Acids, Bases and Salts Practice Questions — JAMB UTME — Page 5

Question 81Question →

A student needs to prepare a pure, dry sample of barium tetraoxosulfate(VI), $BaSO_4$ , in the laboratory. Which of the following methods provides the most suitable procedure for obtaining this salt?

Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Titrating aqueous barium hydroxide with dilute tetraoxosulfate(VI) acid, evaporating the resulting solution to dryness, and crystallizing

Reacting solid barium carbonate with dilute tetraoxosulfate(VI) acid, filtering off the unreacted solid, and evaporating the filtrate to dryness

Heating barium metal directly in a stream of sulfur dioxide gas and collecting the crystalline residue

Show answer & explanation

Answer: Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Answer

Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Barium tetraoxosulfate(VI) (

BaSO_4

) is insoluble in water. The standard laboratory procedure for preparing insoluble salts is double decomposition (precipitation), where two aqueous solutions containing the constituent ions (in this case,

Ba(NO_3)_2(aq)

and

Na_2SO_4(aq)

) are mixed. The insoluble salt precipitates immediately, is collected by filtration, washed with distilled water to eliminate spectator ions, and dried.

Step-by-Step Solution

1

Determine the solubility of the target salt

Barium tetraoxosulfate(VI),

BaSO_4

, is insoluble in water according to general solubility rules.

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

2

Select the appropriate laboratory synthesis technique

Double decomposition (precipitation) between two soluble compounds containing the required cations and anions (

Ba^{2+}

and

SO_4^{2-}

) is required.

Insoluble salts are best prepared by combining solutions of two soluble salts to precipitate the insoluble product.

3

Identify the separation and purification steps

Filter the solid precipitate from the mixture, wash the residue with distilled water to remove soluble byproduct ions (

Na^+

and

NO_3^-

), and dry the solid.

Washing and drying ensure that a pure, dry sample of the insoluble salt is obtained.

Key Concept

Laboratory Preparation of Insoluble Salts by Double Decomposition (Precipitation)

Question 82Question →

Match each aqueous salt solution on the left to the corresponding hydrolyzing ion or hydrolysis behavior on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium chloride matches with

NH_4^+

cation hydrolysis producing

H_3O^+

; Sodium sulfide matches with

S^{2-}

anion hydrolysis producing

OH^-

; Iron(III) nitrate matches with

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

hydrolysis producing

H_3O^+

; Potassium sulfate matches with neither ion undergoing hydrolysis.

Matching each salt depends on identifying which ion hydrolyzes. Ammonium chloride contains the weak acid cation

NH_4^+

which yields hydronium ions; sodium sulfide contains the weak acid conjugate base

S^{2-}

which generates hydroxide ions; iron(III) nitrate contains the hydrated cation

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

which donates a proton to water; and potassium sulfate consists only of non-hydrolyzing spectator ions.

Step-by-Step Solution

1

Identify the parent acid and base for each salt to determine which ions undergo hydrolysis.

Salts derived from weak parents hydrolyze:

NH_4^+

comes from weak base

NH_3

,

S^{2-}

comes from weak acid

H_2S

,

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

is a weak acidic complex ion, whereas

K^+

and

SO_4^{2-}

come from strong parent species.

Only ions derived from weak acids or weak bases are strong enough conjugate species to react significantly with water.

2

Write the hydrolysis equilibrium equations for the reactive species.

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

,

S^{2-} + H_2O \rightleftharpoons HS^- + OH^-

, and

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

.

Cation hydrolysis increases the concentration of hydronium ions (

H_3O^+

), while anion hydrolysis increases the concentration of hydroxide ions (

OH^-

).

3

Pair each salt solution with its correct hydrolyzing species and resulting ionic effect.

Each salt is uniquely matched to its hydrolysis behavior.

Matches strictly conform to Brønsted-Lowry acid-base and salt hydrolysis principles.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Question 83Question →

A $24.4\text{ g}$ sample of hydrated barium chloride, $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ , was heated strongly in a crucible to constant mass. The residue of anhydrous barium chloride obtained weighed $20.8\text{ g}$ . What is the value of $x$ ? $[\text{Ba} = 137, \text{Cl} = 35.5, \text{H} = 1, \text{O} = 16]$

2

1

4

5

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2, giving the formula

\text{BaCl}_2 \cdot 2\text{H}_2\text{O}

.

Heating the hydrated salt removes all water of crystallization, leaving

20.8\text{ g}

of anhydrous

\text{BaCl}_2

(

0.10\text{ mol}

) and releasing

3.6\text{ g}

of water vapor (

0.20\text{ mol}

). The ratio of moles of water to moles of anhydrous salt is

\frac{0.20}{0.10} = 2

, so

x = 2

.

Step-by-Step Solution

1

Calculate the mass of water of crystallization lost during heating.

\text{Mass of H}_2\text{O} = 24.4\text{ g} - 20.8\text{ g} = 3.6\text{ g}

Heating drives off all water of crystallization, leaving only anhydrous salt.

