Acids, Bases and Salts Practice Questions — JAMB UTME — Page 3

Question 41Question →

During a volumetric analysis experiment, $20.0\text{ cm}^3$ of a $0.050\text{ mol dm}^{-3}$ sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3$ ) solution was completely neutralized by $25.0\text{ cm}^3$ of a tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution. What is the mass concentration of the tetraoxosulfate(VI) acid solution in $\text{g dm}^{-3}$ ? $[\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]$

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

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

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

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

Show answer & explanation

Answer: $3.92\text{ g dm}^{-3}$

Answer

The mass concentration of the tetraoxosulfate(VI) acid solution is

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

.

The balanced chemical reaction shows a

1:1

stoichiometric ratio between

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

and

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

. Substituting the given values into the titration equation yields a molar concentration of

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

. Multiplying this molarity by the molar mass of

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

(

98.0\text{ g mol}^{-3}

) correctly yields

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

.

Step-by-Step Solution

1

Write the balanced chemical equation for the neutralization reaction.

\text{Na}_2\text{CO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

Determines the mole ratio between the acid (

n_a = 1

) and base (

n_b = 1

).

2

Apply the volumetric neutralization formula to find the molar concentration of the acid (

C_a

).

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{C_a \times 25.0}{0.050 \times 20.0} = \frac{1}{1} \implies C_a = 0.040\text{ mol dm}^{-3}

Calculates the molarity of the tetraoxosulfate(VI) acid solution.

3

Calculate the molar mass of tetraoxosulfate(VI) acid (

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

).

\text{Molar Mass} = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-3}

Required to convert molar concentration to mass concentration.

4

Convert molar concentration to mass concentration.

\text{Mass concentration} = C_a \times \text{Molar Mass} = 0.040 \times 98.0 = 3.92\text{ g dm}^{-3}

Obtains the final concentration in grams per cubic decimetre.

Key Concept

Volumetric Analysis and Concentration Conversions

Question 42Question →

A solution of a monoprotic acid $HX$ contains $3.65\text{ g dm}^{-3}$ of the acid. If $25.0\text{ cm}^3$ of this acid solution neutralizes $20.0\text{ cm}^3$ of a $0.125\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution, what is the molar mass of the acid $HX$ in $\text{g mol}^{-3}$ ?

Show answer & explanation

Answer: 36.5

Answer

The molar mass of the acid

HX

is

36.5\text{ g mol}^{-3}

.

The reaction of monoprotic acid

HX

with

\text{NaOH}

follows a 1:1 stoichiometric ratio. Applying the titration formula

\frac{C_a V_a}{C_b V_b} = 1

yields an acid concentration of

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

. Dividing the mass concentration of

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

by this molarity gives a molar mass of

36.5\text{ g mol}^{-3}

.

Step-by-Step Solution

1

Determine the stoichiometric mole ratio between acid and base.

The mole ratio

n_a : n_b

for

HX

reacting with

\text{NaOH}

is

1 : 1

.

A monoprotic acid donates one proton per molecule to react with one mole of sodium hydroxide.

2

Calculate the molar concentration of the acid (

C_a

).

C_a = \frac{0.125\text{ mol dm}^{-3} \times 20.0\text{ cm}^3}{25.0\text{ cm}^3} = 0.100\text{ mol dm}^{-3}

.

Using the titration relation

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

allows determination of the acid molarity.

3

Calculate the molar mass of the acid.

\text{Molar mass} = \frac{\text{Mass concentration}}{\text{Molarity}} = \frac{3.65\text{ g dm}^{-3}}{0.100\text{ mol dm}^{-3}} = 36.5\text{ g mol}^{-3}

.

Molar mass is defined as mass of substance per mole.

Key Concept

Determination of molar mass using volumetric titration stoichiometry

Estimated Time:1m 30s

Question 43Question →

An aqueous solution of iron(III) chloride, $FeCl_3$ , exhibits an acidic pH because the $Fe^{3+}$ ion undergoes cation hydrolysis in water.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is correct because iron(III) chloride (

FeCl_3

) is a salt of a weak base (

Fe(OH)_3

) and a strong acid (

HCl

). The

Fe^{3+}

cation undergoes hydrolysis in water to liberate

H^+

ions, resulting in an acidic solution.

Step-by-Step Solution

1

Identify the parent acid and base of iron(III) chloride (

FeCl_3

).

Parent base is iron(III) hydroxide (

Fe(OH)_3

, weak) and parent acid is hydrochloric acid (

HCl

, strong).

