^+

) and conjugate base (

CH_3COO^-

) hydrolyze simultaneously; neutrality arises solely because

K_a

of ethanoic acid equals

K_b

of ammonia.

Step-by-Step Solution

1

Identify the parent acid and parent base of ammonium ethanoate (

CH_3COONH_4

).

Ammonium ethanoate is derived from ethanoic acid (

CH_3COOH

), a weak acid, and aqueous ammonia (

NH_3

), a weak base.

Salts formed from weak acids and weak bases yield conjugate species that are strong enough to react with water.

2

Analyze the hydrolysis behavior of both ions in aqueous solution.

Cation hydrolysis:

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

. Anion hydrolysis:

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

.

Both

NH_4^+

(weak acid) and

CH_3COO^-

(weak base) hydrolyze simultaneously in water.

3

Determine the net effect of mutual hydrolysis on solution pH.

Since

K_a(CH_3COOH) = 1.8 \times 10^{-5}

and

K_b(NH_3) = 1.8 \times 10^{-5}

,

[H_3O^+] = [OH^-]

, leading to

\text{pH} \approx 7.0

.

Equal ionization constants produce balanced amounts of hydronium and hydroxide ions, resulting in a neutral solution despite significant hydrolysis of both ions.

Key Concept

Hydrolysis of salts derived from a weak acid and a weak base

Question 70Question →

When exposed to moist atmospheric air, certain solid compounds absorb sufficient water vapor to completely dissolve in it and form a liquid solution. Which of the following substances exhibits this phenomenon known as deliquescence?

Calcium chloride (

\text{CaCl}_2

)

Concentrated tetraoxosulfate(VI) acid (

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

)

Sodium trioxocarbonate(IV) decahydrate (

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

)

Copper(II) oxide (

\text{CuO}

)

Show answer & explanation

Answer: Calcium chloride ( $\text{CaCl}_2$ )

Answer

Calcium chloride (

\text{CaCl}_2

) is deliquescent because it absorbs atmospheric water vapor to form a liquid solution.

Calcium chloride (

\text{CaCl}_2

) is a deliquescent solid. When exposed to air with sufficient humidity, its surface vapor pressure is much lower than the partial pressure of water vapor in the atmosphere, causing it to absorb water continuously until it fully dissolves into an aqueous solution.

Step-by-Step Solution

1

Define deliquescence

Deliquescence is the process in which a solid substance absorbs moisture from the atmosphere until it completely dissolves in the absorbed water, forming a concentrated solution.

Understanding the precise chemical definition distinguishes deliquescent solids from hygroscopic liquids or efflorescent hydrated salts.

2

Evaluate the options based on their behavior in moist air

Solid calcium chloride (

\text{CaCl}_2

) absorbs water vapor and turns into a liquid solution, making it deliquescent.

Calcium chloride has a very low saturated solution vapor pressure compared to atmospheric water vapor pressure.

Key Concept

Atmospheric Phenomena of Salts: Deliquescence vs Hygroscopy vs Efflorescence

Question 71Question →

When potassium sulfite ( $K_2SO_3$ ) is dissolved in distilled water at $25^\circ\text{C}$ , the solution turns red litmus paper blue. Which of the following net ionic equations correctly represents the hydrolysis reaction responsible for the basic nature of this solution?

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

K^+(aq) + H_2O(l) \rightleftharpoons KOH(aq) + H^+(aq)

SO_3^{2-}(aq) + 2H_2O(l) \rightleftharpoons H_2SO_3(aq) + 2OH^-(aq)

2K^+(aq) + SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons 2KOH(aq) + SO_2(g)

Show answer & explanation

Answer: $SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)$

Answer

The net ionic equation representing the hydrolysis of potassium sulfite is

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

, which generates hydroxide ions and causes the solution to be alkaline.

