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201. According to Arrhenius, an acid is a substance that
ⓐ. Increases the concentration of \(H^+\) ions in aqueous solution
ⓑ. Increases the concentration of \(OH^-\) ions in aqueous solution
ⓒ. Accepts a proton from water
ⓓ. Neutralizes a base by donating electrons
Correct Answer: Increases the concentration of \(H^+\) ions in aqueous solution
Explanation: The Arrhenius concept defines acids as substances that dissociate in water to produce hydrogen ions \((H^+)\) or hydronium ions \((H_{3}O^+)\). This theory was the earliest formal definition of acids, explaining the acidic nature of compounds like (HCl) and \(HNO_{3}\) through ion formation in water.
202. According to Arrhenius, a base is a substance that
ⓐ. Accepts a proton from an acid
ⓑ. Produces \(OH^-\) ions in aqueous solution
ⓒ. Neutralizes \(H^+\) ions in gas phase
ⓓ. Increases hydrogen ion concentration
Correct Answer: Produces \(OH^-\) ions in aqueous solution
Explanation: Arrhenius defined bases as substances that release hydroxide ions \((OH^-)\) when dissolved in water. For example, \(NaOH \rightarrow Na^+ + OH^-\). The generation of these ions increases the basic nature of the solution by reducing the hydrogen ion concentration.
203. Which of the following is an example of an Arrhenius acid?
ⓐ. \(NH_{3}\)
ⓑ. (NaOH)
ⓒ. \(CH_{3}OH\)
ⓓ. (HCl)
Correct Answer: (HCl)
Explanation: Hydrochloric acid dissociates completely in water as \(HCl \rightarrow H^+ + Cl^-\). The increase in hydrogen ion concentration makes the solution acidic. This behavior perfectly aligns with Arrhenius’s definition of acids, which emphasizes ionization in aqueous medium.
204. Which of the following is an example of an Arrhenius base?
ⓐ. \(NH_{4}Cl\)
ⓑ. \(H_{2}SO_{4}\)
ⓒ. (NaOH)
ⓓ. \(CO_{2}\)
Correct Answer: (NaOH)
Explanation: Sodium hydroxide dissociates in water to yield \(Na^+\) and \(OH^-\) ions. The presence of hydroxide ions imparts basicity to the solution. Arrhenius’s theory uses this property to classify substances like (NaOH), (KOH), and \(Ca(OH)_2\) as bases.
205. According to Arrhenius theory, neutralization occurs when
ⓐ. \(H^+\) ions react with \(OH^-\) ions to form water
ⓑ. \(H^+\) reacts with base molecules directly
ⓒ. A salt ionizes completely
ⓓ. Two acids react to form hydrogen gas
Correct Answer: \(H^+\) ions react with \(OH^-\) ions to form water
Explanation: The neutralization process is the combination of hydrogen and hydroxide ions to form water molecules: \(H^+ + OH^- \rightarrow H_{2}O\). This reaction eliminates both acidic and basic properties, producing a neutral solution, which is the foundation of acid-base titration analysis.
206. Which of the following compounds cannot be classified as an Arrhenius acid or base?
ⓐ. (HCl)
ⓑ. \(HNO_{3}\)
ⓒ. (NaOH)
ⓓ. \(NH_{3}\)
Correct Answer: \(NH_{3}\)
Explanation: Ammonia does not contain hydroxide ions yet exhibits basic behavior. It reacts with water to form \(NH_{4}^+\) and \(OH^-\), showing its proton-accepting ability. Thus, while it behaves as a base, it does not directly fit the Arrhenius definition, which is limited to aqueous hydroxide ion producers.
207. The Arrhenius concept is limited to
ⓐ. Aqueous solutions only
ⓑ. Gaseous acid–base systems
ⓒ. Non-electrolytic solvents
ⓓ. Solid-state compounds
Correct Answer: Aqueous solutions only
Explanation: Arrhenius’s theory assumes dissociation into ions in water. It cannot explain acid–base behavior in non-aqueous or gaseous systems, such as \(NH_{3} + HCl \rightarrow NH_{4}Cl\). Later models like Brønsted–Lowry and Lewis extended the concept beyond aqueous ionization.
208. Which of the following best represents the neutralization of an acid and a base according to Arrhenius?
ⓐ. \(NaOH + HCl \rightarrow NaCl + H_{2}O\)
ⓑ. \(NH_{3} + HCl \rightarrow NH_{4}Cl\)
ⓒ. \(H_{2}O + H_{2} \rightarrow H_{3}O^+\)
ⓓ. \(CO_{2} + H_{2}O \rightarrow H_{2}CO_{3}\)
Correct Answer: \(NaOH + HCl \rightarrow NaCl + H_{2}O\)
Explanation: In this neutralization, hydrogen ions from (HCl) combine with hydroxide ions from (NaOH) to form water, while the remaining ions produce salt. This is a classic demonstration of the Arrhenius concept, where \(H^+\) and \(OH^-\) ions neutralize each other.
209. According to Arrhenius, which of the following aqueous solutions has the highest concentration of \(OH^-\) ions?
ⓐ. \(NH_{4}OH\)
ⓑ. \(CH_{3}COOH\)
ⓒ. (HCl)
ⓓ. (NaOH)
Correct Answer: (NaOH)
Explanation: Sodium hydroxide dissociates almost completely in water, yielding a high concentration of \(OH^-\) ions. This strong base has high ionic conductivity and causes a significant rise in pH, making it one of the strongest Arrhenius bases known.
210. Which of the following statements correctly describes a limitation of the Arrhenius concept?
ⓐ. It cannot explain basicity of substances without \(OH^-\) ions
ⓑ. It explains reactions in all solvents
ⓒ. It defines acids based on electron acceptance
ⓓ. It includes gaseous systems effectively
Correct Answer: It cannot explain basicity of substances without \(OH^-\) ions
Explanation: The Arrhenius concept cannot classify compounds like \(NH_{3}\), which act as bases by accepting protons from water but lack \(OH^-\) groups. This limitation led to broader definitions such as the Brønsted–Lowry theory, which considers proton transfer rather than ion release as the defining feature of acidity or basicity.
211. According to the Brønsted–Lowry concept, an acid is a substance that
ⓐ. Produces \(H^+\) ions in water
ⓑ. Produces \(OH^-\) ions in water
ⓒ. Accepts a proton
ⓓ. Donates a proton to another species
Correct Answer: Donates a proton to another species
Explanation: In the Brønsted–Lowry theory, acids are defined as proton donors. When an acid reacts, it transfers a proton \((H^+)\) to another molecule or ion. For example, (HCl) donates a proton to \(H_{2}O\) to form \(H_{3}O^+\). This definition broadens acid–base behavior beyond aqueous systems, covering gaseous and non-aqueous reactions too.
212. According to the Brønsted–Lowry concept, a base is a substance that
ⓐ. Produces \(H^+\) in water
ⓑ. Produces \(OH^-\) directly
ⓒ. Releases electrons
ⓓ. Accepts a proton from another substance
Correct Answer: Accepts a proton from another substance
Explanation: A Brønsted–Lowry base acts as a proton acceptor. For instance, ammonia \((NH_{3})\) reacts with water by accepting a proton to form \(NH_{4}^+\). This concept helps explain basic behavior even in the absence of hydroxide ions and extends the Arrhenius definition to more chemical systems.
213. In the reaction \(NH_{3} + H_{2}O \rightleftharpoons NH_{4}^+ + OH^-\), the Brønsted–Lowry base is
ⓐ. \(NH_{3}\)
ⓑ. \(NH_{4}^+\)
ⓒ. \(OH^-\)
ⓓ. \(H_{2}O\)
Correct Answer: \(NH_{3}\)
Explanation: Ammonia accepts a proton \((H^+)\) from water to form ammonium ions \((NH_{4}^+)\). Hence, \(NH_{3}\) acts as the Brønsted–Lowry base. Water, which donates the proton, behaves as the acid. This reaction also shows that water can act as both an acid and a base (amphoteric behavior).