Salts composed of weak base cations and strong acid anions produce acidic solutions upon hydrolysis.

2

Determine which ion undergoes salt hydrolysis.

Fe^{3+}

reacts with water molecules, while

Cl^-

remains an unreactive spectator ion.

Anions of strong acids do not hydrolyze, but cations of weak bases undergo cation hydrolysis to yield excess

H^+

ions.

3

Conclude the acid-base nature of the resulting solution.

The concentration of

H^+

increases, resulting in a solution pH less than 7.

Since cation hydrolysis releases

H^+

ions, the statement that

FeCl_3

forms an acidic solution is True.

Key Concept

Cation hydrolysis of salts formed from a weak base and a strong acid

Question 44Question →

A weak monobasic acid, $\text{HA}$ , has an acid dissociation constant ( $K_a$ ) of $4.5 \times 10^{-5}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the percentage ionization of a $0.05\text{ mol dm}^{-3}$ solution of this acid.

Show answer & explanation

Answer: 3

Answer

The percentage ionization of the weak monobasic acid solution is 3.0%.

According to Ostwald's dilution law for weak monobasic acids,

K_a = \alpha^2 C

. Rearranging to solve for the degree of ionization gives

\alpha = \sqrt{K_a / C} = \sqrt{(4.5 \times 10^{-5}) / 0.05} = \sqrt{9.0 \times 10^{-4}} = 0.03

. Expressed as a percentage,

0.03 \times 100\% = 3.0\%

.

Step-by-Step Solution

1

Relate acid dissociation constant (

K_a

), initial molar concentration (

C

), and degree of ionization (

\alpha

)

K_a = \alpha^2 C

For a weak monobasic acid undergoing partial ionization (

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

), Ostwald's dilution law simplifies to

K_a = \frac{\alpha^2 C}{1 - \alpha} \approx \alpha^2 C

because

\alpha \ll 1

.

2

Substitute the given values into the simplified expression and calculate

\alpha

\alpha = \sqrt{\frac{4.5 \times 10^{-5}}{0.05}} = \sqrt{9.0 \times 10^{-4}} = 0.03

Rearranging the equation yields

\alpha = \sqrt{K_a / C}

. Dividing the acid dissociation constant by the molar concentration gives

9.0 \times 10^{-4}

, and taking the square root yields

0.03

.

3

Convert the fractional degree of ionization to percentage ionization

3.0%

Multiplying the fractional degree of ionization by 100% gives the percentage of acid molecules ionized in solution.

Key Concept

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

Estimated Time:1m 30s

Question 45Question →

An aqueous solution of hydrocyanic acid ( $\text{HCN}$ ), a weak monobasic acid, has a concentration of $0.40\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Given that the acid dissociation constant ( $K_a$ ) for $\text{HCN}$ is $4.9 \times 10^{-10}\text{ mol dm}^{-3}$ , calculate the hydrogen ion concentration, $[\text{H}^+]$ , in $\text{mol dm}^{-3}$ . Express your answer as the coefficient $A$ in the form $A \times 10^{-5}\text{ mol dm}^{-3}$ .

Show answer & explanation

Answer: 1.4

Answer

The coefficient A is 1.4, which corresponds to a hydrogen ion concentration of

1.4 \times 10^{-5}\text{ mol dm}^{-3}

.

For a weak monobasic acid, the hydrogen ion concentration is determined using the weak acid ionization relationship

[\text{H}^+] = \sqrt{K_a \cdot c}

. Substituting

K_a = 4.9 \times 10^{-10}\text{ mol dm}^{-3}

and

c = 0.40\text{ mol dm}^{-3}

yields

[\text{H}^+] = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}

, giving a coefficient of 1.4.

Step-by-Step Solution

1

Write the ionization reaction and equilibrium constant expression

\text{HCN}_{(aq)} \rightleftharpoons \text{H}^+_{(aq)} + \text{CN}^-_{(aq)}

, giving

K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}

Hydrocyanic acid is a weak monobasic acid that ionizes partially in water.

2

Apply weak acid approximations

Since

[\text{H}^+] = [\text{CN}^-]

and

K_a

is extremely small,

[\text{HCN}] \approx c = 0.40\text{ mol dm}^{-3}

. Thus,

K_a = \frac{[\text{H}^+]^2}{c}

.

The negligible ionization degree allows the equilibrium concentration of un-ionized acid to be approximated as its initial concentration.

