Acids, Bases and Salts Practice Questions — JAMB UTME

Match each of the following chemical compounds to its characteristic behavior when exposed to moist atmospheric air.

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

matches with losing water of crystallization to form a powdery residue (efflorescence);

\text{CaCl}_2

matches with absorbing moisture to form a solution (deliquescence);

\text{CaO}

matches with absorbing moisture without dissolving (hygroscopy).

Each compound exhibits a specific behavior upon exposure to air: hydrated sodium carbonate loses water of crystallization (efflorescence), calcium chloride absorbs moisture and dissolves (deliquescence), and calcium oxide absorbs moisture without dissolving (hygroscopy).

Step-by-Step Solution

1

Identify the atmospheric phenomenon characteristic of hydrated sodium carbonate (

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

).

It undergoes efflorescence, losing water molecules to the surrounding dry air.

Hydrated salts with higher vapor pressure than the atmospheric water vapor pressure tend to lose hydration water.

2

Identify the atmospheric behavior of anhydrous calcium chloride (

\text{CaCl}_2

).

It undergoes deliquescence, absorbing atmospheric moisture until it completely dissolves in the absorbed water.

Deliquescent compounds have an extremely high affinity for water and lower saturated solution vapor pressure than ambient air.

3

Identify the behavior of quicklime (

\text{CaO}

).

It undergoes hygroscopy, absorbing atmospheric moisture without dissolving into a liquid solution.

Hygroscopic substances absorb moisture into their structure without forming a liquid solution.

Key Concept

Differentiation between efflorescence, deliquescence, and hygroscopy.

Question 2Question →

An aqueous solution at $25^\circ\text{C}$ has a hydroxide ion concentration ( $[OH^-]$ ) of $1.0 \times 10^{-11}\text{ mol dm}^{-3}$ . What is the pH of this solution?

3.0

11.0

7.0

1.0 \times 10^{-11}

Show answer & explanation

Answer: $3.0$

Answer

The pH of the solution is

3.0

.

Taking the negative logarithm of the hydroxide ion concentration gives a pOH value of

11.0

. Subtracting this from

14.0

yields a pH of

3.0

, accurately indicating an acidic solution.

Step-by-Step Solution

1

Calculate the pOH from the hydroxide ion concentration

\text{pOH} = -\log_{10}(1.0 \times 10^{-11}) = 11.0

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

2

Calculate the pH using the water ionic product relationship at

25^\circ\text{C}

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

The sum of pH and pOH in aqueous solutions at

25^\circ\text{C}

is equal to 14.0.

Key Concept

Relationship between

[OH^-]

, pOH, and pH in aqueous solutions

Estimated Time:45s

Question 3Question →

Which of the following salts dissolves in water to form an alkaline solution that turns red litmus paper blue?

Sodium trioxocarbonate(IV)

Ammonium chloride

Sodium chloride

Potassium tetraoxosulfate(VI)

Show answer & explanation

Answer: Sodium trioxocarbonate(IV)

Answer

Sodium trioxocarbonate(IV) forms an alkaline solution in water due to anion hydrolysis.

Sodium trioxocarbonate(IV) dissolves in water to form sodium ions and trioxocarbonate(IV) ions. Because trioxocarbonate(IV) is the conjugate base of a weak acid, it reacts with water to release hydroxide ions (

OH^-

), creating an alkaline solution that turns red litmus paper blue.

Step-by-Step Solution

1

Identify the parent acid and base for the salt.

Sodium trioxocarbonate(IV) (

Na_2CO_3

) is formed from sodium hydroxide (

NaOH

, a strong base) and carbonic acid (

H_2CO_3

, a weak acid).

The hydrolysis behavior of a salt is determined by the relative strengths of its parent acid and base.

2

Analyze the ion hydrolysis reaction.

The carbonate anion (

CO_3^{2-}

) reacts with water according to:

CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-

.

Anions derived from weak acids act as conjugate bases and hydrolyze in water to generate hydroxide ions (

OH^-

).

3

Determine the solution acidity/alkalinity and indicator response.

The production of excess

OH^-

ions makes the solution basic (

pH > 7

), causing red litmus paper to turn blue.

Alkaline solutions contain a higher concentration of hydroxide ions than hydrogen ions, turning red litmus blue.

Key Concept

Salt Hydrolysis

Estimated Time:45s

Question 4Question →

What is the pH of an aqueous solution prepared by dissolving $0.49\text{ g}$ of tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) in distilled water to make $500\text{ cm}^3$ of solution? [Molar mass of $\text{H}_2\text{SO}_4 = 98\text{ g mol}^{-1}$ , $\log_{10} 2 = 0.301$ ]

1.70

2.00

2.30

12.30

Show answer & explanation

Answer: $1.70$

Answer

The pH of the solution is

1.70

.

Dissolving

0.49\text{ g}

of

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

(molar mass

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

) in

0.5\text{ dm}^3

yields a

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

solution. Because tetraoxosulfate(VI) acid fully ionizes into

2\text{H}^+

and

\text{SO}_4^{2-}

, the hydrogen ion concentration

[\text{H}^+]

is

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

. Taking the negative logarithm base 10 gives

\text{pH} = -\log_{10}(0.02) = 1.70

.