214. In the reaction \(HCl + H_{2}O \rightarrow H_{3}O^+ + Cl^-\), the conjugate base of the acid (HCl) is
ⓐ. \(H_{3}O^+\)
ⓑ. \(H_{2}O\)
ⓒ. (HCl)
ⓓ. \(Cl^-\)
Correct Answer: \(Cl^-\)
Explanation: When (HCl) donates a proton to water, it forms its conjugate base \(Cl^-\). In the Brønsted–Lowry framework, conjugate pairs differ by one proton. Here, \(HCl/Cl^-\) is an acid–base pair, illustrating reversible proton transfer essential to dynamic acid–base equilibria.
215. The conjugate acid of the base \(NH_{3}\) is
ⓐ. \(H_{2}O\)
ⓑ. \(NH_{4}^+\)
ⓒ. \(NH_{2}^-\)
ⓓ. \(OH^-\)
Correct Answer: \(NH_{4}^+\)
Explanation: A conjugate acid forms when a base gains a proton. In this case, ammonia \((NH_{3})\) accepts a proton \((H^+)\) to yield ammonium ion \((NH_{4}^+)\). The conjugate acid–base relationship helps predict equilibrium positions in acid–base systems and quantify their relative strengths.
216. The conjugate base of \(H_{2}O\) is
ⓐ. \(H_{3}O^+\)
ⓑ. \(H^-\)
ⓒ. \(OH^-\)
ⓓ. \(O^{2-}\)
Correct Answer: \(OH^-\)
Explanation: When water donates a proton, it forms a hydroxide ion. Thus, \(H_{2}O\) acts as an acid, and \(OH^-\) is its conjugate base. This reaction underlies the self-ionization of water: \(2H_{2}O \rightleftharpoons H_{3}O^+ + OH^-\), which demonstrates that water is both a weak acid and a weak base.
217. In the reaction \(H_{2}CO_{3} \rightleftharpoons H^+ + HCO_{3}^-\), the conjugate base of \(H_{2}CO_{3}\) is
ⓐ. \(HCO_{3}^-\)
ⓑ. \(H_{2}O\)
ⓒ. \(CO_{3}^{2-}\)
ⓓ. \(OH^-\)
Correct Answer: \(HCO_{3}^-\)
Explanation: When carbonic acid donates one proton, it forms bicarbonate \((HCO_{3}^-)\), which can further act as a weak acid. The presence of multiple ionizable hydrogens makes \(H_{2}CO_{3}\) a polyprotic acid, capable of forming a series of conjugate bases by successive proton losses.
218. The pair \(NH_{4}^+ / NH_{3}\) is an example of
ⓐ. Oxidation–reduction pair
ⓑ. Electrolytic couple
ⓒ. Conjugate acid–base pair
ⓓ. Isomeric pair
Correct Answer: Conjugate acid–base pair
Explanation: The two species differ by a single proton. \(NH_{4}^+\) can donate a proton to regenerate \(NH_{3}\), and \(NH_{3}\) can accept a proton to form \(NH_{4}^+\). Such reversible relationships characterize conjugate acid–base pairs central to the Brønsted–Lowry theory.
219. In the Brønsted–Lowry theory, the strength of an acid is inversely related to
ⓐ. Concentration of \(OH^-\)
ⓑ. Basicity of its conjugate base
ⓒ. Solubility in water
ⓓ. Ionization energy
Correct Answer: Basicity of its conjugate base
Explanation: A strong acid donates protons easily, leaving behind a weak conjugate base. Conversely, weak acids have strong conjugate bases that readily accept protons. This inverse relationship allows comparison of acid and base strengths through their equilibrium constants.
220. Which of the following reactions illustrates amphoteric behavior of water?
ⓐ. \(H_{2}O + H_{2} \rightarrow H_{3}O^+\)
ⓑ. \(H_{2}O + CO_{2} \rightarrow H_{2}CO_{3}\)
ⓒ. \(H_{2}O + HCl \rightarrow H_{3}O^+ + Cl^-\) and \(H_{2}O + NH_{3} \rightarrow NH_{4}^+ + OH^-\)
ⓓ. \(2H_{2}O \rightarrow 2H_{2} + O_{2}\)
Correct Answer: \(H_{2}O + HCl \rightarrow H_{3}O^+ + Cl^-\) and \(H_{2}O + NH_{3} \rightarrow NH_{4}^+ + OH^-\)
Explanation: Water can act both as an acid (proton donor) and as a base (proton acceptor), depending on the reacting species. With (HCl), it accepts a proton, while with \(NH_{3}\), it donates one. This dual ability is called amphoteric behavior, making water a self-neutral substance under the Brønsted–Lowry framework.
221. According to the Lewis concept, a Lewis acid is defined as a species that
ⓐ. Accepts an electron pair to form a coordinate bond
ⓑ. Donates an electron pair to form a coordinate bond
ⓒ. Releases \(H^+\) in water
ⓓ. Produces \(OH^-\) in water
Correct Answer: Accepts an electron pair to form a coordinate bond
Explanation: A Lewis acid is an electron‐pair acceptor with an empty orbital (or the ability to expand its valence shell) that can accommodate a donated pair. Typical examples include \(BF_{3}\), \(AlCl_{3}\), and \(H^+\). When a base donates its lone pair, the acid–base interaction forms a dative (coordinate) \(\sigma\)-bond, creating an adduct. This electron‐pair view generalizes acid–base behavior beyond aqueous media and proton transfer.
222. Which of the following acts as a Lewis base in the reaction \(BF_{3} + :NH_{3} \rightarrow F_{3}B!\leftarrow NH_{3}\)?
ⓐ. \(BF_{3}\)
ⓑ. \(H^+\)
ⓒ. \(NH_{3}\)
ⓓ. \(BF_{4}^-\)
Correct Answer: \(NH_{3}\)
Explanation: Ammonia has a lone pair on nitrogen that it can donate. In the adduct \(F_{3}B!\leftarrow NH_{3}\), the arrow points from the donor to the acceptor, showing that the nitrogen lone pair fills the vacant (p)-orbital of boron. This donation stabilizes boron, producing a classic Lewis acid–base complex. The interaction is independent of proton transfer.
223. Which statement best explains why \(AlCl_{3}\) is a Lewis acid?
ⓐ. It donates an electron pair from aluminum
ⓑ. It accepts a proton to form \(AlCl_{3}H^+\)
ⓒ. It has an incomplete octet at aluminum and can accept a lone pair
ⓓ. It releases \(OH^-\) in aqueous solution
Correct Answer: It has an incomplete octet at aluminum and can accept a lone pair
Explanation: In \(AlCl_{3}\), aluminum is electron‐deficient (six valence electrons around Al), allowing it to accept a pair from a donor such as \(Cl^-\), \(H_{2}O\), or \(NH_{3}\). This electron deficiency drives dimerization to \(Al_{2}Cl_{6}\) or formation of adducts with bases. The Lewis definition focuses on orbital occupancy rather than the presence of \(H^+\) or \(OH^-\).
224. Identify the Lewis acid in the reaction \(Ag^+ + 2:NH_{3} \rightleftharpoons [Ag(NH_{3})_2]^+\).
ⓐ. \(NH_{3}\)
ⓑ. \(Cl^-\)
ⓒ. \(H_{2}O\)
ⓓ. \(Ag^+\)
Correct Answer: \(Ag^+\)
Explanation: The silver ion accepts electron pairs from two ammonia molecules to form linear \([Ag(NH_{3})_2]^+\). Each \(NH_{3}\) donates its lone pair into the empty (5s/5p) acceptor orbitals of \(Ag^+\), establishing coordinate bonds. The metal center functions as the electron‐pair acceptor, fitting the Lewis acid role precisely.
225. Which of the following is the best Lewis base toward \(BF_{3}\)?
ⓐ. \(CO_{2}\)
ⓑ. \(BF_{4}^-\)
ⓒ. \(F^-\)
ⓓ. \(H^+\)
Correct Answer: \(F^-\)
Explanation: Fluoride possesses high electron density and a strong donating ability toward the electron‐deficient boron center. Formation of \(BF_{4}^-\) is a classic stabilization pathway where the lone pair on \(F^-\) fills boron’s vacant orbital. The resulting tetrafluoroborate anion is a stable Lewis acid–base adduct widely encountered in inorganic chemistry.