2

Determine the molar masses of anhydrous

\text{BaCl}_2

and

\text{H}_2\text{O}

.

\text{Molar mass of BaCl}_2 = 137 + 2(35.5) = 208\text{ g/mol}

;

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

Molar masses are needed to convert mass quantities into mole quantities.

3

Calculate the number of moles of anhydrous

\text{BaCl}_2

and water.

\text{Moles of BaCl}_2 = \frac{20.8\text{ g}}{208\text{ g/mol}} = 0.10\text{ mol}

;

\text{Moles of H}_2\text{O} = \frac{3.6\text{ g}}{18\text{ g/mol}} = 0.20\text{ mol}

The stoichiometric coefficient

x

represents the mole ratio of water molecules per mole of salt.

4

Find the mole ratio

x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of BaCl}_2}

.

x = \frac{0.20\text{ mol}}{0.10\text{ mol}} = 2

Simplifying the mole ratio determines the integer coefficient

x

in the hydrated salt formula.

Key Concept

Determination of Water of Crystallization by Stoichiometric Gravimetric Heating

Question 84Question →

The solubility product ( $K_{sp}$ ) of calcium tetraoxosulfate(VI), $\text{CaSO}_4$ , is $2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}$ at $25^\circ\text{C}$ . What is the molar solubility of $\text{CaSO}_4$ in water at this temperature?

( $\text{Molar mass of CaSO}_4 = 136\text{ g mol}^{-1}$ )

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

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

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

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

Show answer & explanation

Answer: $5.0 \times 10^{-3}\text{ mol dm}^{-3}$

Answer

The molar solubility of calcium tetraoxosulfate(VI) is

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

.

For a 1:1 sparingly soluble salt like

\text{CaSO}_4

, dissociation yields equal molar concentrations of

\text{Ca}^{2+}

and

\text{SO}_4^{2-}

ions (

s

). The solubility product expression is

K_{sp} = s^2

. Taking the square root of

2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}

gives

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

.

Step-by-Step Solution

1

Write the solubility equilibrium equation and

K_{sp}

expression for

\text{CaSO}_4

.

\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)

, so

K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}]

.

Establishing the stoichiometric relationship between dissolved ions and molar solubility.

2

Substitute molar solubility

s

into the

K_{sp}

expression.

If

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

and

[\text{SO}_4^{2-}] = s

, then

K_{sp} = s \times s = s^2

.

Expressing

K_{sp}

in terms of a single variable

s

.

3

Calculate the value of

s

by taking the square root of

K_{sp}

.

s = \sqrt{2.5 \times 10^{-5}} = \sqrt{25 \times 10^{-6}} = 5.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Solving the algebraic equation for molar solubility.

Key Concept

Solubility product constant (

K_{sp}

) and molar solubility relation for AB-type binary ionic salts.

Question 85Question →

Match each salt in aqueous solution to the correct chemical description of its hydrolysis behavior and resulting pH at $25^\circ\text{C}$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium sulfate pairs with the acidic cation-hydrolyzing description; Sodium propanoate pairs with the alkaline anion-hydrolyzing description; Potassium nitrate pairs with the neutral non-hydrolyzing description; Ammonium cyanide pairs with the alkaline dual-hydrolyzing description where

K_b > K_a

.

The solution pH resulting from salt hydrolysis depends directly on the relative strengths of the parent acid and base. Salts derived from weak bases and strong acids yield acidic solutions via cation hydrolysis. Salts derived from strong bases and weak acids yield alkaline solutions via anion hydrolysis. Salts of strong acids and strong bases do not undergo net hydrolysis, remaining neutral. For salts derived from both a weak acid and a weak base, both ions undergo hydrolysis, and the solution acidity or alkalinity is determined by comparing the

K_b

of the weak base to the

K_a

of the weak acid.

Step-by-Step Solution

1

Identify the parent acid and parent base for each salt.

(NH_4)_2SO_4

comes from

NH_3

(weak base) and

H_2SO_4

(strong acid).

CH_3CH_2COONa

comes from

NaOH

(strong base) and

CH_3CH_2COOH

(weak acid).

KNO_3

comes from

KOH

(strong base) and

HNO_3

(strong acid).

NH_4CN

comes from

NH_3

(weak base) and

HCN

(weak acid).

The strength of parent acids and bases determines which ions undergo hydrolysis in water.

2

Determine which ion hydrolyzes for single-weak component salts.

In

(NH_4)_2SO_4

,

NH_4^+

hydrolyzes to produce

H_3O^+

(acidic,

pH < 7

). In

CH_3CH_2COONa

,

CH_3CH_2COO^-

hydrolyzes to produce

OH^-

(alkaline,

pH > 7

). In

KNO_3

, neither ion hydrolyzes (neutral,

pH = 7

).

Conjugate ions of weak species react with water, whereas conjugate ions of strong species do not hydrolyze.