3

Substitute values and solve for

[\text{H}^+]

[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{4.9 \times 10^{-10} \times 0.40} = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}

Multiplying

K_a

by the molar concentration gives the square of the hydrogen ion concentration.

4

Determine the coefficient A

A = 1.4

Matching

1.4 \times 10^{-5}\text{ mol dm}^{-3}

to the requested standard scientific notation form

A \times 10^{-5}\text{ mol dm}^{-3}

yields

A = 1.4

.

Key Concept

Weak Acid Ionization Equilibrium and Ka Calculations

Question 46Question →

If $50.0\text{ cm}^3$ of a saturated solution of potassium chloride ( $\text{KCl}$ ) contains $14.9\text{ g}$ of the salt at $298\text{ K}$ , what is the solubility of potassium chloride at this temperature in $\text{mol dm}^{-3}$ ? (Molar mass of $\text{KCl} = 74.5\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 4

Answer

The solubility of potassium chloride at

298\text{ K}

is

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

.

First, the mass concentration is determined by scaling the

14.9\text{ g}

in

50.0\text{ cm}^3

to

1000\text{ cm}^3

, giving

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

. Dividing

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

by the molar mass of

\text{KCl}

(

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

) yields

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

.

Step-by-Step Solution

1

Calculate the mass of solute present per cubic decimetre (1000 cm³) of saturated solution.

Mass concentration = 298.0 g dm⁻³

Solubility is expressed relative to 1 dm³ of solution volume.

2

Divide the mass concentration in g dm⁻³ by the molar mass of KCl.

Solubility = 4.0 mol dm⁻³

Molar solubility equals mass concentration divided by molar mass (M).

Key Concept

Solubility Determination in Moles per Decimetre Cubed

Estimated Time:1m 30s

Question 47Question →

An aqueous solution of sodium fluoride ( $NaF$ ) is basic to litmus. Which species undergoes hydrolysis to cause this alkalinity?

Fluoride ion (

F^-

)

Sodium ion (

Na^+

)

Hydroxide ion (

OH^-

)

Hydrogen ion (

H^+

)

Show answer & explanation

Answer: Fluoride ion ( $F^-$ )

Answer

The fluoride ion (

F^-

) undergoes anion hydrolysis in water to produce hydroxide ions (

OH^-

), resulting in a basic solution.

Sodium fluoride (

NaF

) dissociates completely in water to yield

Na^+

and

F^-

. Because

HF

is a weak acid, its conjugate base (

F^-

) reacts with water molecules (anion hydrolysis) via

F^- + H_2O \rightleftharpoons HF + OH^-

, generating hydroxide ions which make the solution alkaline.

Step-by-Step Solution

1

Identify the parent acid and base of the salt.

NaF

is formed from a strong base (

NaOH

) and a weak acid (

HF

).

Salts of strong bases and weak acids produce basic solutions upon dissolving in water.

2

Determine which ion reacts with water (hydrolyzes).

The fluoride ion (

F^-

) reacts with water according to

F^- + H_2O \rightleftharpoons HF + OH^-

.

The conjugate base of a weak acid is strong enough to accept a proton from water.

3

Relate the generated ion to the acidity/alkalinity of the solution.

The production of extra

OH^-

ions increases the pH above 7, turning the solution basic.

An excess of hydroxide ions relative to hydronium ions causes basic/alkaline behavior.

Key Concept

Anion Hydrolysis of Weak Acid-Strong Base Salts

Question 48Question →

Complete the statement below regarding indicator selection and color transition during an acid-base titration.

Fill in the blanks below

In the volumetric titration of a strong acid against a weak base, is the most suitable indicator, and it changes color from to red at the acidic equivalence point when acid is added to the base.

Show answer & explanation

Answer

In the volumetric titration of a strong acid against a weak base, methyl orange is the most suitable indicator, and it changes color from yellow to red at the acidic equivalence point when acid is added to the base.

Titrations involving a strong acid and a weak base yield an acidic equivalence point (

\text{pH } 3 - 5

). Methyl orange is ideal because its transition range is

\text{pH } 3.1 - 4.4

. In the initial basic environment of the conical flask, methyl orange is yellow, and it shifts to red/pink as the acid endpoint is reached.

Step-by-Step Solution

1

Determine the pH of the equivalence point for the titration system.

Titrating a strong acid against a weak base produces a salt that undergoes hydrolysis, yielding an acidic equivalence point with

\text{pH} < 7

(typically between

\text{pH } 3

and

5

).