226. In the adduct formation \(H^+ + :H_{2}O \rightarrow H_{3}O^+\), water behaves as a
ⓐ. Lewis base donating its lone pair to \(H^+\)
ⓑ. Lewis acid accepting a pair from \(H^+\)
ⓒ. Reducing agent transferring electrons to oxygen
ⓓ. Spectator that does not participate
Correct Answer: Lewis base donating its lone pair to \(H^+\)
Explanation: Oxygen in water has two lone pairs; one is donated to the empty (1s) orbital of \(H^+\) to create the coordinate bond in hydronium. This interpretation shows Arrhenius/Brønsted behavior as a special case of the Lewis framework: protonation is electron‐pair donation to a proton, a quintessential base action.
227. Which interaction most clearly illustrates Lewis acidity of a metal cation in solution?
ⓐ. \(Na^+ + Cl^- \rightarrow NaCl(s)\) lattice formation
ⓑ. \(HCl \rightarrow H^+ + Cl^-\) ionization in water
ⓒ. \(Cu^{2+} + 4NH_{3} \rightleftharpoons [Cu(NH_{3})_4]^{2+}\) complexation
ⓓ. \(CH_{3}COOH \rightleftharpoons H^+ + CH_{3}COO^-\) dissociation
Correct Answer: \(Cu^{2+} + 4NH_{3} \rightleftharpoons [Cu(NH_{3})_4]^{2+}\) complexation
Explanation: Transition‐metal cations accept electron pairs from ligands (Lewis bases) to form coordination complexes. Ammonia donates its lone pair to \(Cu^{2+}\), yielding a tetraammine complex with dative bonds. This electron‐pair acceptance is the hallmark of Lewis acidity and underlies coordination chemistry and bioinorganic binding.
228. Which pair correctly represents a Lewis acid–base adduct?
ⓐ. \(H_{2} + Cl_{2} \rightarrow 2HCl\) (radical path)
ⓑ. \(BF_{3} + :OEt_{2} \rightarrow F_{3}B!\leftarrow OEt_{2}\)
ⓒ. \(Na^+ + OH^- \rightarrow NaOH\) (ionic association)
ⓓ. \(H^+ + e^- \rightarrow H\cdot\) (reduction)
Correct Answer: \(BF_{3} + :OEt_{2} \rightarrow F_{3}B!\leftarrow OEt_{2}\)
Explanation: Diethyl ether donates a lone pair on oxygen into the vacant (p)-orbital of boron, producing a stable ether–borane adduct. The arrow notation indicates donation from base to acid. Such adducts are pervasive in catalysis and synthesis where Lewis acids activate substrates via electron‐pair acceptance.
229. Which statement best captures the scope advantage of the Lewis concept?
ⓐ. It applies only to aqueous solutions
ⓑ. It requires the presence of \(H^+\) or \(OH^-\)
ⓒ. It explains acid–base behavior through electron‐pair transfer, covering non-protic and gas-phase reactions
ⓓ. It excludes coordination compounds
Correct Answer: It explains acid–base behavior through electron‐pair transfer, covering non-protic and gas-phase reactions
Explanation: By focusing on electron‐pair donation and acceptance, the Lewis idea unifies protonic acid–base chemistry with coordination and organic electrophile–nucleophile reactions. It accommodates non-aqueous media, metal–ligand bonding, and adduct formation, providing a broad, mechanism-oriented framework.
230. According to Lewis acidity/basicity trends, which order of acid strength is most reasonable toward a hard donor like \(H_{2}O\)?
ⓐ. \(I_{2} > Br_{2} > Cl_{2}\)
ⓑ. \(BF_{3} > BCl_{3} > BBr_{3}\)
ⓒ. \(BBr_{3} > BCl_{3} > BF_{3}\)
ⓓ. \(NH_{3} > H_{2}O > F^-\)
Correct Answer: \(NH_{3} > H_{2}O > F^-\)
Explanation: Hard–hard interactions favor small, highly charged, less polarizable partners; conversely, as donors become harder (more basic charge density), their binding to very hard acids strengthens. Among the listed donors, \(F^-\) is the hardest and binds most strongly to hard acids; however, as a donor basicity scale toward hard acids, the decreasing polarizability and increasing charge density favor \(F^-\) over \(H_{2}O\) and \(NH_{3}\). Therefore, the most reasonable acid-strength ordering toward a hard donor is best reflected oppositely in donor basicity; matching hard–hard preferences leads to strongest interaction with \(F^-\), then \(H_{2}O\), then \(NH_{3}\).
231. A strong acid is one that
ⓐ. Completely ionizes in aqueous solution
ⓑ. Partially ionizes in aqueous solution
ⓒ. Does not ionize at all
ⓓ. Forms insoluble salts in water
Correct Answer: Completely ionizes in aqueous solution
Explanation: Strong acids like (HCl), \(HNO_{3}\), and \(H_{2}SO_{4}\) dissociate almost 100% in water, producing a high concentration of \(H^+\) or \(H_{3}O^+\) ions. Their ionization is irreversible under normal conditions, which gives them very low (pH) values and high electrical conductivity.
232. A weak acid is characterized by
ⓐ. High electrical conductivity
ⓑ. Complete ionization in water
ⓒ. Partial ionization in water
ⓓ. No dissociation at all
Correct Answer: Partial ionization in water
Explanation: Weak acids such as \(CH_{3}COOH\) and \(H_{2}CO_{3}\) ionize only slightly in water, establishing an equilibrium between undissociated and ionized forms. Their equilibrium constants \((K_a)\) are small, resulting in higher (pH) values compared to strong acids of similar concentration.
233. Which of the following is a strong acid?
ⓐ. (HCl)
ⓑ. \(CH_{3}COOH\)
ⓒ. \(H_{2}CO_{3}\)
ⓓ. \(H_{3}PO_{4}\)
Correct Answer: (HCl)
Explanation: Hydrochloric acid dissociates completely in water into \(H^+\) and \(Cl^-\) ions. Its ionization is nearly complete, resulting in high acidity and low (pH). This makes (HCl) a prototypical strong acid used widely in titrations and chemical analysis.
234. Which of the following is a weak base?
ⓐ. (NaOH)
ⓑ. (KOH)
ⓒ. \(NH_{3}\)
ⓓ. \(Ca(OH)_2\)
Correct Answer: \(NH_{3}\)
Explanation: Ammonia reacts with water incompletely to produce \(NH_{4}^+\) and \(OH^-\). The equilibrium constant \((K_b)\) for ammonia is small, so the concentration of hydroxide ions is limited. This partial ionization behavior is the hallmark of weak bases.
235. Strong bases are those that
ⓐ. Partially dissociate to give \(OH^-\) ions
ⓑ. Completely dissociate to give \(OH^-\) ions
ⓒ. Do not dissolve in water
ⓓ. Accept \(H^+\) slowly
Correct Answer: Completely dissociate to give \(OH^-\) ions
Explanation: Bases like (NaOH), (KOH), and \(Ba(OH)_2\) ionize fully in water to release hydroxide ions. Their dissociation produces high (pH) and strong alkaline behavior. They also conduct electricity efficiently because of the high concentration of mobile ions.
236. Which of the following acids is weak?
ⓐ. \(HNO_{3}\)
ⓑ. \(HClO_{4}\)
ⓒ. \(H_{2}SO_{4}\)
ⓓ. (HF)
Correct Answer: (HF)
Explanation: Although hydrofluoric acid contains hydrogen, it does not ionize completely due to strong hydrogen bonding between \(H^+\) and \(F^-\). This makes (HF) a weak acid despite the electronegativity of fluorine. It exhibits moderate conductivity and partial dissociation equilibrium in water.
237. The strength of an acid is determined by the magnitude of its
ⓐ. Ionization constant \((K_a)\)
ⓑ. Solubility product \((K_{sp})\)
ⓒ. Partition coefficient
ⓓ. Vapor pressure
Correct Answer: Ionization constant \((K_a)\)
Explanation: The acid dissociation constant \(K_a\) quantifies the degree of ionization of an acid. Larger \(K_a\) values indicate stronger acids, as more \(H^+\) ions are produced. Conversely, weak acids have smaller \(K_a\) values due to limited proton release in aqueous solution.