3

Compare ionization constants for the weak acid-weak base salt.

For

NH_4CN

, compare

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

with

K_a(HCN) = 6.2 \times 10^{-10}

. Since

K_b > K_a

,

CN^-

anion hydrolysis produces more

OH^-

than

NH_4^+

cation hydrolysis produces

H_3O^+

, resulting in an alkaline solution (

pH > 7

).

When both ions hydrolyze, the relative magnitudes of

K_a

and

K_b

govern whether

OH^-

or

H_3O^+

is in excess.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Estimated Time:1m 30s

Question 86Question →

A standard solution is prepared by dissolving $4.0\text{ g}$ of pure solid sodium hydroxide ( $\text{NaOH}$ ) in distilled water to make $1.0\text{ dm}^3$ of solution. What is the molar concentration of this solution? $[\text{Na} = 23,\text{ O} = 16,\text{ H} = 1]$

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

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

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

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

Show answer & explanation

Answer: $0.10\text{ mol dm}^{-3}$

Answer

The molar concentration of the sodium hydroxide solution is

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

.

The solution containing

4.0\text{ g}

of

\text{NaOH}

per

\text{dm}^3

is converted to molar concentration by dividing by the molar mass of

\text{NaOH}

(

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

), yielding

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

.

Step-by-Step Solution

1

Calculate the molar mass of sodium hydroxide (

\text{NaOH}

).

Molar mass

= 23 + 16 + 1 = 40\text{ g mol}^{-1}

.

Molar mass is required to convert mass to moles.

2

Calculate the concentration in

\text{mol dm}^{-3}

by dividing mass concentration by molar mass.

\text{Molar concentration} = \frac{4.0\text{ g dm}^{-3}}{40\text{ g mol}^{-1}} = 0.10\text{ mol dm}^{-3}.

Molar concentration is defined as moles of solute per unit volume of solution in

\text{dm}^3

.

Key Concept

Conversion from mass concentration (g/dm³) to molar concentration (mol/dm³)

Estimated Time:45s

Question 87Question →

A $16.1\text{ g}$ sample of hydrated sodium tetraoxosulfate(VI), $\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}$ , is heated in a crucible until all the water of crystallization is driven off. The mass of the remaining anhydrous salt is $7.1\text{ g}$ . Calculate the value of $x$ . [Relative atomic masses: $\text{Na} = 23$ , $\text{S} = 32$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 10

Answer

The integer value of

x

is 10.

Heating

16.1\text{ g}

of hydrated sodium tetraoxosulfate(VI) yields

7.1\text{ g}

of anhydrous

\text{Na}_2\text{SO}_4

(

0.05\text{ mol}

) and releases

9.0\text{ g}

of water (

0.5\text{ mol}

). The mole ratio of

\text{H}_2\text{O}

to

\text{Na}_2\text{SO}_4

is

0.5 / 0.05 = 10

, giving

x = 10

.

Step-by-Step Solution

1

Calculate the mass of water lost upon heating

Mass of

\text{H}_2\text{O} = 16.1\text{ g} - 7.1\text{ g} = 9.0\text{ g}

The difference between the initial hydrated mass and final anhydrous mass represents the driven-off water of crystallization.

2

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{Na}_2\text{SO}_4 = 142\text{ g/mol}

, Molar mass of

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

Required to convert sample masses to mole quantities.

3

Calculate moles of anhydrous salt and water, and find the mole ratio

Moles of

\text{Na}_2\text{SO}_4 = 0.05\text{ mol}

, Moles of

\text{H}_2\text{O} = 0.5\text{ mol}

,

x = \frac{0.5}{0.05} = 10

The subscript

x

gives the ratio of moles of water to moles of anhydrous salt per mole of compound.

Key Concept

Stoichiometric determination of water of crystallization in hydrated salts

Question 88Question →

During a titration experiment, $25.0\text{ cm}^3$ of a potassium hydroxide ( $\text{KOH}$ ) solution of unknown concentration required $20.0\text{ cm}^3$ of a $0.050\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution for complete neutralization. What is the mass concentration of the potassium hydroxide solution in $\text{g dm}^{-3}$ ?

[ $\text{K} = 39, \text{O} = 16, \text{H} = 1$ ]

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

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

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

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

Show answer & explanation

Answer: $4.48\text{ g dm}^{-3}$

Answer

The mass concentration of the potassium hydroxide solution is

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

.

The reaction between tetraoxosulfate(VI) acid and potassium hydroxide has a

1:2

mole ratio (

\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}

). Substituting the given values into

\frac{C_a V_a}{C_b V_b} = \frac{1}{2}

gives

C_b = 0.080\text{ mol dm}^{-3}

. Multiplying this molarity by the molar mass of

\text{KOH}

(

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

) gives the mass concentration of

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

.

Step-by-Step Solution

1

Write the balanced chemical equation for the neutralization reaction.