The conjugate acid of the weak base hydrolyzes in water to release hydrogen ions

\text{H}^+

.

2

Match the equivalence point pH range with the appropriate indicator.

Methyl orange has an effective indicator range of

\text{pH } 3.1\text{ to } 4.4

, which overlaps with the sharp pH drop of this titration curve.

An indicator is suitable only if its working range coincides with the steep vertical region of the titration curve.

3

Identify the initial and endpoint colors of methyl orange.

Before titration begins (in basic solution), methyl orange is yellow. At the endpoint after excess acid is introduced, it turns red.

Methyl orange is yellow in alkaline media (

\text{pH} > 4.4

) and turns pink/red in acidic media (

\text{pH} < 3.1

).

Key Concept

Indicator Selection and Endpoint Color Changes in Volumetric Analysis

Question 49Question →

Match each chemical species or system on the left with its correct theoretical acid-base role or behavior on the right according to the Arrhenius, Brønsted-Lowry, or Lewis concepts.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

BF_3

matches Lewis acid (electron-pair acceptor);

HSO_4^-

acting to produce

H_3O^+

matches Brønsted-Lowry acid (proton donor);

HCO_3^-

matches amphoteric/amphiprotic species; and

NH_2^-

from liquid ammonia matches conjugate base in a non-aqueous solvent system.

Each species is correctly categorized based on fundamental acid-base definitions: Lewis theory accounts for electron-pair transfer (

BF_3

), Brønsted-Lowry theory accounts for proton donor/acceptor roles (

HSO_4^-

and

HCO_3^-

), and solvent self-ionization describes non-aqueous acid-base equilibria (

NH_2^-

in liquid

NH_3

).

Step-by-Step Solution

1

Analyze

BF_3

in

BF_3 + :NH_3 \rightarrow F_3B:NH_3

Boron has six valence electrons and accepts an electron pair from nitrogen.

According to Lewis theory, an electron-pair acceptor is defined as a Lewis acid.

2

Analyze

HSO_4^-

converting to

SO_4^{2-}

in water

HSO_4^-

transfers a proton (

H^+

) to

H_2O

to form

H_3O^+

.

According to Brønsted-Lowry theory, a proton donor is an acid.

3

Analyze

HCO_3^-

double-behavior in water

HCO_3^-

can accept

H^+

to become

H_2CO_3

or donate

H^+

to become

CO_3^{2-}

.

Species that can act as either a proton donor or proton acceptor are termed amphiprotic/amphoteric.

4

Analyze

NH_2^-

in liquid ammonia auto-ionization

Ammonia undergoes auto-protolysis:

2NH_3 \rightleftharpoons NH_4^+ + NH_2^-

.

The amide ion (

NH_2^-

) is formed when

NH_3

loses a proton, acting as the characteristic conjugate base of the liquid ammonia solvent system.

Key Concept

Distinction and application of Arrhenius, Brønsted-Lowry, Lewis, and solvent-system theories of acids and bases.

Question 50Question →

Match each chemical reaction or behavior in Column A with the acid-base theory in Column B that uniquely or best explains it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Dissociation yielding H+ as the only positive ion matches Arrhenius Theory. 2. Proton transfer forming conjugate base SO42- matches Brønsted-Lowry Theory. 3. Electron-pair donation forming a dative bond with AlCl3 matches Lewis Theory.

Each statement matches its respective historical acid-base theory based on foundational criteria: Arrhenius requires aqueous H+ generation, Brønsted-Lowry centers on proton transfer and conjugate species, and Lewis broadens the scope to electron-pair donation and coordinate bond formation.

Step-by-Step Solution

1

Analyze the first item regarding HCl yielding H+ as the only positive ion in water.

Identified as Arrhenius Theory.

Arrhenius strictly defined acids by their ability to ionize in water to yield hydrogen ions as sole positive ions.

2

Analyze the second item involving HSO4- donating a proton to H2O to generate SO42-.

Identified as Brønsted-Lowry Theory.

Brønsted-Lowry focuses on proton transfer, where the donor is the acid and the resulting species is its conjugate base.

3

Analyze the third item involving NH3 donating a lone pair of electrons to AlCl3 in a non-protic coordinate covalent bond context.

Identified as Lewis Theory.

Lewis theory encompasses electron pair transfer, defining electron pair donors as bases and acceptors as acids.