Potassium sulfite (

K_2SO_3

) dissociates in water into

K^+

and

SO_3^{2-}

ions. Because

KOH

is a strong base, the

K^+

ion does not react with water. However, sulfurous acid (

H_2SO_3

) is a weak acid, meaning its conjugate anion (

SO_3^{2-}

) acts as a Bronsted-Lowry base. It hydrolyzes by removing a proton from water to form

HSO_3^-

and

OH^-

. The excess hydroxide ions render the solution alkaline, turning red litmus paper blue.

Step-by-Step Solution

1

Identify the parent acid and parent base of the salt

K_2SO_3

.

Potassium sulfite (

K_2SO_3

) is formed from a strong base (

KOH

) and a weak diprotic acid (

H_2SO_3

).

Determining the strength of parent species dictates which ion undergoes hydrolysis in aqueous solution.

2

Determine which constituent ion undergoes hydrolysis in water.

Potassium ions (

K^+

) do not hydrolyze because they are cations of a strong base. Sulfite ions (

SO_3^{2-}

), being conjugate bases of a weak acid, undergo anion hydrolysis.

Conjugate bases of weak acids are sufficiently strong to abstract protons from water molecules.

3

Write the net ionic proton-transfer equation for anion hydrolysis.

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

The sulfite anion abstracts one proton (

H^+

) from water, yielding hydrogen sulfite (

HSO_3^-

) and releasing a free hydroxide ion (

OH^-

), which increases solution pH above 7.

Key Concept

Salt Hydrolysis of Weak Acid-Strong Base Salts

Estimated Time:1m 30s

Question 72Question →

A $4.99\text{ g}$ sample of hydrated copper(II) tetraoxosulfate(VI), $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$ , is heated to constant mass, yielding $3.19\text{ g}$ of anhydrous $\text{CuSO}_4$ . Separately, a salt whose saturated solution vapor pressure is greater than the partial pressure of water vapor in the surrounding atmosphere loses its water of crystallization to become an anhydrous powder upon exposure to air. $[\text{Cu} = 63.5, \text{S} = 32, \text{O} = 16, \text{H} = 1]$ . What is the value of $x$ in the hydrated salt, and what term describes the atmospheric behavior of the second salt?

x = 5

, and the second salt undergoes efflorescence

x = 5

, and the second salt undergoes deliquescence

x = 7

, and the second salt undergoes efflorescence

x = 7

, and the second salt undergoes hygroscopy

Show answer & explanation

Answer: $x = 5$ , and the second salt undergoes efflorescence

Answer

The value of

x

is 5, and the atmospheric behavior described is efflorescence.

Calculating the moles of anhydrous

\text{CuSO}_4

(

3.19\text{ g} / 159.5\text{ g/mol} = 0.020\text{ mol}

) and water (

1.80\text{ g} / 18\text{ g/mol} = 0.100\text{ mol}

) yields a mole ratio

x = 0.100 / 0.020 = 5

. Additionally, when a hydrate's vapor pressure is higher than the atmospheric partial pressure of water vapor, it loses water of crystallization to the surrounding air, which is defined as efflorescence.

Step-by-Step Solution

1

Calculate the molar masses of anhydrous

\text{CuSO}_4

and

\text{H}_2\text{O}

.

\text{Molar mass of CuSO}_4 = 63.5 + 32 + (4 \times 16) = 159.5\text{ g/mol}

;

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

.

Molar masses are needed to convert the experimental masses into mole quantities.

2

Determine the mass and moles of anhydrous

\text{CuSO}_4

and water of crystallization lost.

\text{Mass of CuSO}_4 = 3.19\text{ g} \Rightarrow n(\text{CuSO}_4) = \frac{3.19}{159.5} = 0.020\text{ mol}

.

\text{Mass of H}_2\text{O} = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g} \Rightarrow n(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.100\text{ mol}

.

Subtracting the anhydrous mass from the total hydrated mass gives the mass of lost water.

3

Calculate the stoichiometric integer ratio

x

.

x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.100}{0.020} = 5

.

The value of

x

is the mole ratio of water to anhydrous salt.

4

Identify the atmospheric phenomenon based on vapor pressure conditions.

The phenomenon is efflorescence.