\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}

The stoichiometry shows that

1\text{ mole}

of

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

reacts with

2\text{ moles}

of

\text{KOH}

(

n_a = 1, n_b = 2

).

2

Calculate the molarity (

C_b

) of the potassium hydroxide solution using the titration equation.

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{0.050 \times 20.0}{C_b \times 25.0} = \frac{1}{2} \implies C_b = 0.080\text{ mol dm}^{-3}

Equating the mole ratio allows determination of the concentration of the base in moles per cubic decimetre.

3

Calculate the molar mass of

\text{KOH}

and convert the concentration to

\text{g dm}^{-3}

.

Molar mass of

\text{KOH} = 39 + 16 + 1 = 56\text{ g mol}^{-1}

. Mass concentration

= 0.080\text{ mol dm}^{-3} \times 56\text{ g mol}^{-1} = 4.48\text{ g dm}^{-3}

.

Mass concentration is obtained by multiplying molar concentration by the relative molar mass.

Key Concept

Determination of mass concentration from volumetric analysis data using stoichiometric mole ratios.

Question 89Question →

A $1.50\text{ g}$ sample of an impure hydrated dibasic acid, $\text{H}_2\text{X}\cdot 2\text{H}_2\text{O}$ (molar mass of anhydrous $\text{H}_2\text{X} = 90.0\text{ g mol}^{-1}$ ), was dissolved in distilled water and made up to $250.0\text{ cm}^3$ of solution in a volumetric flask. A $25.0\text{ cm}^3$ portion of this acid solution required $20.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the percentage purity of the hydrated acid sample?

60.0\%

42.0\%

84.0\%

8.4\%

Show answer & explanation

Answer: $84.0\%$

Answer

The percentage purity of the hydrated acid sample is

84.0\%

.

The correct answer is

84.0\%

. Each mole of dibasic acid reacts with 2 moles of

\text{NaOH}

. The

20.0\text{ cm}^3

of

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

\text{NaOH}

contains

0.0020\text{ mol}

of base, neutralizing

0.0010\text{ mol}

of acid in

25.0\text{ cm}^3

. Scaling to the total

250.0\text{ cm}^3

gives

0.010\text{ mol}

of pure acid in the flask. Multiplying by the hydrated molar mass of

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

(

90.0 + 36.0

) yields

1.26\text{ g}

of pure acid, which corresponds to

(1.26 / 1.50) \times 100\% = 84.0\%

purity.

Step-by-Step Solution

1

Calculate molar mass of the hydrated acid and write balanced neutralization equation

Molar mass of

\text{H}_2\text{X}\cdot 2\text{H}_2\text{O} = 90.0 + 2(18.0) = 126.0\text{ g mol}^{-1}

. Reaction equation:

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

, so mole ratio

n_a : n_b = 1 : 2

.

The acid is dibasic, meaning each mole of acid reacts with two moles of sodium hydroxide, and the molar mass must include the water of crystallization.

2

Calculate the moles of base reacted and corresponding moles of acid in the

25.0\text{ cm}^3

aliquot

\text{Moles of NaOH} = \frac{20.0\text{ cm}^3}{1000} \times 0.100\text{ mol dm}^{-3} = 0.0020\text{ mol}

. Moles of acid in

25.0\text{ cm}^3 = \frac{0.0020}{2} = 0.0010\text{ mol}

.

Applying the stoichiometric ratio

n_a/n_b = 1/2

converts the moles of base used to moles of dibasic acid neutralized.

3

Scale the moles of pure acid to the total

250.0\text{ cm}^3

solution volume and determine pure mass

\text{Moles of pure acid in } 250.0\text{ cm}^3 = 0.0010\text{ mol} \times \left(\frac{250.0}{25.0}\right) = 0.010\text{ mol}

.

\text{Mass of pure hydrated acid} = 0.010\text{ mol} \times 126.0\text{ g mol}^{-1} = 1.26\text{ g}

.

The entire sample was dissolved to make 250 cm³, so multiplying by the dilution factor (10) gives the total moles of pure acid in the sample.

4

Calculate percentage purity of the sample

\text{Percentage purity} = \left(\frac{1.26\text{ g}}{1.50\text{ g}}\right) \times 100\% = 84.0\%

.

Percentage purity is the ratio of mass of pure substance to total mass of impure sample, expressed as a percentage.

Key Concept

Volumetric Analysis and Percentage Purity Calculation of a Hydrated Dibasic Acid

Question 90Question →

Complete the statement regarding indicator color changes during an acid-base titration by filling in the blanks.

Fill in the blanks below

When titrating ethanoic acid (

\text{CH}_3\text{COOH}

) with sodium hydroxide (

\text{NaOH}

) using phenolphthalein as the indicator, the solution in the conical flask changes from to at the end point.

Show answer & explanation

Answer

The solution in the conical flask changes from colorless (or colourless) to pink at the end point.