Key Concept

Definitions and Theories of Acids and Bases (Arrhenius, Brønsted-Lowry, and Lewis)

Estimated Time:1m 0s

Question 51Question →

In liquid ammonia as a solvent, ammonium ions react with amide ions according to the following ionic equilibrium equation:

NH_4^+ + NH_2^- \rightleftharpoons 2NH_3

Based on the Brønsted-Lowry theory of acids and bases, which of the following statements correctly identifies the acid and its corresponding conjugate base in the forward reaction?

NH_4^+

is the acid, and

NH_3

is its conjugate base.

NH_4^+

is the acid, and

NH_2^-

is its conjugate base.

NH_2^-

is the acid, and

NH_3

is its conjugate base.

NH_3

is the acid, and

NH_4^+

is its conjugate base.

Show answer & explanation

Answer: $NH_4^+$ is the acid, and $NH_3$ is its conjugate base.

Answer

The species

NH_4^+

acts as the acid, and

NH_3

is its corresponding conjugate base.

According to the Brønsted-Lowry concept, an acid is a proton (

H^+

) donor, and a conjugate base is the species formed when an acid loses a proton. In the given reaction,

NH_4^+

donates a proton to

NH_2^-

, producing

NH_3

. Therefore,

NH_4^+

is the acid and

NH_3

is its conjugate base.

Step-by-Step Solution

1

Analyze the forward reaction and track proton transfer

In

NH_4^+ + NH_2^- \rightleftharpoons 2NH_3

,

NH_4^+

loses a proton (

H^+

) to form one molecule of

NH_3

.

The Brønsted-Lowry definition specifies that an acid is a proton (

H^+

) donor.

2

Identify the Brønsted-Lowry acid

NH_4^+

is the proton donor, so it is the Brønsted-Lowry acid.

It donates

H^+

to the amide ion (

NH_2^-

).

3

Determine the conjugate base formed from the acid

When

NH_4^+

loses a proton, it becomes

NH_3

.

A conjugate base is the species remaining after a Brønsted-Lowry acid donates a proton.

Key Concept

Brønsted-Lowry Acid-Base Theory and Conjugate Pairs

Estimated Time:1m 30s

Question 52Question →

A concentrated solution of ethanoic acid ( $CH_3COOH$ ) exhibits higher electrical conductivity than a dilute solution of hydrochloric acid ( $HCl$ ) because the higher concentration provides a greater total number of free mobile ions.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Dilute hydrochloric acid conducts electricity more effectively than concentrated ethanoic acid because hydrochloric acid is a strong acid that ionizes completely to yield a higher concentration of free mobile ions.

The statement is false because the electrical conductivity of an electrolyte solution depends strictly on the concentration of free mobile ions available to carry electric current. Hydrochloric acid (

HCl

) is a strong acid that ionizes completely in water, yielding a high density of

H^+

and

Cl^-

ions. Ethanoic acid (

CH_3COOH

) is a weak acid that ionizes only slightly in water, leaving most molecules unionized. Consequently, even a dilute solution of

HCl

has a much higher concentration of mobile ions and a higher electrical conductivity than a concentrated solution of ethanoic acid.

Step-by-Step Solution

1

Analyze the acid strength and degree of ionization of both compounds.

Hydrochloric acid (

HCl

) is a strong acid that undergoes complete ionization (

HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)

), whereas ethanoic acid (

CH_3COOH

) is a weak acid that undergoes partial ionization (

CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)

).

Electrical conductivity in aqueous solutions depends directly on the concentration of free, mobile charge-carrying ions.

2

Compare the concentration of mobile ions in concentrated weak acid vs. dilute strong acid.

Despite having a high molecular concentration, concentrated ethanoic acid contains mostly neutral, unionized molecules and very few free ions. Dilute

HCl

, even at lower molarity, yields a significantly higher concentration of free ions due to complete dissociation.

Only ionized species conduct electricity in solution; neutral molecules do not carry current.

3

Determine the relative electrical conductivity and evaluate the statement.

Dilute

HCl

is a far better conductor of electricity than concentrated

CH_3COOH

. Thus, the claim in the statement is false.

High acid concentration does not guarantee high conductivity if the acid is weak.

Key Concept

Electrical Conductivity and Ionization of Strong vs. Weak Acids

Question 53Question →

An acid salt is produced when the replaceable hydrogen ions of a polybasic acid are only partially replaced by a metallic or ammonium ion. Which of the following chemical equations represents the preparation of an acid salt?

NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O

2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O

NH_3 + HCl \rightarrow NH_4Cl

CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O

Show answer & explanation

Answer: $NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O$

Answer

The equation

NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O

represents the preparation of sodium hydrogensulfate (

NaHSO_4

), an acid salt.

The equation producing sodium hydrogensulfate (

NaHSO_4

) represents partial neutralization of dibasic tetraoxosulfate(VI) acid (

H_2SO_4

), leaving a replaceable hydrogen ion within the salt structure.

Step-by-Step Solution

1

Define an acid salt in terms of neutralization.

Acid salts are formed when polybasic acids undergo partial neutralization, retaining at least one replaceable hydrogen ion in the anion.

Monobasic acids (like

HCl

) cannot form acid salts, while dibasic acids (like

H_2SO_4

) can form both acid salts and normal salts depending on stoichiometry.

2

Analyze the stoichiometry and products of each reaction equation.

Reacting 1 mole of

NaOH

with 1 mole of

H_2SO_4

replaces only one of the two hydrogen atoms in

H_2SO_4

, giving

NaHSO_4

.

Because

NaHSO_4

contains ionizable hydrogen (

H^+

), it forms an acidic aqueous solution, satisfying the definition of an acid salt.

Key Concept

Classification and Synthesis of Acid Salts

Estimated Time:1m 0s

Question 54Question →

Two aqueous solutions, one of ethanoic acid ( $\text{CH}_3\text{COOH}$ ) and the other of hydrochloric acid ( $\text{HCl}$ ), have the exact same concentration of $0.10\text{ mol dm}^{-3}$ . Which of the following statements correctly accounts for why hydrochloric acid is classified as a stronger acid than ethanoic acid?

Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

Hydrochloric acid has a higher molar concentration of acid molecules in the solution than ethanoic acid.

Ethanoic acid undergoes complete dissociation in water while hydrochloric acid forms a reversible equilibrium.

Ethanoic acid produces a higher hydrogen ion concentration because it contains four hydrogen atoms per molecule.

Show answer & explanation

Answer: Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

Answer

Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

The strength of an acid is defined by its degree of ionization in aqueous solution. Hydrochloric acid (

\text{HCl}

) is a strong acid because it ionizes completely in water to yield hydrogen ions. In contrast, ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak acid that ionizes only partially, setting up a dynamic equilibrium mixture of molecules and ions.

Step-by-Step Solution

1

Define acid strength in terms of degree of ionization.

Acid strength is determined by the extent to which an acid dissociates into hydrogen ions (

\text{H}^+

) in water, independent of solution concentration.

Strong acids dissociate completely, whereas weak acids dissociate only partially.

2

Compare the ionization behavior of hydrochloric acid and ethanoic acid.

Hydrochloric acid (

\text{HCl}

) is a strong monobasic acid that ionizes virtually

100\%

in water. Ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak monobasic acid that ionizes only slightly, establishing a dynamic equilibrium.

Because both solutions have equal molar concentrations, the completely ionized strong acid yields a substantially greater concentration of

\text{H}^+

ions.

Key Concept

Relative Strength and Ionization of Acids and Bases

Question 55Question →

Fill in the missing numerical values in the statement below regarding the pH and pOH scale at $25^\circ\text{C}$ .

Fill in the blanks below

For an aqueous solution at

25^\circ\text{C}

, a hydrogen ion concentration of

1.0 \times 10^{-4}\text{ mol dm}^{-3}

corresponds to a pH of and a pOH of .

Show answer & explanation

Answer

The pH of the solution is 4 (or 4.0) and the pOH is 10 (or 10.0).

Taking the negative logarithm of

[H^+] = 1.0 \times 10^{-4}\text{ mol dm}^{-3}

gives a pH of 4. Since

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

at

25^\circ\text{C}

, subtracting 4 from 14 gives a pOH of 10.

Step-by-Step Solution

1

Calculate the pH from the given hydrogen ion concentration

[H^+]

.

\text{pH} = -\log_{10}(1.0 \times 10^{-4}) = 4

pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration.

2

Calculate the pOH using the relationship

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

.

\text{pOH} = 14 - 4 = 10

At

25^\circ\text{C}

, the sum of pH and pOH for an aqueous solution equals 14.