Step-by-Step Solution

1

Calculate the amount of

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

in moles

\text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.49\text{ g}}{98\text{ g mol}^{-1}} = 0.005\text{ mol}

Converting mass of solute to moles using molar mass.

2

Determine the molar concentration of the acid solution

\text{Volume} = 500\text{ cm}^3 = 0.5\text{ dm}^3

;

\text{Molarity} = \frac{0.005\text{ mol}}{0.5\text{ dm}^3} = 0.01\text{ mol dm}^{-3}

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

3

Calculate the hydrogen ion concentration

[\text{H}^+]

[\text{H}^+] = 2 \times 0.01\text{ mol dm}^{-3} = 0.02\text{ mol dm}^{-3} = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

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

is a strong dibasic acid that ionizes completely to produce two

\text{H}^+

ions per molecule.

4

Calculate the pH of the solution

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.0 \times 10^{-2}) = 2 - \log_{10} 2 = 2 - 0.301 = 1.699 \approx 1.70

Applying the logarithmic definition of pH.

Key Concept

pH Calculation of Dibasic Strong Acids

Question 5Question →

A $6.95\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , is dissolved in dilute tetraoxosulfate(VI) acid and made up to $250\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution requires $25.0\text{ cm}^3$ of $0.020\text{ mol dm}^{-3}$ acidified potassium tetraoxomanganate(VII), $\text{KMnO}_4$ , for complete titration. Given the relative atomic masses $\text{Fe} = 56$ , $\text{S} = 32$ , $\text{O} = 16$ , and $\text{H} = 1$ , what is the integer value of $x$ , the number of molecules of water of crystallization per formula unit?

Show answer & explanation

Answer: 7

Answer

The integer value of x is 7.

By using the titration volume and concentration of acidified potassium tetraoxomanganate(VII), the amount of iron(II) ions in the sample is calculated. Knowing the total mole amount of anhydrous iron(II) tetraoxosulfate(VI) allows determination of the mass of the anhydrous salt component (3.80 g). Subtracting this from the initial hydrated sample mass (6.95 g) gives the mass of water of crystallization (3.15 g). Dividing the moles of water (0.175 mol) by the moles of anhydrous salt (0.025 mol) yields exactly 7 water molecules of crystallization per formula unit.

Step-by-Step Solution

1

Calculate the moles of

\text{KMnO}_4

consumed in the titration

n(\text{KMnO}_4) = 0.020 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.00050 \text{ mol}

Concentration and volume of titrant are provided.

2

Determine moles of

\text{Fe}^{2+}

present in the

25.0\text{ cm}^3

aliquot using the redox reaction stoichiometry

n(\text{Fe}^{2+})_{25\text{cm}^3} = 5 \times 0.00050 \text{ mol} = 0.0025 \text{ mol}

The mole ratio of

\text{MnO}_4^-

to

\text{Fe}^{2+}

in acidic redox titration is

1:5

according to

\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

.

3

Calculate the total moles of

\text{FeSO}_4

in the original

250\text{ cm}^3

volumetric flask

n(\text{FeSO}_4)_{\text{total}} = 0.0025 \text{ mol} \times \left(\frac{250\text{ cm}^3}{25.0\text{ cm}^3}\right) = 0.025 \text{ mol}

The aliquot represents one-tenth of the total solution volume.

4

Calculate the mass of anhydrous

\text{FeSO}_4

in the sample

Molar mass of

\text{FeSO}_4 = 56 + 32 + (4 \times 16) = 152 \text{ g mol}^{-1}

. Mass

= 0.025 \text{ mol} \times 152 \text{ g mol}^{-1} = 3.80 \text{ g}

.

Converting moles of anhydrous salt to mass using molar mass.

5

Determine the mass and moles of water of crystallization

Mass of

\text{H}_2\text{O} = 6.95 \text{ g} - 3.80 \text{ g} = 3.15 \text{ g}

. Moles of

\text{H}_2\text{O} = \frac{3.15 \text{ g}}{18 \text{ g mol}^{-1}} = 0.175 \text{ mol}

.

Subtracting anhydrous mass from initial mass gives water of crystallization mass, converted to moles using molar mass of

\text{H}_2\text{O} = 18 \text{ g mol}^{-1}

.

6

Compute the hydration coefficient

x = \frac{n(\text{H}_2\text{O})}{n(\text{FeSO}_4)}

x = \frac{0.175 \text{ mol}}{0.025 \text{ mol}} = 7

The coefficient

x

represents the mole ratio of water of crystallization to anhydrous salt.

Key Concept

Quantitative determination of water of crystallization in hydrated salts via redox volumetric analysis

Question 6Question →

A $0.04\text{ mol dm}^{-3}$ aqueous solution of a weak monobasic acid, $\text{HA}$ , is $2.0\%$ ionized at $25^\circ\text{C}$ . If distilled water is added to this solution until its total volume is quadrupled, what is the degree of ionization of the acid in the diluted solution?