When the vapor pressure of a hydrated salt's water of crystallization exceeds the atmospheric vapor pressure, the salt spontaneously loses water to the atmosphere, becoming anhydrous or lower-hydrated powder.

Key Concept

Water of crystallization stoichiometry and vapor pressure behavior in efflorescence vs deliquescence

Question 73Question →

An aqueous solution of $0.01\text{ mol dm}^{-3}$ ethanoic acid ( $\text{CH}_3\text{COOH}$ ) exhibits a higher degree of ionization ( $\alpha$ ) than a $0.10\text{ mol dm}^{-3}$ ethanoic acid solution at the same temperature. Which of the following statements correctly explains this observation?

Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Diluting a weak acid increases its acid dissociation constant (

K_a

), causing it to ionize more completely.

Ethanoic acid converts into a strong acid upon dilution because its concentration decreases significantly.

Concentrated ethanoic acid solutions have a higher percentage ionization because more acid molecules are present per unit volume.

Show answer & explanation

Answer: Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Answer

Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Dilution increases the volume of the solvent relative to solute molecules, which shifts the position of the dynamic ionization equilibrium toward the side with a greater number of individual particles (ions) to maintain equilibrium, thereby increasing the degree of ionization.

Step-by-Step Solution

1

Write the ionization equilibrium expression for ethanoic acid

\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq})

Ethanoic acid is a weak monobasic acid that ionizes partially in aqueous medium.

2

Apply Ostwald's Dilution Law equation for weak electrolytes

\alpha \approx \sqrt{\frac{K_a}{C}}

, where

\alpha

is degree of ionization,

K_a

is acid dissociation constant, and

C

is molar concentration.

As concentration

C

decreases (dilution),

\alpha

increases, shifting equilibrium toward ion formation.

Key Concept

Ostwald's Dilution Law and Extent of Ionization of Weak Acids

Estimated Time:1m 0s

Question 74Question →

A $17.2\text{ g}$ sample of hydrated calcium tetraoxosulfate(VI), $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , is heated at $120^\circ\text{C}$ until it partially dehydrates, losing $2.70\text{ g}$ of water vapor to form plaster of Paris, $\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}$ . What is the value of $x$ , the number of molecules of water of crystallization per formula unit in the original hydrated salt? $[\text{Ca} = 40, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 2

Answer

The value of

x

in the hydrated salt formula is 2.

Applying mass conservation, the mass of plaster of Paris (

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

) remaining is

17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

, corresponding to

0.10\text{ mol}

. The mass of evolved water is

2.70\text{ g}

, which equals

0.15\text{ mol}

. The mole ratio of evolved water to salt formula units is

\frac{0.15}{0.10} = 1.5

. Since the reaction is

\text{CaSO}_4 \cdot x\text{H}_2\text{O} \rightarrow \text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} + (x - 0.5)\text{H}_2\text{O}

,

x - 0.5 = 1.5 \implies x = 2

.

Step-by-Step Solution

1

Calculate the molar masses of

\text{CaSO}_4

,

\text{H}_2\text{O}

, and the residue

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

.

Molar mass of

\text{CaSO}_4 = 40 + 32 + (4 \times 16) = 136\text{ g/mol}

;

\text{H}_2\text{O} = 18\text{ g/mol}

;

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = 136 + (0.5 \times 18) = 145\text{ g/mol}

.

Molar masses are necessary to perform mole calculations from mass measurements.

2

Determine the mass and amount in moles of plaster of Paris formed.

Mass of residue

= 17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

. Moles of

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = \frac{14.50\text{ g}}{145\text{ g/mol}} = 0.10\text{ mol}

.

Subtracting the mass of lost water gives the mass of solid product remaining.

3

Calculate the moles of water vapor driven off.

Moles of

\text{H}_2\text{O} = \frac{2.70\text{ g}}{18\text{ g/mol}} = 0.15\text{ mol}

.

Determining the quantity of lost water allows finding the mole ratio of lost water to salt units.

4

Relate the moles of lost water to the stoichiometry of partial dehydration to solve for

x

.