Ethanoic acid is a weak acid and sodium hydroxide is a strong base. The equivalence point occurs above pH 7 in the alkaline region. Phenolphthalein is the most suitable indicator for this titration because its color transition range is pH 8.3 to 10.0. Initially, the solution in the conical flask is acidic, rendering phenolphthalein colorless. At the end point, the solution turns pink as it transitions into the basic pH range of the indicator.

Step-by-Step Solution

1

Identify the nature of the titrant and analyte.

Ethanoic acid is a weak acid, and sodium hydroxide is a strong base.

The equivalence point for a weak acid-strong base titration occurs in the alkaline region (pH 8 to 10).

2

Determine the color states of phenolphthalein.

Phenolphthalein is colorless in acidic solutions (pH < 8.3) and turns pink in basic solutions (pH 8.3–10.0).

Before the end point, ethanoic acid predominates so the flask is colorless. At the end point, excess hydroxide ions cause the solution to turn pink.

Key Concept

Indicator Choice and End Point Colors in Acid-Base Titrations

Estimated Time:1m 0s

Question 91Question →

A saturated solution of potassium trioxochlorate(V), $\text{KClO}_3$ , contains $7.35\text{ g}$ of the salt dissolved in $250\text{ cm}^3$ of solution at $25^\circ\text{C}$ . What is the solubility of $\text{KClO}_3$ at $25^\circ\text{C}$ in $\text{mol dm}^{-3}$ ?
$[\text{K} = 39.0,\text{ Cl} = 35.5,\text{ O} = 16.0]$

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

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

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

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

Show answer & explanation

Answer: $0.24\text{ mol dm}^{-3}$

Answer

The solubility of potassium trioxochlorate(V) at

25^\circ\text{C}

is

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

.

The correct answer is

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

because

7.35\text{ g}

of

\text{KClO}_3

corresponds to

0.06\text{ mol}

. Dissolving

0.06\text{ mol}

in

0.25\text{ dm}^3

gives a molar concentration of

\frac{0.06}{0.25} = 0.24\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KClO}_3

\text{Molar Mass} = 39.0 + 35.5 + (3 \times 16.0) = 122.5\text{ g mol}^{-1}

Molar mass is required to convert mass of solute to amount in moles.

2

Calculate the number of moles of

\text{KClO}_3

in

7.35\text{ g}

\text{Moles of } \text{KClO}_3 = \frac{7.35\text{ g}}{122.5\text{ g mol}^{-1}} = 0.060\text{ mol}

Solubility in

\text{mol dm}^{-3}

measures moles of solute per unit volume of solution.

3

Convert solution volume to

\text{dm}^3

and calculate molar solubility

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.250\text{ dm}^3

;

\text{Solubility} = \frac{0.060\text{ mol}}{0.250\text{ dm}^3} = 0.24\text{ mol dm}^{-3}

Dividing total moles by total volume in cubic decimeters yields concentration in

\text{mol dm}^{-3}

.

Key Concept

Solubility concentration calculations in mol/dm³

Estimated Time:1m 30s

Question 92Question →

What is the molar concentration (in $\text{mol dm}^{-3}$ ) of a tetraoxosulfate(VI) acid solution if $20.0\text{ cm}^3$ of the acid is required to completely neutralize $25.0\text{ cm}^3$ of a $0.08\text{ mol dm}^{-3}$ sodium hydroxide solution?

Show answer & explanation

Answer: 0.05

Answer

The correct molar concentration of the tetraoxosulfate(VI) acid solution is

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

.

From 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}

,

1\text{ mole}

of tetraoxosulfate(VI) acid reacts with

2\text{ moles}

of sodium hydroxide (

n_a = 1, n_b = 2

). Substituting the given values into the titration equation

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

gives

C_a = \frac{0.08 \times 25.0 \times 1}{20.0 \times 2} = 0.05\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Write the balanced chemical equation for the reaction to find the mole ratio of acid to base.

\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

.

Tetraoxosulfate(VI) acid is a dibasic acid and requires two moles of sodium hydroxide for complete neutralization.

2

Apply the volumetric analysis formula relating concentration, volume, and stoichiometry.

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{C_a \times 20.0}{0.08 \times 25.0} = \frac{1}{2}

The standard titration formula relates acid concentration (

C_a

), acid volume (

V_a

), base concentration (

C_b

), base volume (

V_b

), and their mole coefficients.

3

Rearrange the equation to solve for

C_a

.

C_a = \frac{C_b \times V_b \times n_a}{V_a \times n_b} = \frac{0.08 \times 25.0 \times 1}{20.0 \times 2} = \frac{2.0}{40.0} = 0.05\text{ mol dm}^{-3}

.

Simplifying the arithmetic gives the precise molar concentration of the acid.

Key Concept

Volumetric analysis stoichiometry and stoichiometric concentration calculations for acid-base neutralization.

Question 93Question →

Complete the following statement on acid-base conjugate pairs under the Brønsted-Lowry theory.