0.5\%

1.0\%

4.0\%

8.0\%

Show answer & explanation

Answer: $4.0\%$

Answer

The degree of ionization of the acid in the diluted solution is

4.0\%

.

By Ostwald's dilution law for weak monobasic acids,

K_a = \alpha^2 C

, which rearranges to

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

. Since

K_a

is constant at a fixed temperature, the degree of ionization

\alpha

is inversely proportional to the square root of the concentration (

\alpha \propto \frac{1}{\sqrt{C}}

). Quadrupling the volume decreases the concentration by a factor of 4 (

C_2 = \frac{C_1}{4}

), which increases the degree of ionization by a factor of

\sqrt{4} = 2

. Therefore, the new degree of ionization is

2.0\% \times 2 = 4.0\%

.

Step-by-Step Solution

1

Calculate the acid dissociation constant (

K_a

) using the initial concentration and degree of ionization.

Initial concentration

C_1 = 0.04\text{ mol dm}^{-3}

, initial degree of ionization

\alpha_1 = 2.0\% = 0.02

. Using

K_a \approx \alpha_1^2 C_1

, we get

K_a = (0.02)^2 \times 0.04 = 4.0 \times 10^{-4} \times 0.04 = 1.6 \times 10^{-5}\text{ mol dm}^{-3}

.

The value of

K_a

depends only on temperature and remains constant upon dilution.

2

Determine the new concentration (

C_2

) after quadrupling the solution volume.

C_2 = \frac{C_1}{4} = \frac{0.04\text{ mol dm}^{-3}}{4} = 0.01\text{ mol dm}^{-3}

.

Diluting a solution to 4 times its original volume reduces its molar concentration by a factor of 4.

3

Calculate the new degree of ionization (

\alpha_2

) in the diluted solution.

\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} = \sqrt{16 \times 10^{-4}} = 4.0 \times 10^{-2} = 0.04 = 4.0\%$.

Applying Ostwald's dilution law to find the updated degree of dissociation at the lower concentration.

Key Concept

Ostwald's Dilution Law and the relationship between dilution, concentration, and degree of ionization of weak electrolytes.

Estimated Time:2m 30s

Question 7Question →

Which of the following gases is evolved when dilute hydrochloric acid ( $HCl$ ) reacts with calcium trioxocarbonate(IV) ( $CaCO_3$ )?

Carbon(IV) oxide (

CO_2

)

Hydrogen (

H_2

)

Chlorine (

Cl_2

)

Carbon(II) oxide (

CO

)

Show answer & explanation

Answer: Carbon(IV) oxide ( $CO_2$ )

Answer

Carbon(IV) oxide (

CO_2

)

A characteristic chemical property of acids is that they react with trioxocarbonate(IV) salts (such as

CaCO_3

) to produce a salt, water, and carbon(IV) oxide gas (

CO_2

).

Step-by-Step Solution

1

Identify the reactants and the class of reaction

Dilute hydrochloric acid (

HCl

) reacts with calcium trioxocarbonate(IV) (

CaCO_3

), which is an acid-carbonate reaction.

Acids characteristically react with trioxocarbonates to form salt, water, and carbon dioxide.

2

Write the balanced chemical equation

2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)

The equation shows that carbon(IV) oxide gas (

CO_2

) is liberated during the chemical process.

Key Concept

Action of acids on trioxocarbonate(IV) salts

Estimated Time:45s

Question 8Question →

A weak monobasic acid has a concentration of $0.20\text{ mol dm}^{-3}$ in aqueous solution. At equilibrium, the hydrogen ion concentration, $[H^+]$ , is measured to be $1.0 \times 10^{-3}\text{ mol dm}^{-3}$ . What is the percentage degree of ionization of the acid in this solution?

Show answer & explanation

Answer: 0.5

Answer

The percentage degree of ionization of the weak acid is 0.5%.

The percentage degree of ionization measures the fraction of acid molecules that ionize in water, expressed as a percentage. By substituting

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

and initial concentration

C = 0.20\text{ mol dm}^{-3}

into

\text{Percentage ionization} = \left(\frac{[H^+]}{C}\right) \times 100\%

, we obtain

\left(\frac{0.001}{0.20}\right) \times 100\% = 0.5\%

.

Step-by-Step Solution

1

Identify the relationship between degree of ionization (\alpha), hydrogen ion concentration (

[H^+]

), and initial acid concentration (

C

).

\alpha = \frac{[H^+]}{C}

For a weak monobasic acid ionizing according to

HA \rightleftharpoons H^+ + A^-

, the concentration of ionized hydrogen ions equals

\alpha C

.

2

Calculate the fractional degree of ionization (\alpha).

\alpha = \frac{1.0 \times 10^{-3}\text{ mol dm}^{-3}}{0.20\text{ mol dm}^{-3}} = 0.005

Dividing the equilibrium hydrogen ion concentration by the initial acid concentration gives the fraction of acid molecules that ionized.

3

Convert the fractional degree of ionization into a percentage.

\text{Percentage ionization} = 0.005 \times 100\% = 0.5\%

Multiplying the decimal fraction by 100 converts the degree of ionization into percentage form.