Moles of water lost per mole of salt

= \frac{0.15\text{ mol}}{0.10\text{ mol}} = 1.5\text{ mol}

. Since partial dehydration yields

(x - 0.5)

moles of lost water,

x - 0.5 = 1.5 \implies x = 2

.

Connecting empirical mole ratios to chemical formula coefficients yields the integer hydration number

x

.

Key Concept

Stoichiometric determination of water of crystallization from mass loss during partial dehydration.

Question 75Question →

Match each aqueous salt solution to its characteristic effect on litmus paper at $25^\circ\text{C}$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium chloride solution (

NH_4Cl(aq)

) matches with 'Turns blue litmus paper red (

pH < 7

)'; Sodium ethanoate solution (

CH_3COONa(aq)

) matches with 'Turns red litmus paper blue (

pH > 7

)'; Sodium chloride solution (

NaCl(aq)

) matches with 'Has no effect on either red or blue litmus paper (

pH = 7

)'.

Salt hydrolysis determines the acidity or alkalinity of an aqueous salt solution based on the strengths of the parent acid and base. Ammonium chloride (

NH_4Cl

) yields acidic solutions (

pH < 7

) turning blue litmus red due to

NH_4^+

cation hydrolysis. Sodium ethanoate (

CH_3COONa

) produces alkaline solutions (

pH > 7

) turning red litmus blue due to

CH_3COO^-

anion hydrolysis. Sodium chloride (

NaCl

) consists of spectator ions from a strong acid and strong base, undergoing no hydrolysis and remaining neutral (

pH = 7

).

Step-by-Step Solution

1

Analyze the parent acid and base for Ammonium chloride (

NH_4Cl

).

NH_4Cl

forms from

HCl

(strong acid) and

NH_3

(weak base). Cation hydrolysis occurs:

NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)

. Excess

H_3O^+

turns blue litmus red.

Salts of strong acids and weak bases yield acidic solutions.

2

Analyze the parent acid and base for Sodium ethanoate (

CH_3COONa

).

CH_3COONa

forms from

CH_3COOH

(weak acid) and

NaOH

(strong base). Anion hydrolysis occurs:

CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)

. Excess

OH^-

turns red litmus blue.

Salts of weak acids and strong bases yield alkaline solutions.

3

Analyze the parent acid and base for Sodium chloride (

NaCl

).

NaCl

forms from

HCl

(strong acid) and

NaOH

(strong base). Neither ion undergoes hydrolysis. The solution remains neutral (

pH = 7

) and does not change litmus color.

Salts of strong acids and strong bases do not undergo hydrolysis.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Estimated Time:45s

Question 76Question →

What is the pH of an aqueous solution prepared by dissolving $0.365\text{ g}$ of pure hydrogen chloride gas ( $\text{HCl}$ ) in distilled water to make a total solution volume of $10.0\text{ dm}^3$ at $25^\circ\text{C}$ ? [Molar mass: $\text{H} = 1.0\text{ g mol}^{-1}, \text{Cl} = 35.5\text{ g mol}^{-1}$ ]

1.0

2.0

3.0

11.0

Show answer & explanation

Answer: 3.0

Answer

The pH of the resulting solution is 3.0.

Dissolving

0.365\text{ g}

of

\text{HCl}

(molar mass

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

) yields

0.01\text{ mol}

of

\text{HCl}

. In a

10.0\text{ dm}^3

solution, the hydrogen ion concentration is

\frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. Taking the negative logarithm gives a pH of

3.0

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{HCl}

and find the number of moles dissolved.

\text{Molar mass of HCl} = 1.0 + 35.5 = 36.5\text{ g mol}^{-1}

. Moles of

\text{HCl} = \frac{0.365\text{ g}}{36.5\text{ g mol}^{-1}} = 0.01\text{ mol} = 1.0 \times 10^{-2}\text{ mol}

.

pH depends on the molar concentration of hydrogen ions, which requires knowing the total moles of solute.

2

Determine the molarity (

[\text{H}^+]

) of the solution.