238. The conjugate base of a strong acid is always
ⓐ. Strong
ⓑ. Weak
ⓒ. Neutral
ⓓ. Amphoteric
Correct Answer: Weak
Explanation: When a strong acid donates a proton completely, its conjugate base has negligible tendency to reaccept it. Thus, the conjugate base of a strong acid is extremely weak. For example, \(Cl^-\) (from (HCl)) is a weak base with very low proton affinity.
239. The conjugate acid of a strong base is
ⓐ. Strong
ⓑ. Weak
ⓒ. Neutral
ⓓ. Amphiprotic
Correct Answer: Weak
Explanation: Strong bases like (NaOH) and (KOH) dissociate completely, and their conjugate acids \((Na^+), (K^+)\) have no proton-donating ability. Hence, they are considered extremely weak acids, showing no significant acidity in aqueous environments.
240. Among the following, which has the highest (pH) value for 0.1 M solution?
ⓐ. \(NH_{3}\)
ⓑ. (KOH)
ⓒ. \(Ca(OH)_2\)
ⓓ. (NaOH)
Correct Answer: (NaOH)
Explanation: Sodium hydroxide completely ionizes in water, producing a high concentration of \(OH^-\) ions. The resulting solution has a (pH) close to 13 for 0.1 M concentration. Its full dissociation and strong ionic character make it one of the strongest bases known in aqueous chemistry.
241. The ionization constant \((K_a)\) of an acid is a measure of its
ⓐ. Basic strength
ⓑ. Degree of ionization in water
ⓒ. Solubility in organic solvents
ⓓ. Tendency to donate electrons
Correct Answer: Degree of ionization in water
Explanation: The acid dissociation constant \(K_a = \dfrac{[H^+][A^-]}{[HA]}\) quantifies how much an acid ionizes in aqueous solution. A larger \(K_a\) value means the acid releases more \(H^+\) ions, making it stronger. Smaller \(K_a\) values indicate weaker acids that dissociate only slightly.
242. The base ionization constant \((K_b)\) expresses
ⓐ. Strength of a base in accepting protons
ⓑ. Solubility of a base in organic solvents
ⓒ. Neutralization capacity with strong acid
ⓓ. Conductivity of a salt solution
Correct Answer: Strength of a base in accepting protons
Explanation: For a weak base (B), \(K_b = \dfrac{[BH^+][OH^-]}{[B]}\). The larger the \(K_b\) value, the stronger the base and the more \(OH^-\) ions it generates. Weak bases have small \(K_b\) values, establishing equilibrium between the base and its conjugate acid.
243. The ionization constant of acetic acid is \(1.8 \times 10^{-5}\). This indicates it is
ⓐ. A strong acid
ⓑ. A very strong base
ⓒ. A weak acid
ⓓ. A neutral molecule
Correct Answer: A weak acid
Explanation: Small \(K_a\) values like \(10^{-5}\) reflect limited ionization and partial dissociation of the acid. Acetic acid establishes equilibrium between \(CH_{3}COOH\) and \(CH_{3}COO^-\), resulting in moderate \(H^+\) concentration and a (pH) around 3 for typical molar solutions.
244. The relationship between \(K_a\) and \(K_b\) for a conjugate acid–base pair is
ⓐ. \(K_a + K_b = K_w\)
ⓑ. \(K_a \times K_b = K_w\)
ⓒ. \(K_a / K_b = K_w\)
ⓓ. \(K_a = K_b\)
Correct Answer: \(K_a \times K_b = K_w\)
Explanation: For a conjugate acid–base pair in water, the product of their ionization constants equals the ionic product of water \((K_w = 1.0 \times 10^{-14}) at 25°C\). This relationship links the strengths of acids and their conjugate bases—if one is strong, the other must be weak.
245. If the \(K_b\) of \(NH_{3}\) is \(1.8 \times 10^{-5}\), then the \(K_a\) of its conjugate acid \(NH_{4}^+\) is
ⓐ. \(1.8 \times 10^{5}\)
ⓑ. \(1.0 \times 10^{-19}\)
ⓒ. \(5.6 \times 10^{-14}\)
ⓓ. \(5.6 \times 10^{-10}\)
Correct Answer: \(5.6 \times 10^{-10}\)
Explanation: From the relation \(K_a \times K_b = K_w = 1.0 \times 10^{-14}\), we get \(K_a = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}\). Thus, the conjugate acid \(NH_{4}^+\) is weak because its ionization constant is small, showing limited proton donation ability.
246. The smaller the value of \(K_a\), the
ⓐ. Stronger the acid
ⓑ. Weaker the acid
ⓒ. More neutral the acid
ⓓ. More basic the acid
Correct Answer: Weaker the acid
Explanation: A small \(K_a\) indicates minimal ionization and a low \(H^+\) concentration in solution. Weak acids exist mainly in undissociated form, with their conjugate bases being relatively stronger. Therefore, \(K_a\) values directly measure acid strength numerically.
247. If \(K_b = 4.0 \times 10^{-10}\) for a base, the strength of the base is
ⓐ. Strong
ⓑ. Moderate
ⓒ. Very strong
ⓓ. Weak
Correct Answer: Weak
Explanation: Small \(K_b\) values signify that the base does not ionize completely. It accepts protons or produces \(OH^-\) ions only to a limited extent, resulting in low alkalinity. Examples include pyridine and aniline, which are weak bases due to partial equilibrium in aqueous solution.
248. The expression for the acid dissociation constant of \(HA \rightleftharpoons H^+ + A^-\) is
ⓐ. \(K_a = \dfrac{[HA]}{[H^+][A^-]}\)
ⓑ. \(K_a = [H^+][A^-]\)
ⓒ. \(K_a = \dfrac{[H^+][A^-]}{[HA]}\)
ⓓ. \(K_a = \dfrac{[A^-]}{[H^+]}\)
Correct Answer: \(K_a = \dfrac{[H^+][A^-]}{[HA]}\)
Explanation: The equilibrium constant for acid ionization measures the ratio of product concentrations \((H^+) and (A^-)\) to the undissociated acid concentration. It reflects the tendency of an acid to donate a proton in solution, which is fundamental to determining (pH) and buffer properties.
249. For a weak acid with small \(K_a\), the degree of ionization \((\alpha)\) is approximately proportional to
ⓐ. \(\sqrt{K_a}\)
ⓑ. \(\sqrt{\dfrac{K_a}{C}}\)
ⓒ. \(\dfrac{K_a}{C}\)
ⓓ. \(C \times K_a\)
Correct Answer: \(\sqrt{\dfrac{K_a}{C}}\)
Explanation: For a weak acid (HA), \(\alpha = \sqrt{\dfrac{K_a}{C}}\) when ionization is small. Here (C) is the initial concentration. This relationship shows that ionization increases with decreasing acid concentration and larger \(K_a\) values. It is derived from the equilibrium expression assuming \((1-\alpha) \approx 1\).
250. Which of the following statements about \(K_a\) and \(K_b\) is true?
ⓐ. Strong acids have large \(K_a\) and small \(K_b\) values for their conjugate bases
ⓑ. Strong acids and their conjugate bases both have large (K) values
ⓒ. Weak bases have large \(K_b\) values
ⓓ. Strong acids always have \(K_a = K_b\)
Correct Answer: Strong acids have large \(K_a\) and small \(K_b\) values for their conjugate bases
Explanation: Because \(K_a \times K_b = K_w\) is constant, if an acid is strong \(large (K_a)\), its conjugate base must have a small \(K_b\). This reciprocal relationship explains why strong acids form very weak bases—once the acid dissociates completely, its conjugate base cannot effectively accept a proton again.
251. The degree of ionization \( \alpha \) of a weak monoprotic acid (HA) in water is defined as
ⓐ. The fraction of initial acid molecules that dissociate into ions
ⓑ. The total number of ions formed per molecule
ⓒ. The inverse of the equilibrium constant
ⓓ. The fraction of solvent molecules ionized
Correct Answer: The fraction of initial acid molecules that dissociate into ions
Explanation: For \(HA \rightleftharpoons H^+ + A^-\), if the initial concentration is (C) and the equilibrium concentrations of ions are \(C\alpha\), then \( \alpha = \dfrac{\text{moles dissociated}}{\text{initial moles}} \). It is a number between 0 and 1 and depends on acid strength, concentration, temperature, and medium. Larger \( \alpha \) means greater ionization and stronger acidic behavior.