Fill in the blanks below

When the hydrogen carbonate ion (

HCO_3^-

) acts as a Brønsted-Lowry base by accepting a proton, it forms as its conjugate acid. Conversely, when it acts as a Brønsted-Lowry acid by donating a proton, it forms the ion as its conjugate base.

Show answer & explanation

Answer

The conjugate acid formed when

HCO_3^-

accepts a proton is carbonic acid (

H_2CO_3

), and the conjugate base formed when it donates a proton is the carbonate ion (

CO_3^{2-}

).

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

H^+

) donor and a base is a proton acceptor. The hydrogen carbonate ion (

HCO_3^-

) is amphiprotic. When acting as a base by accepting

H^+

, it forms its conjugate acid, carbonic acid (

H_2CO_3

). When acting as an acid by donating

H^+

, it leaves behind its conjugate base, the carbonate ion (

CO_3^{2-}

).

Step-by-Step Solution

1

Identify the species formed when

HCO_3^-

acts as a proton acceptor (base).

According to the Brønsted-Lowry definition, a base accepts a proton (

H^+

). Adding

H^+

to

HCO_3^-

yields

H_2CO_3

(carbonic acid).

Accepting a proton increases the number of hydrogen atoms by 1 and raises the net electric charge by +1 (from -1 to 0).

2

Identify the species formed when

HCO_3^-

acts as a proton donor (acid).

According to the Brønsted-Lowry definition, an acid donates a proton (

H^+

). Removing

H^+

from

HCO_3^-

yields

CO_3^{2-}

(carbonate ion).

Donating a proton decreases the number of hydrogen atoms by 1 and reduces the net electric charge by 1 (from -1 to -2).

Key Concept

Brønsted-Lowry Acid-Base Theory and Amphiprotic Conjugate Pairs

Estimated Time:1m 30s

Question 94Question →

When dilute tetraoxosulfate(VI) acid ( $H_2SO_4$ ) reacts with zinc metal, hydrogen gas ( $H_2$ ) is liberated. However, when concentrated trioxonitrate(V) acid ( $HNO_3$ ) reacts with zinc metal, hydrogen gas is not evolved. Which property of concentrated trioxonitrate(V) acid accounts for this difference in behavior?

Its strong oxidizing property, which oxidizes the evolved hydrogen to water

Its dehydrating property, which removes water molecules from the metal surface

Its weak acidic strength when present in concentrated solution

Its high volatility, which causes the acid to evaporate before reacting with the metal

Show answer & explanation

Answer: Its strong oxidizing property, which oxidizes the evolved hydrogen to water

Answer

Concentrated trioxonitrate(V) acid does not liberate hydrogen gas with zinc because of its strong oxidizing property, which oxidizes hydrogen to water.

Unlike typical mineral acids that liberate hydrogen gas when reacting with electropositive metals, concentrated trioxonitrate(V) acid (

HNO_3

) is a powerful oxidizing agent. It oxidizes hydrogen to water (

H_2O

) as soon as it is produced, while the acid itself undergoes reduction to nitrogen dioxide (

NO_2

).

Step-by-Step Solution

1

Analyze the typical reaction of metals with dilute acids

Dilute acids such as

H_2SO_4

or

HCl

act as typical acids where reactive metals displace hydrogen:

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

.

This is a standard displacement reaction driven by hydrogen ion reduction.

2

Examine the behavior of concentrated trioxonitrate(V) acid (

HNO_3

)

Concentrated

HNO_3

is a powerful oxidizing acid. Instead of releasing

H_2

gas, any formed hydrogen is immediately oxidized to water (

H_2O

), while

HNO_3

is reduced to brown nitrogen dioxide gas (

NO_2

).

The equation is:

Zn_{(s)} + 4HNO_{3(aq)} \rightarrow Zn(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}

.

3

Identify the chemical property responsible

The absence of hydrogen gas is due to the strong oxidizing nature of concentrated

HNO_3

.

The acid acts primarily as an oxidizing agent rather than a simple proton donor.

Key Concept

Oxidizing property of trioxonitrate(V) acid

Question 95Question →

Equal masses of zinc granules are added separately to beaker P containing $100\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}\text{ H}_2\text{SO}_4(aq)$ and beaker Q containing $100\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}\text{ HNO}_3(aq)$ at room temperature. Which of the following statements correctly identifies and explains the primary gaseous product liberated in each beaker?

Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Both beakers liberate equal volumes of hydrogen gas because both acids are strong electrolytes that fully ionize in aqueous solution.

Beaker Q liberates twice the volume of hydrogen gas produced in beaker P because trioxonitrate(V) acid undergoes complete reduction to hydrogen ions.

Beaker P liberates sulfur(IV) oxide gas due to the dehydrating action of tetraoxosulfate(VI) acid, while beaker Q liberates hydrogen gas.