[\text{H}^+] = \frac{\text{Moles of solute}}{\text{Volume in dm}^3} = \frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

\text{HCl}

is a strong monoprotic acid and ionizes completely in water to yield equal moles of

\text{H}^+

ions.

3

Calculate the pH using the pH definition formula.

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

The negative logarithm of the hydrogen ion concentration gives the pH value of the solution.

Key Concept

pH Calculation of a Strong Acid from Mass and Volume

Estimated Time:1m 30s

Question 77Question →

Match each chemical term or phenomenon on the left with its correct defining characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Deliquescence matches the spontaneous absorption of atmospheric moisture until dissolving into a solution; Efflorescence matches the loss of water vapor to dry air forming a powdery residue; Hygroscopy matches the absorption of water vapor without dissolving; Water of crystallization matches the definite ratio of water molecules chemically bound in a crystal lattice.

Deliquescence describes solids absorbing moisture until they dissolve into a solution. Efflorescence describes hydrated crystals spontaneously losing water of crystallization into dry air to become powdery. Hygroscopy describes substances absorbing water vapor without turning into a solution. Water of crystallization is the fixed stoichiometric quantity of water built into the salt crystal structure.

Step-by-Step Solution

1

Define deliquescence and match it with its defining process.

Deliquescence is matched with spontaneous absorption of moisture until the solid completely dissolves into a liquid solution.

Deliquescent solids absorb so much water from moist air that they form a liquid solution.

2

Define efflorescence and match it with its defining process.

Efflorescence is matched with the process in which a crystalline salt loses water vapor to dry air, forming a powdery residue.

This occurs because the vapor pressure of the hydrated crystal exceeds the ambient atmospheric water vapor pressure.

3

Define hygroscopy and distinguish it from deliquescence.

Hygroscopy is matched with the absorption of water vapor from the surrounding atmosphere without dissolving or forming a liquid solution.

Hygroscopic materials absorb water but remain in their original state without liquefying into a solution.

4

Define water of crystallization.

Water of crystallization is matched with the definite ratio of water molecules stoichiometrically locked inside a salt's crystal framework.

It represents the chemically bound water necessary for maintaining the specific crystal structure of hydrated salts.

Key Concept

Atmospheric Behavior and Hydration of Salts

Estimated Time:1m 15s

Question 78Question →

The solubility product ( $K_{sp}$ ) of lead(II) chloride ( $\text{PbCl}_2$ ) at $25^\circ\text{C}$ is $3.2 \times 10^{-5}\text{ mol}^3\text{ dm}^{-9}$ . What is the concentration of chloride ions ( $\text{Cl}^-$ ) in $\text{mol dm}^{-3}$ in a saturated solution of lead(II) chloride at $25^\circ\text{C}$ ?

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

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

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

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

Show answer & explanation

Answer: $4.0 \times 10^{-2}\text{ mol dm}^{-3}$

Answer

The concentration of chloride ions in the saturated solution is

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

.

For the dissolution equilibrium

\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)

, the solubility product expression is

K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2

. Letting

s

equal the molar solubility of

\text{PbCl}_2

,

[\text{Pb}^{2+}] = s

and

[\text{Cl}^-] = 2s

. Substituting into

K_{sp}

gives

K_{sp} = s(2s)^2 = 4s^3

. Given

K_{sp} = 3.2 \times 10^{-5}

, solving

4s^3 = 3.2 \times 10^{-5}

yields

s^3 = 8.0 \times 10^{-6}

and

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

. The chloride ion concentration is

[\text{Cl}^-] = 2s = 4.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Write the solubility equilibrium equation and express

K_{sp}

in terms of molar solubility (

s

).

\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)

, so

K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s(2s)^2 = 4s^3

.

Dissolution of one mole of

\text{PbCl}_2

produces one mole of

\text{Pb}^{2+}

ions and two moles of

\text{Cl}^-

ions.

2

Substitute the given

K_{sp}

value and calculate the molar solubility (

s

).