252. For a weak monoprotic acid of initial concentration (C) with \( \alpha \ll 1 \), Ostwald’s dilution law gives
ⓐ. \(K_a \approx C\alpha\)
ⓑ. \(K_a \approx \dfrac{\alpha}{C}\)
ⓒ. \(K_a \approx C\alpha^{2}\)
ⓓ. \(K_a \approx \dfrac{C}{\alpha^{2}}\)
Correct Answer: \(K_a \approx C\alpha^{2}\)
Explanation: From \(K_a=\dfrac{C\alpha \cdot C\alpha}{C(1-\alpha)}=\dfrac{C\alpha^{2}}{1-\alpha}\), and for weak acids \( \alpha \) is small so \(1-\alpha \approx 1\). Thus \(K_a \approx C\alpha^{2}\). Equivalently, \( \alpha \approx \sqrt{\dfrac{K_a}{C}} \), showing that dilution (smaller (C)) increases the degree of ionization.
253. For a weak base (B) of concentration (C), the degree of ionization satisfies \(for ( \alpha \ll 1)\)
ⓐ. \(K_b \approx \dfrac{C}{\alpha^{2}}\)
ⓑ. \(K_b \approx C\alpha\)
ⓒ. \(K_b \approx \dfrac{\alpha}{C}\)
ⓓ. \(K_b \approx C\alpha^{2}\)
Correct Answer: \(K_b \approx C\alpha^{2}\)
Explanation: Writing \(B + H_{2}O \rightleftharpoons BH^+ + OH^-\) with equilibrium amounts \(C(1-\alpha),,C\alpha,,C\alpha\), one obtains \(K_b=\dfrac{(C\alpha)^{2}}{C(1-\alpha)}\). For small \( \alpha \), \(1-\alpha \approx 1\), so \(K_b \approx C\alpha^{2}\), or \( \alpha \approx \sqrt{\dfrac{K_b}{C}} \). Hence, dilution increases ionization of weak bases as well.
254. For a weak acid solution of concentration (C), the hydrogen ion concentration is approximately
ⓐ. \([H^+] \approx C\)
ⓑ. \([H^+] \approx \sqrt{K_a C}\)
ⓒ. \([H^+] \approx \dfrac{K_a}{C}\)
ⓓ. \([H^+] \approx K_a\)
Correct Answer: \([H^+] \approx \sqrt{K_a C}\)
Explanation: Using \(K_a \approx C\alpha^{2}\) and \([H^+] \approx C\alpha\) for a monoprotic weak acid, substitution gives \([H^+] \approx \sqrt{K_a C}\). This relation is central for quick pH estimates of weak acid solutions and shows how both intrinsic strength \((K_a)\) and concentration control acidity.
255. The degree of ionization of a weak electrolyte generally
ⓐ. Decreases with dilution
ⓑ. Is independent of temperature
ⓒ. Increases with dilution according to Ostwald’s law
ⓓ. Becomes unity for any concentration
Correct Answer: Increases with dilution according to Ostwald’s law
Explanation: Decreasing (C) makes \( \alpha \approx \sqrt{\dfrac{K}{C}} \) larger. Physically, dilution reduces ionic recombination, shifting the equilibrium toward more ions. This trend holds well for weak electrolytes where activity effects are modest and \( \alpha \) remains well below 1.
256. In the presence of a common ion, the degree of ionization of a weak acid (HA)
ⓐ. Becomes independent of \(K_a\)
ⓑ. Equals 1 at all concentrations
ⓒ. Decreases because added product suppresses dissociation
ⓓ. Increases because ionic strength rises
Correct Answer: Decreases because added product suppresses dissociation
Explanation: Adding a source of \(A^-\) (e.g., a soluble salt) increases the product term in the equilibrium expression, so the system shifts to the left to maintain \(K_a\). Consequently, fewer (HA) molecules ionize and \( \alpha \) drops. This is the common-ion effect, widely used in buffer and precipitation control.
257. For a weak monoprotic acid, the relation between \(pK_a\), (C), and the degree of ionization is
ⓐ. \(pK_a = \log C – 2\log \alpha\)
ⓑ. \(pK_a = 2\log \alpha – \log C\)
ⓒ. \(pK_a = -\log (C\alpha^{2})\)
ⓓ. \(pK_a = \log \dfrac{C}{\alpha^{2}}\)
Correct Answer: \(pK_a = \log \dfrac{C}{\alpha^{2}}\)
Explanation: From \(K_a \approx C\alpha^{2}\), taking negative logarithms gives \(pK_a = -\log K_a = -\log(C\alpha^{2}) = \log \dfrac{1}{C} + \log \dfrac{1}{\alpha^{2}} = \log \dfrac{C}{\alpha^{2}}\). This linearizes weak-acid ionization data and is useful for quick back-calculations of \( \alpha \) or (C).
258. For a weak base solution where \( \alpha \ll 1 \), the hydroxide ion concentration is
ⓐ. \([OH^-] \approx K_b\)
ⓑ. \([OH^-] \approx \sqrt{K_b C}\)
ⓒ. \([OH^-] \approx \dfrac{K_b}{C}\)
ⓓ. \([OH^-] \approx C\)
Correct Answer: \([OH^-] \approx \sqrt{K_b C}\)
Explanation: With \(K_b \approx C\alpha^{2}\) and \([OH^-] \approx C\alpha\), substitution yields \([OH^-] \approx \sqrt{K_b C}\). This mirrors the weak-acid case and allows estimating (pOH) (and hence (pH)) directly from \(K_b\) and the formal concentration.
259. For a weak electrolyte whose limiting molar conductivity at infinite dilution is \( \Lambda_m^\infty \) and measured molar conductivity at concentration (C) is \( \Lambda_m \), the degree of ionization is
ⓐ. \( \alpha = \dfrac{\Lambda_m^\infty}{\Lambda_m} \)
ⓑ. \( \alpha = \sqrt{\dfrac{\Lambda_m}{\Lambda_m^\infty}} \)
ⓒ. \( \alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty} \)
ⓓ. \( \alpha = \dfrac{C,\Lambda_m}{\Lambda_m^\infty} \)
Correct Answer: \( \alpha = \dfrac{\Lambda_m}{\Lambda_m^\infty} \)
Explanation: Kohlrausch’s law gives \( \Lambda_m = \alpha,\Lambda_m^\infty \) for weak electrolytes at low concentrations where ion–ion interactions are minimal. Hence the ratio of observed to limiting molar conductivity directly equals the degree of ionization, providing an experimental route to \( \alpha \).
260. Temperature generally affects degree of ionization of weak acids/bases by
ⓐ. Always decreasing \( \alpha \) for any electrolyte
ⓑ. Leaving \( \alpha \) unchanged due to kinetic control only
ⓒ. Increasing \( \alpha \) when ionization is endothermic
ⓓ. Forcing \( \alpha = 1 \) at high (T) regardless of (K)
Correct Answer: Increasing \( \alpha \) when ionization is endothermic
Explanation: If the ionization enthalpy is positive, raising temperature increases (K) by the van’t Hoff relation, and consequently \( \alpha \) rises. Many weak electrolyte ionizations are endothermic because energy is needed to separate charges and hydrate ions. Thus higher temperature typically enhances ionization for such systems.
261. Which of the following is a weak electrolyte?
ⓐ. ( NaOH )
ⓑ. ( HCl )
ⓒ. \( CH_{3}COOH \)
ⓓ. \( KNO_{3} \)
Correct Answer: \( CH_{3}COOH \)
Explanation: Acetic acid \(( CH_{3}COOH )\) ionizes only partially in water into \( CH_{3}COO^- \) and \( H^+ \). Its ionization constant \(( K_a = 1.8 \times 10^{-5} )\) indicates weak dissociation. Because few ions form, it has low electrical conductivity and follows Ostwald’s dilution law.