Key Concept

Logarithmic pH and pOH calculations from hydrogen ion concentration

Question 56Question →

What mass of calcium hydroxide, $\text{Ca(OH)}_2$ , must be dissolved in distilled water to make $250\text{ cm}^3$ of an aqueous solution with a pH of $12.00$ at $25^\circ\text{C}$ ? $[\text{Molar mass of Ca(OH)}_2 = 74.0\text{ g mol}^{-1}]$

0.0925\text{ g}

0.185\text{ g}

0.370\text{ g}

0.740\text{ g}

Show answer & explanation

Answer: $0.0925\text{ g}$

Answer

0.0925\text{ g}

At

25^\circ\text{C}

,

\text{pOH} = 14 - 12 = 2

, giving

[\text{OH}^-] = 10^{-2} = 0.01\text{ mol dm}^{-3}

. Because

\text{Ca(OH)}_2

produces two

\text{OH}^-

ions per formula unit upon dissociation, the molar concentration of

\text{Ca(OH)}_2

is

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

. In

250\text{ cm}^3

(

0.25\text{ dm}^3

), the number of moles is

0.005 \times 0.25 = 0.00125\text{ mol}

. Multiplying by the molar mass (

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

) gives

0.0925\text{ g}

.

Step-by-Step Solution

1

Calculate the pOH and hydroxide ion concentration

[\text{OH}^-]

from the given pH.

\text{pOH} = 14.00 - 12.00 = 2.00 \implies [\text{OH}^-] = 10^{-2.00} = 0.01\text{ mol dm}^{-3}

.

Water ion product relationship

\text{pH} + \text{pOH} = 14.00

at

25^\circ\text{C}

connects pH to hydroxide ion concentration.

2

Determine the molar concentration of

\text{Ca(OH)}_2

.

[\text{Ca(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{0.01\text{ mol dm}^{-3}}{2} = 0.005\text{ mol dm}^{-3}

.

Calcium hydroxide dissociates completely according to

\text{Ca(OH)}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq)

, yielding two moles of

\text{OH}^-

per mole of base.

3

Calculate the number of moles of

\text{Ca(OH)}_2

required for

250\text{ cm}^3

of solution.

\text{Moles} = 0.005\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 0.00125\text{ mol}.

Volume must be converted from

\text{cm}^3

to

\text{dm}^3

by dividing by 1000.

4

Calculate the required mass of

\text{Ca(OH)}_2

.

\text{Mass} = 0.00125\text{ mol} \times 74.0\text{ g mol}^{-1} = 0.0925\text{ g}.

Mass equals number of moles multiplied by molar mass.

Key Concept

Calculating required solute mass from pH for a dibasic base by converting pH to pOH, adjusting for stoichiometry, and scaling by volume and molar mass.

Question 57Question →

A solution is formed by mixing $300\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution with $700\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution at $25^\circ\text{C}$ . Assuming complete dissociation of both electrolytes, what is the pH of the resulting mixture?

Show answer & explanation

Answer: 12

Answer

The pH of the resulting mixture is 12.0.

The mixture contains excess hydroxide ions (0.010 mol in 1.0 dm³ solution), resulting in a pOH of 2.0. Subtracting this from 14.0 gives a pH of 12.0.

Step-by-Step Solution

1

Calculate the moles of hydrogen ions (H⁺) contributed by the acid solution.

n(\text{H}^+) = 0.060\text{ mol}

Tetraoxosulfate(VI) acid is diprotic (dibasic), releasing 2 moles of

\text{H}^+

per mole of acid:

0.300\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} \times 2 = 0.060\text{ mol}

.

2

Calculate the moles of hydroxide ions (OH⁻) contributed by the base solution.

n(\text{OH}^-) = 0.070\text{ mol}

Sodium hydroxide is a monobasic base:

0.700\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.070\text{ mol}

.

3

Determine the unneutralized excess ions and calculate their molar concentration in the total volume.

[\text{OH}^-] = 0.010\text{ mol dm}^{-3}

The neutralization reaction is

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

. Excess

\text{OH}^- = 0.070 - 0.060 = 0.010\text{ mol}

. Divided by the total mixture volume of

1.0\text{ dm}^3

,

[\text{OH}^-] = 0.010\text{ mol dm}^{-3}

.

4

Calculate pOH and convert it to pH using the water autoionization relation.

\text{pH} = 12.0

\text{pOH} = -\log_{10}(0.010) = 2.0

. Since

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

at

25^\circ\text{C}

,

\text{pH} = 14.0 - 2.0 = 12.0

.

Key Concept

Neutralization stoichiometry of diprotic acids and strong bases followed by pH determination from excess hydroxide concentration.