3.2 \times 10^{-5} = 4s^3 \implies s^3 = 8.0 \times 10^{-6} \implies s = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Dividing

K_{sp}

by 4 gives

s^3

, and taking the cube root yields

s

.

3

Determine the concentration of chloride ions (

[\text{Cl}^-]

).

[\text{Cl}^-] = 2s = 2 \times (2.0 \times 10^{-2}\text{ mol dm}^{-3}) = 4.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Since two moles of chloride ions are released per mole of salt dissolved,

[\text{Cl}^-]

equals

2s

.

Key Concept

Solubility product (

K_{sp}

) calculation for

MX_2

type salts and stoichiometric determination of ion concentrations.

Question 79Question →

Which of the following indicators is most suitable for detecting the end point in a volumetric titration of ethanoic acid ( $\text{CH}_3\text{COOH}$ ) against sodium hydroxide ( $\text{NaOH}$ ) solution?

Phenolphthalein

Methyl orange

Methyl red

Screened methyl orange

Show answer & explanation

Answer: Phenolphthalein

Answer

Phenolphthalein is the most suitable indicator because the equivalence point of a weak acid (ethanoic acid) titrated against a strong base (sodium hydroxide) occurs in the basic pH range.

In the titration of ethanoic acid (a weak acid) with sodium hydroxide (a strong base), the salt formed at neutralization undergoes hydrolysis to yield an alkaline solution with a pH greater than 7. Phenolphthalein is the correct choice because its color change interval (pH 8.3–10.0) coincides with this basic equivalence point.

Step-by-Step Solution

1

Identify the nature of the acid and base involved in the titration.

Ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak acid, while sodium hydroxide (

\text{NaOH}

) is a strong base.

The relative strengths of the acid and base determine the pH of the salt solution at the equivalence point.

2

Determine the expected pH at the equivalence point.

The resulting salt (sodium ethanoate) undergoes hydrolysis to produce hydroxide ions, giving an alkaline equivalence point with a pH between 8 and 9.

Anions of weak acids hydrolyze in water to form basic solutions.

3

Match the equivalence point pH with the transition range of an indicator.

Phenolphthalein has a pH transition range of 8.3–10.0.

A suitable indicator must undergo a sharp color change in the pH region corresponding to the equivalence point of the reaction.

Key Concept

Indicator Selection for Acid-Base Titrations

Question 80Question →

Match each salt listed in Column A with its appropriate laboratory preparation method from Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Lead(II) sulfate pairs with Precipitation (Double Decomposition); Sodium nitrate pairs with Titration of an acid with a soluble alkali; Copper(II) sulfate pairs with Action of dilute acid on an insoluble base; Anhydrous iron(III) chloride pairs with Direct combination of constituent elements.

Matching each salt to its preparation method requires analyzing solubility and chemical properties: insoluble salts like lead(II) sulfate are formed by precipitation/double decomposition; soluble sodium salts require neutralization by titration; soluble copper salts are synthesized using an insoluble oxide; and volatile anhydrous halides like iron(III) chloride are synthesized by direct combination of elements.

Step-by-Step Solution

1

Classify the salts by solubility in water.

Lead(II) sulfate is insoluble, while sodium nitrate, copper(II) sulfate, and iron(III) chloride are soluble.

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

2

Determine the preparation method for the insoluble salt.

Lead(II) sulfate is prepared by precipitation (double decomposition) combining soluble aqueous reactants such as lead(II) nitrate and sodium sulfate.

Precipitation is the standard method for preparing insoluble salts.

3

Determine preparation methods for the soluble salts based on reactant nature.

Sodium nitrate requires titration because both sodium hydroxide and sodium nitrate are soluble; Copper(II) sulfate uses an insoluble base (copper(II) oxide) reacted with dilute acid; Anhydrous iron(III) chloride requires direct combination to avoid hydrolysis by water.

Group 1/ammonium soluble salts require titration, while anhydrous iron(III) chloride cannot be prepared by evaporation from aqueous solution due to hydration and hydrolysis.

Key Concept

Laboratory Preparation Methods of Soluble and Insoluble Salts

Acids, Bases and Salts

Answer

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