262. The weak base among the following is
ⓐ. ( NaOH )
ⓑ. \( NH_{4}OH \)
ⓒ. ( KOH )
ⓓ. \( Ca(OH)_2 \)
Correct Answer: \( NH_{4}OH \)
Explanation: Ammonium hydroxide \(( NH_{4}OH )\) ionizes slightly in water to form \( NH_{4}^+ \) and \( OH^- \). The equilibrium lies far to the left, so it’s a weak electrolyte. Its \( K_b \) is small \(( 1.8 \times 10^{-5} )\), resulting in a limited production of hydroxide ions.
263. Which of the following is not a weak electrolyte?
ⓐ. ( HCl )
ⓑ. \( NH_{4}OH \)
ⓒ. \( CH_{3}COOH \)
ⓓ. \( H_{2}CO_{3} \)
Correct Answer: ( HCl )
Explanation: Hydrochloric acid is a strong electrolyte because it ionizes completely into \( H^+ \) and \( Cl^- \) in aqueous solution. Weak electrolytes like \( CH_{3}COOH \) or \( NH_{4}OH \) ionize only partially, whereas strong acids and bases produce almost full dissociation.
264. The ionization equation for acetic acid is
ⓐ. \( CH_{3}COOH \rightarrow CH_{3}COOH \)
ⓑ. \( CH_{3}COOH \rightleftharpoons CH_{3}COO^- + H^+ \)
ⓒ. \( CH_{3}COOH + H_{2}O \rightarrow CH_{3}COO^- + H_{3}O^+ \)
ⓓ. \( CH_{3}COOH \rightarrow CO_{2} + H_{2} \)
Correct Answer: \( CH_{3}COOH + H_{2}O \rightarrow CH_{3}COO^- + H_{3}O^+ \)
Explanation: In water, acetic acid donates a proton to a water molecule, forming hydronium \(( H_{3}O^+ )\) and acetate \(( CH_{3}COO^- )\). This reversible process represents its partial ionization and weak electrolyte behavior.
265. Which of the following statements is true about \( NH_{4}OH \)?
ⓐ. It dissociates completely in water
ⓑ. It is a weak base and a weak electrolyte
ⓒ. It is a strong base with full ionization
ⓓ. It does not form any ions in solution
Correct Answer: It is a weak base and a weak electrolyte
Explanation: Ammonium hydroxide ionizes slightly: \( NH_{4}OH \rightleftharpoons NH_{4}^+ + OH^- \). Due to low ionization, it exhibits low conductivity and a moderate pH. It’s widely used in analytical chemistry for buffer and precipitation reactions.
266. Which of the following pairs contains both weak electrolytes?
ⓐ. ( HCl ) and ( NaOH )
ⓑ. \( CH_{3}COOH \) and \( NH_{4}OH \)
ⓒ. \( H_{2}SO_{4} \) and ( KOH )
ⓓ. \( HNO_{3} \) and \( NH_{3} \)
Correct Answer: \( CH_{3}COOH \) and \( NH_{4}OH \)
Explanation: Both acetic acid and ammonium hydroxide dissociate partially in aqueous solution, forming limited ions. Their equilibrium constants \(( K_a ) and ( K_b )\) are small, and they obey Ostwald’s dilution law, showing increased ionization with dilution.
267. Weak electrolytes have
ⓐ. High molar conductivity at all concentrations
ⓑ. Constant degree of ionization
ⓒ. Low molar conductivity at high concentration
ⓓ. Complete dissociation at all dilutions
Correct Answer: Low molar conductivity at high concentration
Explanation: At higher concentrations, weak electrolytes have small ionization fractions, so fewer ions are present to carry current. As the solution is diluted, \( \alpha \) increases, and molar conductivity rises toward \( \Lambda_m^\infty \).
268. Which of the following has the smallest degree of ionization in water?
ⓐ. \( NH_{4}OH \)
ⓑ. \( CH_{3}COOH \)
ⓒ. ( NaOH )
ⓓ. ( HCN )
Correct Answer: ( HCN )
Explanation: Hydrocyanic acid is a very weak acid with \( K_a \approx 6.2 \times 10^{-10} \). It dissociates negligibly in water, producing an extremely low \( H^+ \) concentration. Thus, it has one of the smallest degrees of ionization among common weak electrolytes.
269. Which of the following statements correctly describes acetic acid solution?
ⓐ. It has very high conductivity
ⓑ. It is neutral and contains no ions
ⓒ. It shows low conductivity due to partial ionization
ⓓ. It dissociates completely like ( HCl )
Correct Answer: It shows low conductivity due to partial ionization
Explanation: Because \( CH_{3}COOH \) ionizes slightly, the number of charge carriers (ions) is small, resulting in low electrical conductivity. The conductivity increases with dilution as more molecules ionize, demonstrating the relationship between \( \alpha \) and molar conductivity.
270. The main difference between a weak acid and a weak electrolyte is that
ⓐ. Every weak acid is necessarily a weak electrolyte
ⓑ. Weak electrolytes are always strong acids
ⓒ. Weak acids do not ionize at all
ⓓ. Weak electrolytes ionize completely in water
Correct Answer: Every weak acid is necessarily a weak electrolyte
Explanation: A weak acid like \( CH_{3}COOH \) ionizes only partially, producing few ions, hence behaving as a weak electrolyte. Weak electrolytes in general (including weak acids and bases) conduct electricity poorly because of incomplete dissociation and reversible ionization equilibria.
271. Ostwald’s dilution law applies to
ⓐ. Strong electrolytes
ⓑ. Weak electrolytes
ⓒ. Non-electrolytes
ⓓ. Metallic conductors
Correct Answer: Weak electrolytes
Explanation: Ostwald’s dilution law explains how the degree of ionization \(( \alpha )\) of a weak electrolyte varies inversely with concentration. For weak acids and bases, \( K = \dfrac{C\alpha^{2}}{1-\alpha} \approx C\alpha^{2} \). As the solution is diluted, ( C ) decreases and \( \alpha \) increases, resulting in higher ionization and conductivity.
272. According to Ostwald’s dilution law, for a weak acid ( HA ), the dissociation constant is given by
ⓐ. \( K_a = \dfrac{[HA]}{[H^+][A^-]} \)
ⓑ. \( K_a = C(1-\alpha)^{2} \)
ⓒ. \( K_a = \dfrac{C\alpha^{2}}{1-\alpha} \)
ⓓ. \( K_a = \dfrac{1-\alpha}{C\alpha^{2}} \)
Correct Answer: \( K_a = \dfrac{C\alpha^{2}}{1-\alpha} \)
Explanation: For a weak acid \( HA \rightleftharpoons H^+ + A^- \), at equilibrium, the concentration of ions is \( C\alpha \) each and of undissociated acid is \( C(1-\alpha) \). Substituting these values gives the above formula, linking \( K_a \), ( C ), and \( \alpha \).
273. For a weak acid where \( \alpha \) is very small, Ostwald’s dilution law simplifies to
ⓐ. \( K_a = \dfrac{C}{\alpha^{2}} \)
ⓑ. \( K_a = C(1-\alpha)^{2} \)
ⓒ. \( K_a = \dfrac{\alpha}{C} \)
ⓓ. \( K_a = C\alpha^{2} \)
Correct Answer: \( K_a = C\alpha^{2} \)
Explanation: When \( \alpha \) is small, \( 1-\alpha \approx 1 \). Thus, the denominator in \( K_a = \dfrac{C\alpha^{2}}{1-\alpha} \) becomes nearly unity, giving \( K_a \approx C\alpha^{2} \). This approximation holds well for weak acids and bases at low concentrations.
274. Ostwald’s dilution law is not applicable to
ⓐ. Weak acids
ⓑ. Weak bases
ⓒ. Strong electrolytes
ⓓ. Organic acids
Correct Answer: Strong electrolytes
Explanation: Strong electrolytes like ( HCl ) and ( NaOH ) ionize almost completely, and their ionization does not depend on dilution. Ostwald’s law assumes equilibrium between dissociated and undissociated molecules, which doesn’t exist for strong electrolytes.
275. According to Ostwald’s law, the degree of ionization \( \alpha \) is related to
ⓐ. \( \alpha = \sqrt{\dfrac{K_a}{C}} \)
ⓑ. \( \alpha = \sqrt{K_a C} \)
ⓒ. \( \alpha = K_a C \)
ⓓ. \( \alpha = \dfrac{C}{K_a} \)
Correct Answer: \( \alpha = \sqrt{\dfrac{K_a}{C}} \)
Explanation: From \( K_a = C\alpha^{2} \), we get \( \alpha = \sqrt{\dfrac{K_a}{C}} \). This shows that the degree of ionization increases with dilution (lower ( C )) and with increasing acid strength \(higher ( K_a )\).
276. Ostwald’s dilution law can also be written in terms of
ⓐ. ( pH )
ⓑ. \( pK_a \)
ⓒ. Molar conductivity
ⓓ. Vapor pressure
Correct Answer: Molar conductivity
Explanation: Since molar conductivity \(( \Lambda_m )\) is directly proportional to the degree of ionization \(( \alpha )\), we can write \( \Lambda_m = \alpha \Lambda_m^\infty \). Substituting \( \alpha \) from Ostwald’s law allows experimental determination of dissociation constants using conductivity data.
277. For a weak base ( BOH ), Ostwald’s dilution law is expressed as
ⓐ. \( K_b = \dfrac{C\alpha^{2}}{1-\alpha} \)
ⓑ. \( K_b = \dfrac{1-\alpha}{C\alpha^{2}} \)
ⓒ. \( K_b = C(1-\alpha)^{2} \)
ⓓ. \( K_b = C\alpha \)
Correct Answer: \( K_b = \dfrac{C\alpha^{2}}{1-\alpha} \)
Explanation: The derivation for weak bases parallels weak acids. For \( BOH \rightleftharpoons B^+ + OH^- \), equilibrium concentrations yield the same form of equation, showing that dilution increases ionization for both weak acids and bases.
278. Which of the following graphs best represents Ostwald’s dilution law?
ⓐ. \( \alpha \) decreases linearly with \( \sqrt{C} \)
ⓑ. \( \alpha \) decreases with dilution
ⓒ. \( \alpha \) remains constant with ( C )
ⓓ. \( \alpha \) increases as \( 1/\sqrt{C} \)
Correct Answer: \( \alpha \) increases as \( 1/\sqrt{C} \)
Explanation: From \( \alpha = \sqrt{\dfrac{K}{C}} \), \( \alpha \) is inversely proportional to \( \sqrt{C} \). Therefore, dilution (decreasing ( C )) increases the degree of ionization. The plot of \( \alpha \) versus \( 1/\sqrt{C} \) gives a straight line through the origin.
279. Ostwald’s dilution law helps to determine
ⓐ. \( K_a \) or \( K_b \) of weak electrolytes
ⓑ. \( K_{sp} \) of sparingly soluble salts
ⓒ. Solubility of gases in liquids
ⓓ. Reaction rate constants
Correct Answer: \( K_a \) or \( K_b \) of weak electrolytes
Explanation: By measuring the degree of ionization or molar conductivity at known concentration, one can calculate the dissociation constant using Ostwald’s equation. It is especially useful for experimentally determining the strength of weak acids and bases.
280. Ostwald’s dilution law fails for strong electrolytes because
ⓐ. \( \alpha \) becomes independent of ( C )
ⓑ. Ionization is complete and not governed by equilibrium
ⓒ. \( K_a \) becomes infinite
ⓓ. Activity coefficients are negligible
Correct Answer: Ionization is complete and not governed by equilibrium
Explanation: For strong electrolytes, dissociation is nearly 100%, and ionic interactions dominate over equilibrium effects. The assumption of reversible ionization with undissociated molecules no longer holds, so Ostwald’s dilution law cannot describe their behavior accurately.
281. The degree of ionization \(( \alpha )\) of a weak acid is related to its concentration (( C )) as
ⓐ. \( \alpha \propto C \)
ⓑ. \( \alpha \propto \dfrac{1}{C} \)
ⓒ. \( \alpha \propto \dfrac{1}{\sqrt{C}} \)
ⓓ. \( \alpha \propto \sqrt{C} \)
Correct Answer: \( \alpha \propto \dfrac{1}{\sqrt{C}} \)
Explanation: From Ostwald’s dilution law, \( \alpha = \sqrt{\dfrac{K_a}{C}} \). Since \( K_a \) is constant for a given acid, the degree of ionization increases as the concentration decreases, i.e., with dilution. This inverse square-root relationship explains why weak acids ionize more in dilute solutions.
282. When the concentration of a weak acid is decreased to one-fourth, the degree of ionization
ⓐ. Doubles
ⓑ. Becomes half
ⓒ. Remains unchanged
ⓓ. Becomes four times
Correct Answer: Doubles
Explanation: \( \alpha = \sqrt{\dfrac{K_a}{C}} \). If ( C ) becomes ( C/4 ), then \( \alpha’ = \sqrt{\dfrac{K_a}{C/4}} = 2\sqrt{\dfrac{K_a}{C}} = 2\alpha \). Thus, halving or reducing concentration increases ionization due to equilibrium shifting toward the ionic side.
283. The variation of the degree of ionization \(( \alpha )\) with concentration (( C )) for a weak electrolyte is represented by
ⓐ. \( \alpha \propto C \)
ⓑ. \( \alpha \propto \sqrt{C} \)
ⓒ. \( \alpha \propto \dfrac{1}{\sqrt{C}} \)
ⓓ. \( \alpha \propto C^{2} \)
Correct Answer: \( \alpha \propto \dfrac{1}{\sqrt{C}} \)
Explanation: As concentration decreases, \( \alpha \) increases according to Ostwald’s law. This relation clearly indicates that ionization depends inversely on the square root of concentration, explaining why conductivity rises with dilution for weak acids and bases.
284. For a weak acid solution, if concentration decreases 100 times, the degree of ionization increases approximately by a factor of
ⓐ. 2
ⓑ. 5
ⓒ. 10
ⓓ. 100
Correct Answer: 10
Explanation: From \( \alpha = \sqrt{\dfrac{K_a}{C}} \), if ( C ) decreases by 100 times, \( \alpha’ = \sqrt{100} \alpha = 10\alpha \). Thus, ionization increases tenfold when concentration decreases hundredfold. This shows a square-root dependence between \( \alpha \) and ( 1/C ).
285. The relation between hydrogen ion concentration \([H^+]\) and the initial concentration ( C ) of a weak acid is
ⓐ. \([H^+] = C\)
ⓑ. \([H^+] = K_a C\)
ⓒ. \([H^+] = \dfrac{K_a}{C}\)
ⓓ. \([H^+] = \sqrt{K_a C}\)
Correct Answer: \([H^+] = \sqrt{K_a C}\)
Explanation: For a weak monoprotic acid, \([H^+] = C\alpha\) and \(K_a = C\alpha^{2}\). Combining gives \([H^+] = \sqrt{K_a C}\). Hence, lower concentration increases \([H^+]\) relatively due to higher ionization, though absolute acidity decreases as total acid decreases.
286. The product of \( \alpha^{2} \) and ( C ) for a weak acid at a given temperature is
ⓐ. Variable
ⓑ. Constant
ⓒ. Always zero
ⓓ. Infinite
Correct Answer: Constant
Explanation: Since \( K_a = C\alpha^{2} \) for small \( \alpha \), their product remains constant at fixed temperature. If ( C ) decreases, \( \alpha \) must increase such that \( C\alpha^{2} \) remains unchanged, reflecting the equilibrium’s self-adjusting nature.
287. For a weak base of concentration ( C ), the hydroxide ion concentration is given by
ⓐ. \([OH^-] = C\)
ⓑ. \([OH^-] = K_b C\)
ⓒ. \([OH^-] = \sqrt{K_b C}\)
ⓓ. \([OH^-] = \dfrac{K_b}{C}\)
Correct Answer: \([OH^-] = \sqrt{K_b C}\)
Explanation: Weak bases ionize partially as \( BOH \rightleftharpoons B^+ + OH^- \). Using Ostwald’s approximation, \( K_b = C\alpha^{2} \) and \( [OH^-] = C\alpha = \sqrt{K_b C} \). Thus, decreasing ( C ) increases ionization and raises the proportion of \( OH^- \) ions.
288. Which statement is true about the relation between \( \alpha \) and ( C )?
ⓐ. \( \alpha \) decreases with dilution
ⓑ. \( \alpha \) increases with dilution
ⓒ. \( \alpha \) is independent of ( C )
ⓓ. \( \alpha \) decreases as ( K ) increases
Correct Answer: \( \alpha \) increases with dilution
Explanation: Lowering concentration weakens interionic interactions, shifting equilibrium toward greater ionization. This means that diluting a weak electrolyte enhances its degree of ionization, which is why conductivity increases with dilution.
289. For a weak acid with \( K_a = 1.6 \times 10^{-5} \) and \( C = 0.1,M \), the degree of ionization is
ⓐ. 0.001
ⓑ. 0.1
ⓒ. 0.04
ⓓ. 0.0126
Correct Answer: 0.0126
Explanation: \( \alpha = \sqrt{\dfrac{K_a}{C}} = \sqrt{\dfrac{1.6 \times 10^{-5}}{0.1}} = 0.0126 \). Hence, only about 1.26% of the acid molecules ionize, confirming it as a weak electrolyte with partial dissociation even at moderate concentration.
290. If \( C_{1} \) and \( C_{2} \) are two concentrations of the same weak acid and \( \alpha_{1}, \alpha_{2} \) are their ionizations, then
ⓐ. \( C_{1} \alpha_{1}^{2} = C_{2} \alpha_{2}^{2} \)
ⓑ. \( C_{1} \alpha_{1} = C_{2} \alpha_{2} \)
ⓒ. \( \alpha_{1} C_{1} = \alpha_{2} C_{2}^{2} \)
ⓓ. \( \alpha_{1}^{2} C_{1}^{2} = \alpha_{2}^{2} C_{2} \)
Correct Answer: \( C_{1} \alpha_{1}^{2} = C_{2} \alpha_{2}^{2} \)
Explanation: Since \( K_a = C\alpha^{2} \) is constant for a given acid at constant temperature, the product \( C\alpha^{2} \) remains unchanged for different concentrations. This equality forms the mathematical basis of the relation between concentration and ionization in weak electrolytes.
291. The dielectric constant of a solvent affects the ionization of an electrolyte because it
ⓐ. Increases the energy of ion formation
ⓑ. Decreases interionic attraction, thus increasing ionization
ⓒ. Makes ions combine faster
ⓓ. Does not affect ionization at all
Correct Answer: Decreases interionic attraction, thus increasing ionization
Explanation: A solvent with a higher dielectric constant reduces the electrostatic force between oppositely charged ions \(( F = \dfrac{1}{D} )\). This weakening of ionic attraction allows more dissociation, increasing the degree of ionization. Hence, water \(high ( D = 80 )\) is an excellent ionizing solvent.
292. The degree of ionization of an electrolyte is higher in solvents having
ⓐ. High dielectric constant
ⓑ. High viscosity
ⓒ. Low dielectric constant
ⓓ. Non-polar character
Correct Answer: High dielectric constant
Explanation: Polar solvents with large dielectric constants, like water and formamide, stabilize ions by reducing electrostatic forces between them. This facilitates dissociation and enhances ionization. Non-polar solvents like benzene or ether have low ( D ) and thus poor ionizing power.
293. If the dielectric constant of a medium increases, the force of attraction between ions
ⓐ. Increases linearly
ⓑ. Decreases
ⓒ. Remains constant
ⓓ. First increases then decreases
Correct Answer: Decreases
Explanation: According to Coulomb’s law, \( F = \dfrac{q_{1} q_{2}}{4\pi \varepsilon_{0} D r^{2}} \). The force ( F ) is inversely proportional to the dielectric constant ( D ). A higher ( D ) reduces attraction between charged particles, promoting ion separation and increasing electrolyte dissociation.
294. The low ionizing power of solvents like benzene is due to
ⓐ. Low dielectric constant
ⓑ. High dielectric constant
ⓒ. High solvation energy
ⓓ. Presence of hydrogen bonds
Correct Answer: Low dielectric constant
Explanation: Benzene \(( D \approx 2.3 )\) is non-polar and cannot stabilize ions effectively. The weak solvation and high interionic attraction prevent dissociation, making benzene a poor medium for ionic compounds. Therefore, ionization is minimal in low-dielectric media.
295. Which of the following solvents would favor maximum ionization of NaCl?
ⓐ. Benzene
ⓑ. Ether
ⓒ. Ethanol
ⓓ. Water
Correct Answer: Water
Explanation: Water has a very high dielectric constant \(( D = 80 ) at 25°C\) and is strongly polar. It reduces electrostatic attractions between \( Na^+ \) and \( Cl^- \) ions, allowing nearly complete dissociation. Hydrogen bonding further stabilizes the hydrated ions, enhancing solubility and ionization.
296. The relation between force of attraction ( F ) and dielectric constant ( D ) is
ⓐ. \( F \propto D \)
ⓑ. \( F \propto D^{2} \)
ⓒ. \( F \propto \dfrac{1}{D} \)
ⓓ. \( F \propto \dfrac{1}{D^{2}} \)
Correct Answer: \( F \propto \dfrac{1}{D} \)
Explanation: Coulomb’s law in a medium is \( F = \dfrac{q_{1} q_{2}}{4\pi \varepsilon_{0} D r^{2}} \). The higher the dielectric constant, the smaller the electrostatic force between ions. Thus, solvents with large ( D ) values favor ionization and stabilize charged species efficiently.
297. Which of the following pairs correctly shows the order of increasing ionization?
ⓐ. Water < Benzene < Ethanol
ⓑ. Benzene < Ethanol < Water
ⓒ. Ethanol < Benzene < Water
ⓓ. Water < Ethanol < Benzene
Correct Answer: Benzene < Ethanol < Water
Explanation: The dielectric constants are: benzene (2.3), ethanol (24.3), and water (80). The ionization capacity increases with ( D ), so benzene shows minimal ionization, ethanol moderate, and water the highest. Solvent polarity thus controls ionic dissociation.
298. The ability of a solvent to dissolve ionic compounds depends mainly on its
ⓐ. Density
ⓑ. Color
ⓒ. Dielectric constant
ⓓ. Molecular weight
Correct Answer: Dielectric constant
Explanation: The dielectric constant measures a solvent’s capacity to weaken electrostatic interactions between ions. A high value implies greater stabilization of separated ions through solvation and electrostatic screening, promoting dissolution and ionization of electrolytes.
299. The dielectric constant of water at 25°C is approximately
ⓐ. 18
ⓑ. 24
ⓒ. 55
ⓓ. 80
Correct Answer: 80
Explanation: Water’s high dielectric constant (around 80) arises from its strong polarity and hydrogen bonding network. This allows it to reduce interionic attractions drastically, making it a universal solvent capable of dissolving many ionic and polar compounds efficiently.
300. Ionization of weak acids like acetic acid increases when
ⓐ. The dielectric constant of the solvent decreases
ⓑ. The dielectric constant of the solvent increases
ⓒ. The solvent is replaced by benzene
ⓓ. Ionic strength of solution decreases
Correct Answer: The dielectric constant of the solvent increases
Explanation: A solvent with higher ( D ) reduces the electrostatic force between ions, thereby shifting equilibrium toward greater ionization. Water, with its high dielectric constant, enhances the dissociation of weak acids, while non-polar solvents suppress it almost completely.
Welcome to Class 11 Chemistry MCQs – Chapter 7: Equilibrium (Part 3). In this section, we explore deeper into the equilibrium expressions for gaseous reactions and solutions. The MCQs here emphasize advanced concepts, including the effect of temperature on equilibrium, the Van’t Hoff factor, and applications of equilibrium in real-world scenarios like the Haber process and contact process.
Topics covered in Part 3 (100 MCQs): Temperature’s effect on equilibrium, the Van’t Hoff factor, and industrial applications like the production of ammonia and sulfuric acid. You will also tackle questions on phase equilibria and the relationship between free energy and equilibrium constant.
How to practice here
- Answer the MCQs and immediately review explanations to identify learning gaps.
- Mark essential MCQs with the ❤️ Heart feature, then toggle Favourite to filter them for easy access.
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- For better recall, use Random to shuffle the questions during your review sessions.
- 👉 Total in chapter: 400 MCQs (100 + 100 + 100 + 100)
- 👉 This page: Third set of 100 solved MCQs
- 👉 Next: Move on to Part 4 for the final set of MCQs