Show answer & explanation

Answer: Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Answer

Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Dilute tetraoxosulfate(VI) acid (

H_2SO_4

) reacts with active metals such as zinc via standard single-replacement to evolve hydrogen gas (

H_2

). Conversely, trioxonitrate(V) acid (

HNO_3

) is a strong oxidizing acid; the nitrate ions (

NO_3^-

) are preferentially reduced to oxides of nitrogen rather than hydrogen ions being reduced to hydrogen gas.

Step-by-Step Solution

1

Analyze the chemical reaction between zinc and dilute tetraoxosulfate(VI) acid in beaker P.

Zinc displaces hydrogen ions from dilute

H_2SO_4

to yield zinc tetraoxosulfate(VI) and hydrogen gas:

\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)

.

Dilute tetraoxosulfate(VI) acid exhibits standard acid behavior with reactive metals above hydrogen in the electrochemical series.

2

Analyze the chemical reaction between zinc and dilute trioxonitrate(V) acid in beaker Q.

Nitrate ions (

NO_3^-

) act as powerful oxidizing agents, undergoing reduction to form nitrogen oxides (such as

N_2O

,

NO

, or

NO_2

) instead of liberating hydrogen gas.

Trioxonitrate(V) acid (

HNO_3

) is a strong oxidizing acid, so hydrogen ions are not reduced to

H_2(g)

during reaction with metals.

3

Synthesize the observations from both beakers to select the correct explanation.

Beaker P produces

H_2(g)

while beaker Q produces oxides of nitrogen.

The distinct chemical property of

HNO_3

as an oxidizing acid alters the gaseous product compared to typical mineral acids.

Key Concept

Oxidizing property of trioxonitrate(V) acid versus typical acid-metal displacement reactions

Estimated Time:2m 0s

Question 96Question →

Match each chemical reaction or treatment involving an acid or base on the left with its corresponding characteristic observation or underlying chemical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming

NH_4Cl

with

NaOH

liberates

NH_3

gas which turns moist red litmus blue; heating

NaCl

with concentrated

H_2SO_4

liberates

HCl

gas which forms white fumes with ammonia vapor; adding excess

NaOH

to

Zn(OH)_2

dissolves the precipitate due to its amphoteric nature; reacting copper with concentrated

HNO_3

yields brown

NO_2

gas due to the acid's strong oxidizing power.

Each pair correctly matches the specific acid/base transformation to its corresponding property: liberation of alkaline

NH_3

gas from ammonium salt displacement, liberation of volatile

HCl

gas by non-volatile

H_2SO_4

, dissolution of amphoteric

Zn(OH)_2

in excess base, and evolution of

NO_2

gas due to the oxidizing nature of concentrated

HNO_3

.

Step-by-Step Solution

1

Examine the reaction between ammonium chloride (

NH_4Cl

) and sodium hydroxide (

NaOH

).

Heating ammonium salts with soluble alkalis yields sodium chloride, water, and ammonia gas (

NH_3

), which is alkaline and turns red litmus blue.

This demonstrates the general chemical property of alkalis reacting with ammonium salts to displace volatile ammonia.

2

Examine the reaction between solid sodium chloride (

NaCl

) and concentrated tetraoxosulfate(VI) acid (

H_2SO_4

).

Concentrated

H_2SO_4

displaces hydrogen chloride (

HCl

) gas because

H_2SO_4

is significantly less volatile (higher boiling point) than

HCl

.

This illustrates the principle that non-volatile acids displace volatile acids from their salts.

3

Examine the effect of adding excess sodium hydroxide (

NaOH

) to zinc hydroxide (

Zn(OH)_2

).

The insoluble precipitate dissolves to form a clear solution containing the soluble complex anion

[Zn(OH)_4]^{2-}

.

This exhibits the amphoteric property of zinc hydroxide, allowing it to react with strong bases.

4

Examine the reaction of copper metal (

Cu

) with concentrated trioxonitrate(V) acid (

HNO_3

).

Copper is oxidized to

Cu^{2+}

while

HNO_3

is reduced to brown nitrogen(IV) oxide gas (

NO_2

), rather than evolving hydrogen gas.

This highlights that

HNO_3

functions primarily as an oxidizing agent when reacting with metals below hydrogen in the reactivity series.

Key Concept

Distinct physical and chemical properties of acids and bases, including acid/base displacement volatility, amphoterism, and oxidizing acid behavior.

Question 97Question →

Hydrogen chloride gas ( $HCl$ ) dissolved in anhydrous methylbenzene turns blue litmus paper red and conducts an electric current. Is this statement true or false?

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Hydrogen chloride dissolved in anhydrous methylbenzene does not ionize and therefore does not show acidic properties or conduct electricity.

The statement is false because hydrogen chloride (

HCl

) exists purely as un-ionized covalent molecules in non-polar solvents like methylbenzene. Water or another polar solvent is strictly required for

HCl

to dissociate into hydroxonium ions (

H_3O^+

) and chloride ions (

Cl^-

), which are responsible for changing indicator colors and conducting electricity.

Step-by-Step Solution

1

Analyze the nature of the solvent and solute interaction.