Question 58Question →

An aqueous solution at $25^\circ\text{C}$ has a hydrogen ion concentration, $[\text{H}^+]$ , of $1.0 \times 10^{-5}\text{ mol dm}^{-3}$ . What is the pOH of this solution?

5.0

9.0

14.0

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

Show answer & explanation

Answer: $9.0$

Answer

The pOH of the solution is

9.0

.

First, find the pH using

\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0

. Then, apply the relationship

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

at

25^\circ\text{C}

to calculate

\text{pOH} = 14.0 - 5.0 = 9.0

. Thus, the option specifying

9.0

is correct.

Step-by-Step Solution

1

Calculate the pH of the solution from the given hydrogen ion concentration

\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0

pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration.

2

Use the relationship between pH and pOH at

25^\circ\text{C}

to find pOH

\text{pOH} = 14.0 - \text{pH} = 14.0 - 5.0 = 9.0

For any aqueous solution at 25°C,

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

.

Key Concept

Relationship between pH, pOH, and ion concentrations in aqueous solutions at 25°C

Estimated Time:45s

Question 59Question →

A student dilutes $100\text{ cm}^3$ of a $0.05\text{ mol dm}^{-3}$ nitric acid ( $\text{HNO}_3$ ) solution with distilled water to achieve a total volume of $500\text{ cm}^3$ at $25^\circ\text{C}$ . What is the pOH of the resulting diluted solution?

2.0

12.0

1.3

12.7

Show answer & explanation

Answer: $12.0$

Answer

The pOH of the resulting diluted solution is

12.0

.

Diluting

100\text{ cm}^3

of

0.05\text{ mol dm}^{-3}\text{ HNO}_3

to

500\text{ cm}^3

decreases the hydrogen ion concentration to

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

. The pH of this solution is

-\log_{10}(0.01) = 2.0

. Since

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

at

25^\circ\text{C}

, the pOH is

14 - 2.0 = 12.0

.

Step-by-Step Solution

1

Calculate the hydrogen ion concentration of the diluted solution using the dilution formula

C_1 V_1 = C_2 V_2

.

C_2 = \frac{0.05\text{ mol dm}^{-3} \times 100\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Nitric acid is a strong monoprotic acid that fully dissociates in water, so

[\text{H}^+] = C_2

.

2

Determine the pH of the diluted solution.

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-2}) = 2.0

.

The negative logarithm of the hydrogen ion concentration defines the pH.

3

Calculate the pOH using the water ion product relationship at

25^\circ\text{C}

.

\text{pOH} = 14 - \text{pH} = 14 - 2.0 = 12.0

.

The sum of pH and pOH equals

14

at standard room temperature.

Key Concept

Dilution effect on hydrogen ion concentration and the relation

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

Estimated Time:1m 30s

Question 60Question →

At $25^\circ\text{C}$ , an aqueous solution of a weak monoacidic base has a concentration of $0.08\text{ mol dm}^{-3}$ and a degree of ionization ( $\alpha$ ) of $0.025$ ( $2.5\%$ ). What is the concentration of hydroxide ions, $[\text{OH}^-]$ , in the solution in $\text{mol dm}^{-3}$ ?

Show answer & explanation

Answer: 0.002

Answer

The concentration of hydroxide ions,

[\text{OH}^-]

, in the solution is

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

.

For a weak monoacidic base in aqueous solution, only a fraction (

\alpha

) of the base molecules ionize to form hydroxide ions. The concentration of hydroxide ions is given by

[\text{OH}^-] = C \alpha

. Substituting the concentration

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

and degree of ionization

0.025

yields

[\text{OH}^-] = 0.08 \times 0.025 = 0.002\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Identify the relationship between degree of ionization and hydroxide ion concentration for a weak monoacidic base

[\text{OH}^-] = C \cdot \alpha

A weak monoacidic base

\text{BOH}

ionizes partially according to

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

, so the concentration of produced hydroxide ions equals the initial concentration multiplied by the degree of ionization.

2

Substitute the given values into the equation

[\text{OH}^-] = 0.08\text{ mol dm}^{-3} \times 0.025

The initial concentration

C = 0.08\text{ mol dm}^{-3}

and the degree of ionization

\alpha = 2.5\% = 0.025

.

3

Perform the multiplication to find the final concentration

[\text{OH}^-] = 0.002\text{ mol dm}^{-3}

Multiplying

0.08

by

0.025

yields

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

(or

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

).

Key Concept

Ionization equilibrium of weak bases and calculation of hydroxide ion concentration

Estimated Time:1m 15s

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