Key Concept

Degree of Ionization of Weak Acids

Estimated Time:1m 30s

Question 9Question →

When potassium ethanoate, $CH_3COOK$ , is dissolved in distilled water, the resulting solution has a pH greater than 7. Which ion is responsible for this alkalinity by undergoing hydrolysis?

The ethanoate ion,

CH_3COO^-

The potassium ion,

K^+

The hydrogen ion,

H^+

The hydroxide ion,

OH^-

Show answer & explanation

Answer: The ethanoate ion, $CH_3COO^-$

Answer

The ethanoate ion,

CH_3COO^-

The ethanoate ion (

CH_3COO^-

) is the conjugate base of the weak acid ethanoic acid (

CH_3COOH

). When potassium ethanoate dissolves in water,

CH_3COO^-

hydrolyzes by accepting a proton from water to form

CH_3COOH

and releasing

OH^-

ions into the solution:

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

. The accumulation of free hydroxide ions makes the solution alkaline (

pH > 7

).

Step-by-Step Solution

1

Identify the ions formed when potassium ethanoate dissolves in water.

Potassium ethanoate dissociates completely into potassium ions (

K^+

) and ethanoate ions (

CH_3COO^-

).

Soluble ionic salts completely dissociate into their constituent cations and anions in aqueous solution.

2

Determine which ion reacts with water (hydrolyzes).

Potassium (

K^+

) is derived from a strong base (

KOH

) and does not hydrolyze. Ethanoate (

CH_3COO^-

) is derived from a weak acid (

CH_3COOH

) and reacts with water.

Anions derived from weak acids act as weak conjugate bases and undergo anion hydrolysis.

3

Write the hydrolysis equation to identify the species giving rise to alkalinity.

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

Hydrolysis of ethanoate generates free

OH^-

ions, making the solution basic (

pH > 7

).

Key Concept

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

Question 10Question →

Complete the statement regarding the choice of indicator and the expected endpoint color change when titrating aqueous ammonia with hydrochloric acid.

Fill in the blanks below

When titrating aqueous ammonia against hydrochloric acid, the appropriate indicator is , which changes from yellow in alkaline solution to at the equivalence point.

Show answer & explanation

Answer

The appropriate indicator is methyl orange, which changes from yellow to red (or pink) at the equivalence point.

In a titration between a strong acid (

\text{HCl}

) and a weak base (

\text{NH}_3

), ammonium chloride (

\text{NH}_4\text{Cl}

) is formed. The hydrolysis of

\text{NH}_4^+

ions makes the solution acidic at the equivalence point (

\text{pH} < 7

). Methyl orange is the correct indicator because its transition range (

\text{pH } 3.1 - 4.4

) coincides with this acidic equivalence point, turning red (or pink) when the endpoint is reached.

Step-by-Step Solution

1

Analyze the nature of the reactants

Hydrochloric acid (

\text{HCl}

) is a strong acid, and aqueous ammonia (

\text{NH}_3(aq)

) is a weak base.

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

2

Determine the pH range at the equivalence point

The salt formed (

\text{NH}_4\text{Cl}

) undergoes hydrolysis, producing an acidic solution with an equivalence point at

\text{pH} < 7

(typically between

\text{pH } 3

and

6

).

Salt hydrolysis of a strong acid and weak base yields ammonium ions (

\text{NH}_4^+

) which hydrolyze to produce

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

ions.

3

Select the suitable indicator and recall its color change

Methyl orange functions in the

\text{pH}

range of

3.1 - 4.4

, changing from yellow in basic solution to red (or pink) at the acidic endpoint.

An indicator must undergo a distinct color transition within the steep

\text{pH}

region of the titration curve near the equivalence point.

Key Concept

Indicator Selection and Endpoint Determination in Volumetric Analysis

Estimated Time:1m 0s

Question 11Question →

What is the pH of an aqueous solution containing $0.365\text{ g dm}^{-3}$ of hydrochloric acid ( $\text{HCl}$ )? [Molar mass of $\text{HCl} = 36.5\text{ g mol}^{-1}$ ]

2.0

12.0

0.44

1.0

Show answer & explanation

Answer: $2.0$

Answer

The pH of the solution is 2.0.

The solution has a molar concentration of

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

(

0.365\text{ g dm}^{-3} / 36.5\text{ g mol}^{-1}

). Taking the negative base-10 logarithm of

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

yields a pH of 2.0.

Step-by-Step Solution

1

Convert mass concentration to molar concentration (molarity)

[\text{HCl}] = \frac{0.365\text{ g dm}^{-3}}{36.5\text{ g mol}^{-1}} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

pH calculations require concentration in units of

\text{mol dm}^{-3}

.

2

Determine the hydrogen ion concentration

[\text{H}^+]

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

Hydrochloric acid is a strong monoprotic acid that fully dissociates in water.

3

Calculate the pH using the pH formula

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

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

Key Concept

Calculating pH from mass concentration of a strong acid

Estimated Time:1m 0s

Question 12Question →

Fill in the blank to complete the statement regarding the Brønsted-Lowry theory of acids and bases.

Fill in the blanks below

According to the Brønsted-Lowry theory, an acid is defined as a species that acts as a proton .

Show answer & explanation

Answer

According to the Brønsted-Lowry theory, an acid is defined as a species that acts as a proton donor.

In the Brønsted-Lowry concept, acids are species capable of donating a hydrogen ion (proton,

H^+

) to another chemical species. Therefore, the term 'donor' correctly completes the definition.

Step-by-Step Solution

1

Recall the fundamental definitions of the Brønsted-Lowry acid-base theory.

The Brønsted-Lowry theory defines an acid as a proton (

H^+

) donor and a base as a proton (

H^+

) acceptor.

Acid-base behavior under this theory is governed by the transfer of protons between chemical species.

Key Concept

Brønsted-Lowry Acid Definition

Question 13Question →

At $60^\circ\text{C}$ , the solubility of a trioxonitrate(V) salt $Y$ (molar mass $= 160\text{ g mol}^{-1}$ ) is $1.25\text{ mol dm}^{-3}$ . When the solution is cooled to $20^\circ\text{C}$ , its solubility decreases to $0.25\text{ mol dm}^{-3}$ . What mass of salt $Y$ will crystallize out when $300\text{ g}$ of a saturated solution of $Y$ at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ? (Assume the density of water is $1.0\text{ g cm}^{-3}$ ).

40.0\text{ g}

48.0\text{ g}

50.0\text{ g}

160.0\text{ g}

Show answer & explanation

Answer: $40.0\text{ g}$

Answer

The mass of salt

Y

that crystallizes out upon cooling is

40.0\text{ g}

.

Converting concentrations to mass per

1000\text{ g}

of water gives

200\text{ g}

at

60^\circ\text{C}

and

40\text{ g}

at

20^\circ\text{C}

. A

300\text{ g}

sample of saturated solution at

60^\circ\text{C}

contains

50\text{ g}

of solute dissolved in

250\text{ g}

of solvent (water). Upon cooling to

20^\circ\text{C}

,

250\text{ g}

of water can only hold

10\text{ g}

of solute. Thus,

50\text{ g} - 10\text{ g} = 40.0\text{ g}

crystallizes out.

Step-by-Step Solution

1

Convert solubility at

60^\circ\text{C}

from

\text{mol dm}^{-3}

to

\text{g dm}^{-3}

\text{Solubility at } 60^\circ\text{C} = 1.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 200\text{ g dm}^{-3}

Solubility calculations involving mass of solution require mass concentration (

\text{g dm}^{-3}

).

2

Determine mass of water in

300\text{ g}

of saturated solution at

60^\circ\text{C}

\text{Mass of solution per dm}^3 = 1000\text{ g (water)} + 200\text{ g (salt)} = 1200\text{ g}

. Therefore,

\text{mass of water in } 300\text{ g solution} = \frac{1000\text{ g}}{1200\text{ g}} \times 300\text{ g} = 250\text{ g}

.

Solubility expresses mass of solute per fixed mass of solvent (

1000\text{ g}

water).

3

Convert solubility at

20^\circ\text{C}

to

\text{g dm}^{-3}

and find mass of salt remaining dissolved in

250\text{ g}

water

\text{Solubility at } 20^\circ\text{C} = 0.25\text{ mol dm}^{-3} \times 160\text{ g mol}^{-1} = 40\text{ g dm}^{-3}

. In

250\text{ g}

of water, dissolved salt

= 40\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 10\text{ g}

.

Determining how much solute stays dissolved at the lower temperature.

4

Calculate the mass of salt precipitated

\text{Original salt in solution} = 200\text{ g} \times \frac{250\text{ g}}{1000\text{ g}} = 50\text{ g}

.

\text{Mass precipitated} = 50\text{ g} - 10\text{ g} = 40.0\text{ g}

.

Mass precipitated is the difference between initial dissolved mass and remaining dissolved mass.

Key Concept

Mass of salt precipitated upon cooling saturated solutions

Estimated Time:2m 30s

Question 14Question →

Which of the following terms describes the phenomenon where a solid substance absorbs moisture from the atmosphere and dissolves in it to form a solution?

Deliquescence

Hygroscopy

Efflorescence

Evaporation

Show answer & explanation

Answer: Deliquescence

Answer

Deliquescence is the phenomenon where a solid absorbs water vapor from the atmosphere until it dissolves in that water to form a saturated solution.

Deliquescence is the spontaneous absorption of atmospheric moisture by a solid until it liquefies into a solution. Typical examples of deliquescent compounds include calcium chloride (

CaCl_2

), iron(III) chloride (

FeCl_3

), and sodium hydroxide (

NaOH

).

Step-by-Step Solution

1

Identify the characteristic atmospheric behavior described in the prompt.

The process involves absorbing moisture from the air to the extent of forming a liquid solution.

This specific property defines deliquescence (e.g., as observed with anhydrous calcium chloride,

CaCl_2

, or sodium hydroxide,

NaOH

pellets).

Key Concept

Distinction between deliquescence, hygroscopy, and efflorescence

Question 15Question →

A $25.0\text{ cm}^3$ portion of a $6.30\text{ g dm}^{-3}$ solution of an impure hydrated diprotic weak acid, $\text{H}_2\text{X}\cdot 2\text{H}_2\text{O}$ , requires $20.0\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the percentage purity of the acid sample, and which indicator is most suitable for this titration?
[Molar mass of $\text{H}_2\text{X}\cdot 2\text{H}_2\text{O} = 126\text{ g mol}^{-1}$ ]

80.0\%

, using phenolphthalein

160.0\%

, using phenolphthalein

80.0\%

, using methyl orange

40.0\%

, using phenolphthalein

Show answer & explanation

Answer: $80.0\%$ , using phenolphthalein

Answer

80.0%, using phenolphthalein

The neutralization reaction requires two moles of sodium hydroxide for every mole of diprotic acid. Using

\frac{C_A V_A}{C_B V_B} = \frac{1}{2}

, the concentration of pure acid is

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

. Multiplying by the molar mass (

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

) gives

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

of pure acid. Dividing by the sample mass concentration (

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

) and multiplying by

100\%

yields

80.0\%

. Because the titration involves a weak acid and a strong base, the solution at the end point is alkaline, which requires phenolphthalein as the indicator.

Step-by-Step Solution

1

Write the balanced chemical equation to determine the mole ratio.

\text{H}_2\text{X}\cdot 2\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{X} + 4\text{H}_2\text{O}

, giving mole ratio

\frac{n_A}{n_B} = \frac{1}{2}

.

Diprotic acid yields two hydrogen ions per molecule, requiring two moles of sodium hydroxide for neutralization.

2

Calculate the molar concentration of the pure acid in solution using the titration formula.

\frac{C_A \times 25.0}{0.10 \times 20.0} = \frac{1}{2} \implies C_A = \frac{0.10 \times 20.0}{25.0 \times 2} = 0.040\text{ mol dm}^{-3}

.

Relating acid and base concentrations via volumes and stoichiometric coefficients.

3

Convert the molar concentration of pure acid to mass concentration.

\text{Mass concentration} = 0.040\text{ mol dm}^{-3} \times 126\text{ g mol}^{-1} = 5.04\text{ g dm}^{-3}.

Mass concentration equals molar concentration multiplied by molar mass.

4

Calculate the percentage purity of the acid sample.

\text{Percentage purity} = \left(\frac{5.04\text{ g dm}^{-3}}{6.30\text{ g dm}^{-3}}\right) \times 100\% = 80.0\%.

Percentage purity is the ratio of pure mass concentration to total sample mass concentration.

5

Select the correct indicator for the titration.

Phenolphthalein is selected.

Titrating a weak acid against a strong base produces a basic salt solution at equivalence (pH > 7), making phenolphthalein (pH range 8.3–10.0) the suitable indicator.

Key Concept

Volumetric Stoichiometry and Indicator Selection in Acid-Base Titrations

Question 16Question →

Match each chemical substance with its characteristic atmospheric behavior or hydration property when exposed to ambient laboratory air or heat:

Click a left item, then click its matching right item

Show answer & explanation

Answer

Washing soda crystals undergo efflorescence; anhydrous calcium chloride undergoes deliquescence; concentrated sulfuric acid is hygroscopic; copper(II) sulfate pentahydrate loses its water of crystallization upon heating.

Washing soda crystals (

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

) undergo efflorescence when exposed to dry air. Anhydrous calcium chloride (

\text{CaCl}_2

) absorbs moisture until it forms a solution, which is deliquescence. Concentrated sulfuric acid (

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

) absorbs moisture without dissolving a solid matrix, illustrating hygroscopic behavior. Copper(II) sulfate pentahydrate (

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

) loses its water of crystallization thermally to yield a white anhydrous salt.

Step-by-Step Solution

1

Analyze the behavior of

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

.

It loses 9 molecules of water of crystallization to dry air, forming a powdery monohydrate

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

(efflorescence).

Efflorescence occurs when the vapor pressure of the hydrated salt is greater than the partial pressure of water vapor in the atmosphere.

2

Analyze anhydrous

\text{CaCl}_2

.

It absorbs moisture from air and eventually dissolves in the absorbed water to form a saturated solution (deliquescence).

Deliquescence occurs when a solid compound's saturated solution has a lower vapor pressure than the atmospheric water vapor pressure.

3

Analyze concentrated

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

.

It absorbs moisture from the atmosphere without changing state to form a new dissolved phase, acting as a liquid hygroscopic substance.

Hygroscopic substances absorb water vapor from the surrounding environment without dissolving into a liquid solution from a solid state.

4

Analyze

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

.

Heating drives off the 5 water molecules of crystallization, converting blue crystalline pentahydrate to white anhydrous copper(II) sulfate.

Water of crystallization is chemically bound in the crystal lattice and can be expelled thermally, altering optical/crystalline properties.

Key Concept

Water of Crystallization, Deliquescence, Efflorescence, and Hygroscopy

Question 17Question →

Two weak monobasic acids, $\text{HX}$ ( $K_a = 1.0 \times 10^{-5}\text{ mol dm}^{-3}$ ) and $\text{HY}$ ( $K_a = 4.0 \times 10^{-5}\text{ mol dm}^{-3}$ ), are prepared as aqueous solutions. Solution X contains $0.40\text{ mol dm}^{-3}$ of $\text{HX}$ , whereas Solution Y contains $0.10\text{ mol dm}^{-3}$ of $\text{HY}$ . Based on Ostwald's dilution law and the principles of acid ionization, which of the following statements correctly compares the two solutions?

Solution X has a higher hydrogen ion concentration than Solution Y because

\text{HX}

is four times more concentrated than

\text{HY}

.

Solution Y has a higher hydrogen ion concentration than Solution X because

\text{HY}

has a larger acid dissociation constant (

K_a

) than

\text{HX}

.

Both solutions have the same hydrogen ion concentration of

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

, but

\text{HY}

has a four-fold greater degree of ionization than

\text{HX}

.

Both solutions have the same degree of ionization, but Solution X contains a higher concentration of hydrogen ions.

Show answer & explanation

Answer: Both solutions have the same hydrogen ion concentration of $2.0 \times 10^{-3}\text{ mol dm}^{-3}$ , but $\text{HY}$ has a four-fold greater degree of ionization than $\text{HX}$ .

Answer

Both solutions have the exact same hydrogen ion concentration of

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

, but

\text{HY}

has a four-fold greater degree of ionization than

\text{HX}

.

Evaluating

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

for both solutions gives

[H^+]_X = \sqrt{1.0 \times 10^{-5} \times 0.40} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

and

[H^+]_Y = \sqrt{4.0 \times 10^{-5} \times 0.10} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

, showing both solutions have identical hydrogen ion concentrations. Calculating the degree of ionization

\alpha = \sqrt{K_a / C}

yields

\alpha_X = 0.005

(

0.5\%

) and

\alpha_Y = 0.020

(

2.0\%

), demonstrating that

\text{HY}

is four times more ionized than

\text{HX}

in these conditions.

Step-by-Step Solution

1

Calculate

[H^+]

for Solution X

[H^+]_X = \sqrt{K_{a,X} \times C_X} = \sqrt{(1.0 \times 10^{-5}) \times 0.40} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

For a weak monobasic acid,

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

derived from Ostwald's dilution law.

2

Calculate

[H^+]

for Solution Y

[H^+]_Y = \sqrt{K_{a,Y} \times C_Y} = \sqrt{(4.0 \times 10^{-5}) \times 0.10} = \sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

Applying the weak acid ionization expression to Solution Y.

3

Calculate the degree of ionization (\alpha) for both acids

\alpha_X = \sqrt{\frac{K_{a,X}}{C_X}} = \sqrt{\frac{1.0 \times 10^{-5}}{0.40}} = 5.0 \times 10^{-3} = 0.5\%

;

\alpha_Y = \sqrt{\frac{K_{a,Y}}{C_Y}} = \sqrt{\frac{4.0 \times 10^{-5}}{0.10}} = 2.0 \times 10^{-2} = 2.0\%

The degree of ionization is given by

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

.

4

Compare the calculated parameters

[H^+]_X = [H^+]_Y = 2.0 \times 10^{-3}\text{ mol dm}^{-3}

and

\frac{\alpha_Y}{\alpha_X} = \frac{2.0\%}{0.5\%} = 4

Comparing

[H^+]

and the ratio of degrees of ionization.

Key Concept

Relative Strength and Ionization of Acids and Bases

Estimated Time:2m 0s

Question 18Question →

A saturated solution contains $12.0\text{ g}$ of sodium hydroxide ( $\text{NaOH}$ ) dissolved in $500\text{ cm}^3$ of solution at $25^\circ\text{C}$ . What is the solubility of sodium hydroxide in $\text{mol dm}^{-3}$ at this temperature?
[Molar mass of $\text{NaOH} = 40.0\text{ g mol}^{-1}$ ]

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

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

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

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

Show answer & explanation

Answer: $0.60\text{ mol dm}^{-3}$

Answer

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

To find the solubility in

\text{mol dm}^{-3}

, first determine the number of moles of

\text{NaOH}

by dividing the mass (

12.0\text{ g}

) by the molar mass (

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

), yielding

0.30\text{ mol}

. Next, convert the volume to cubic decimeters (

500\text{ cm}^3 = 0.50\text{ dm}^3

). Finally, divide the amount in moles by the volume in

\text{dm}^3

to get

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

.

Step-by-Step Solution

1

Calculate the amount of

\text{NaOH}

in moles

\text{Moles} = \frac{12.0\text{ g}}{40.0\text{ g mol}^{-1}} = 0.30\text{ mol}

Solubility in

\text{mol dm}^{-3}

requires expressing the amount of solute in moles.

2

Convert the solution volume from

\text{cm}^3

to

\text{dm}^3

\text{Volume} = \frac{500\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.50\text{ dm}^3

Standard concentration units require volume expressed in cubic decimeters.

3

Divide the number of moles by the volume in

\text{dm}^3

\text{Solubility} = \frac{0.30\text{ mol}}{0.50\text{ dm}^3} = 0.60\text{ mol dm}^{-3}

Solubility is the maximum amount of solute in moles that dissolves per

\text{dm}^3

of solution at a given temperature.

Key Concept

Solubility calculation in mol/dm³

Question 19Question →

A standard solution is prepared by dissolving $1.575\text{ g}$ of hydrated ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}$ ) in distilled water to make $250.0\text{ cm}^3$ of solution. A $25.0\text{ cm}^3$ sample of this acid solution requires $25.0\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the integer value of $x$ in the formula of the hydrated acid? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Show answer & explanation

Answer: 2

Answer

The integer value of

x

is 2.

The value of

x

is calculated as

2

. Based on the reaction stoichiometry,

25.0\text{ cm}^3

of

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

\text{NaOH}

contains

0.0025\text{ mol}

of base, which neutralizes

0.00125\text{ mol}

of the diprotic acid in the

25.0\text{ cm}^3

sample. The entire

250.0\text{ cm}^3

solution therefore contains

0.0125\text{ mol}

of acid. Dividing the mass (

1.575\text{ g}

) by

0.0125\text{ mol}

gives a molar mass of

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

for

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

. Subtracting the molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4

(

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

) gives

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

for water, which corresponds to

x = 36 / 18 = 2

.

Step-by-Step Solution

1

Determine the mole ratio from the balanced chemical neutralization equation.

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + (x+2)\text{H}_2\text{O}

. The stoichiometric ratio of acid to base is

1 : 2

.

Ethanedioic acid is a diprotic acid requiring two moles of hydroxide ions for complete neutralization per mole of acid.

2

Calculate the amount in moles of sodium hydroxide solution used in the titration.

\text{Moles of NaOH} = 0.100\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0025\text{ mol}

.

Number of moles is equal to molar concentration multiplied by volume in cubic decimeters.

3

Calculate the total moles of hydrated acid present in the

250.0\text{ cm}^3

volumetric flask.

\text{Moles in } 25.0\text{ cm}^3 \text{ aliquot} = \frac{0.0025}{2} = 0.00125\text{ mol}

. Total moles in

250.0\text{ cm}^3 = 0.00125 \times \frac{250.0}{25.0} = 0.0125\text{ mol}

.

Using the stoichiometric ratio (

n_a/n_b = 1/2

) and scaling up from the

25.0\text{ cm}^3

aliquot to the full

250.0\text{ cm}^3

solution volume.

4

Compute the molar mass of the hydrated acid and solve for

x

.

\text{Molar mass} = \frac{1.575\text{ g}}{0.0125\text{ mol}} = 126\text{ g mol}^{-1}

. Molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4 = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}

. Mass of

x\text{H}_2\text{O} = 126 - 90 = 36\text{ g mol}^{-1}

. Thus,

x = \frac{36}{18} = 2

.

Subtracting the molar mass of the anhydrous acid from the total molar mass gives the mass of the water of crystallization, which is divided by the molar mass of water (

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

) to find

x

.

Key Concept

Volumetric Analysis and Water of Crystallization Determination

Question 20Question →

Complete the statement below regarding the volumetric analysis of a mixture containing sodium carbonate and sodium hydrogen carbonate using double indicators.

Fill in the blanks below

During the titration of a mixture of sodium carbonate and sodium hydrogen carbonate against standard hydrochloric acid, the first equivalence point marking the half-neutralization of sodium carbonate to sodium hydrogen carbonate is indicated by , which turns from pink to at a pH range of approximately 8.3 to 10.0.

Show answer & explanation

Answer

The first equivalence point is detected using phenolphthalein indicator, which transitions from pink to colorless (or colourless).

Phenolphthalein is the correct indicator for the first equivalence point because its transition interval (pH 8.3–10.0) matches the pH of the sodium hydrogen carbonate solution formed when all carbonate ions are neutralized. Its color changes from pink in alkaline medium to colorless at the endpoint.

Step-by-Step Solution

1

Identify the chemical reaction occurring at the first equivalence point of the carbonate-bicarbonate titration.

At the first endpoint, all

\text{CO}_3^{2-}

ions are protonated to form

\text{HCO}_3^-

according to

\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-

. The pH at this stage is weakly alkaline (around pH 8.3).

Sodium hydrogen carbonate is an amphiprotic salt that forms a solution with a pH near 8.3, requiring an indicator with a pH transition range in the alkaline region.

2

Determine the appropriate indicator and its observed color change at this pH transition.

Phenolphthalein has a pH transition range of 8.3–10.0. In basic solution containing carbonate ions, it displays a pink color, and turns colorless once all carbonate ions are converted to hydrogen carbonate ions.

Methyl orange changes color at pH 3.1–4.4, which corresponds to the second equivalence point where hydrogen carbonate is completely converted to carbon dioxide and water.

Key Concept

Double Indicator Titration of Carbonate and Hydrogen Carbonate Mixtures

Acids, Bases and Salts

Answer

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