Methylbenzene is a non-polar organic solvent, while

HCl

is a covalent compound.

Acids display acidic behavior only when ionized into hydrogen ions (

H^+

/

H_3O^+

), which requires a polar solvent such as water.

2

Determine the state of

HCl

in non-polar methylbenzene.

HCl

remains as un-ionized covalent molecules in methylbenzene.

Non-polar solvents cannot stabilize the separated

H^+

and

Cl^-

ions via solvation.

3

Evaluate the chemical and physical observations.

Without free

H^+

ions, blue litmus paper remains blue. Without mobile ions, the solution cannot conduct electricity.

Acidic indicators and electrical conductivity both depend directly on the presence of mobile ions in solution.

Key Concept

Role of Water in Acidic Properties and Ionization

Question 98Question →

Match each chemical reaction or treatment involving an acid or base on the left with its corresponding characteristic chemical observation or property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heating aqueous sodium hydroxide with ammonium chloride matches the liberation of a pungent alkaline gas that turns damp red litmus paper blue; treating iron(II) sulfide with dilute hydrochloric acid matches the liberation of a gas with a rotten-egg odor that turns lead(II) ethanoate paper black; adding concentrated tetraoxosulfate(VI) acid to cane sugar matches dehydration producing a black spongy mass of carbon and steam; reacting copper turnings with moderately concentrated trioxonitrate(V) acid matches the liberation of a gas that forms reddish-brown fumes on contact with air.

Each acid or base reaction produces a distinct diagnostic observation: strong bases liberate ammonia gas from ammonium salts; dilute acids liberate hydrogen sulfide from sulfides; concentrated sulfuric acid dehydrates carbohydrates to elemental carbon; and nitric acid oxidizes metals to yield nitrogen oxides.

Step-by-Step Solution

1

Identify the reaction between a strong base and an ammonium salt.

Heating

NaOH(aq)

with

NH_4Cl(s)

releases

NH_3(g)

, which turns moist red litmus paper blue.

Ammonium salts react with soluble bases to displace ammonia gas.

2

Identify the reaction of dilute mineral acids with metal sulfides.

Dilute

HCl(aq)

reacts with

FeS(s)

to form

FeCl_2(aq)

and

H_2S(g)

, which turns lead(II) ethanoate paper black due to

PbS

formation.

Acids displace hydrogen sulfide from metal sulfides.

3

Analyze the action of concentrated tetraoxosulfate(VI) acid on carbohydrates.

Concentrated

H_2SO_4

removes water elements from sucrose (

C_{12}H_{22}O_{11}

), leaving a porous black carbonaceous residue.

Concentrated

H_2SO_4

acts as a strong dehydrating agent.

4

Analyze the oxidizing behavior of trioxonitrate(V) acid with unreactive metals.

Moderately concentrated

HNO_3

oxidizes copper metal to produce nitrogen monoxide (

NO

), which reacts with atmospheric oxygen to yield brown

NO_2

fumes.

HNO_3

acts primarily as an oxidizing acid rather than liberating hydrogen gas with metals.

Key Concept

Chemical properties and characteristic reactions of acids and bases

Question 99Question →

When solid ammonium chloride ( $NH_4Cl$ ) is heated with aqueous sodium hydroxide ( $NaOH$ ), a characteristic gas is evolved. Which of the following observations correctly identifies the gas liberated and demonstrates a key chemical property of bases?

Ammonia gas (

NH_3

) is liberated, which turns damp red litmus paper blue.

Hydrogen gas (

H_2

) is liberated, which turns damp blue litmus paper red.

Chlorine gas (

Cl_2

) is liberated, which bleaches damp litmus paper.

Nitrogen(IV) oxide (

NO_2

) is liberated, which turns lime water milky.

Show answer & explanation

Answer: Ammonia gas ( $NH_3$ ) is liberated, which turns damp red litmus paper blue.

Answer

Heating an ammonium salt with a soluble base (alkali) liberates ammonia gas (

NH_3

), an alkaline gas that turns damp red litmus paper blue.

A defining chemical property of bases (alkalis) is their ability to react with ammonium salts when heated to displace and liberate ammonia gas (

NH_3

). Ammonia is alkaline in aqueous medium and turns damp red litmus paper blue.

Step-by-Step Solution

1

Identify the chemical reaction taking place

The reaction between ammonium chloride and sodium hydroxide is represented by:

NH_4Cl(s) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l) + NH_3(g)

Reaction of ammonium salts with strong bases upon warming is a general chemical property of bases used to test for ammonium ions and liberate ammonia.

2

Determine the nature and testing method of the evolved gas

Ammonia (

NH_3

) is a alkaline gas soluble in water forming

NH_4^+(aq)

and

OH^-(aq)

ions.

Because it produces hydroxide ions in moisture, damp red litmus paper turns blue when exposed to the gas.

Key Concept

Chemical reaction of bases with ammonium salts to liberate ammonia gas

Acids, Bases and Salts

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes