Timer: Off
Random: Off
301. The pH of a solution is defined as
ⓐ. \( pH = \log [H^+] \)
ⓑ. \( pH = -\log [H^+] \)
ⓒ. \( pH = [H^+] \times 10 \)
ⓓ. \( pH = \dfrac{1}{[H^+]} \)
Correct Answer: \( pH = -\log [H^+] \)
Explanation: The pH concept, introduced by Sørensen, measures the acidity of a solution based on hydrogen ion concentration. It provides a convenient scale to express very small \( [H^+] \) values. For example, \( [H^+] = 1 \times 10^{-3} \) gives \( pH = 3 \), indicating an acidic solution.
302. The pOH of a solution is defined as
ⓐ. \( pOH = \log [OH^-] \)
ⓑ. \( pOH = -\log [OH^-] \)
ⓒ. \( pOH = 10 \times [OH^-] \)
ⓓ. \( pOH = \dfrac{1}{[OH^-]} \)
Correct Answer: \( pOH = -\log [OH^-] \)
Explanation: Similar to pH, pOH expresses the alkalinity of a solution on a logarithmic scale. Lower pOH values correspond to higher \( [OH^-] \) concentrations, indicating a more basic solution. The logarithmic format compresses large variations in hydroxide concentration into a manageable range (0–14).
303. The relationship between pH and pOH at 25°C is
ⓐ. \( pH + pOH = 10 \)
ⓑ. \( pH + pOH = 7 \)
ⓒ. \( pH + pOH = 12 \)
ⓓ. \( pH + pOH = 14 \)
Correct Answer: \( pH + pOH = 14 \)
Explanation: At 25°C, \( K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \). Taking logarithms gives \( pK_w = 14 = pH + pOH \). This relation links acidity and basicity: if pH decreases, pOH must increase by the same amount, maintaining the sum constant.
304. If the hydrogen ion concentration of a solution is \( 1 \times 10^{-4} , M \), its pH is
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 4
Explanation: Using \( pH = -\log [H^+] \), we find \( pH = -\log(10^{-4}) = 4 \). Thus, the solution is acidic. The logarithmic nature of the scale means each unit change in pH corresponds to a tenfold change in \( [H^+] \).
305. A solution has \( [OH^-] = 1 \times 10^{-3} , M \). The pOH and pH values are
ⓐ. pOH = 2, pH = 12
ⓑ. pOH = 3, pH = 11
ⓒ. pOH = 3, pH = 14
ⓓ. pOH = 3, pH = 10
Correct Answer: pOH = 3, pH = 11
Explanation: \( pOH = -\log [OH^-] = -\log(10^{-3}) = 3 \). From \( pH + pOH = 14 \), we get \( pH = 11 \). The solution is therefore basic, with a hydroxide concentration higher than that of pure water.
306. For a neutral solution at 25°C,
ⓐ. \( [H^+] = [OH^-] = 1 \times 10^{-7} , M \)
ⓑ. \( [H^+] = [OH^-] = 1 \times 10^{-5} , M \)
ⓒ. \( [H^+] = 1 \times 10^{-7} , M, [OH^-] = 1 \times 10^{-5} , M \)
ⓓ. \( [H^+] = 1 \times 10^{-5} , M, [OH^-] = 1 \times 10^{-9} , M \)
Correct Answer: \( [H^+] = [OH^-] = 1 \times 10^{-7} , M \)
Explanation: Pure water undergoes self-ionization to produce equal hydrogen and hydroxide ion concentrations of \( 10^{-7} , M \). Therefore, \( pH = 7 \) and \( pOH = 7 \), marking neutrality at 25°C.
307. A solution with \( pH = 2 \) has hydrogen ion concentration
ⓐ. \( 1 \times 10^{-2} , M \)
ⓑ. \( 1 \times 10^{-12} , M \)
ⓒ. \( 2 \times 10^{-1} , M \)
ⓓ. \( 5 \times 10^{-7} , M \)
Correct Answer: \( 1 \times 10^{-2} , M \)
Explanation: \( pH = -\log [H^+] \) implies \( [H^+] = 10^{-pH} \). Substituting \( pH = 2 \) gives \( [H^+] = 10^{-2} , M \). This concentration corresponds to a strongly acidic solution.
308. If \( pH = 9 \), the hydroxide ion concentration is
ⓐ. \( 1 \times 10^{-9} , M \)
ⓑ. \( 1 \times 10^{-2} , M \)
ⓒ. \( 1 \times 10^{-3} , M \)
ⓓ. \( 1 \times 10^{-5} , M \)
Correct Answer: \( 1 \times 10^{-5} , M \)
Explanation: From \( pH + pOH = 14 \), \( pOH = 5 \). Then, \( [OH^-] = 10^{-pOH} = 10^{-5} \)
309. The product \( [H^+][OH^-] \) at 25°C equals
ⓐ. \( 1 \times 10^{-12} \)
ⓑ. \( 1 \times 10^{-14} \)
ⓒ. \( 1 \times 10^{-10} \)
ⓓ. \( 1 \times 10^{-16} \)
Correct Answer: \( 1 \times 10^{-14} \)
Explanation: The ionic product of water, \( K_w = [H^+][OH^-] = 10^{-14} \) at 25°C, is constant for all aqueous solutions. When one ion concentration increases, the other decreases so that their product remains constant.
310. If a solution has \( pOH = 4 \), what is its ( pH ) value?
ⓐ. 4
ⓑ. 6
ⓒ. 7
ⓓ. 10
Correct Answer: 10
Explanation: Using the relationship \( pH + pOH = 14 \), substituting \( pOH = 4 \) gives \( pH = 10 \). A high pH signifies a basic solution, where hydroxide ions predominate over hydrogen ions in concentration.
311. The pH of a (0.01,M) solution of (HCl) is
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 2
Explanation: (HCl) is a strong acid and dissociates completely in water. Hence, \([H^+] = 0.01 = 10^{-2}\). Using \(pH = -\log[H^+]\), we get \(pH = 2\). The complete ionization of strong acids makes their pH directly related to molarity.
312. The pH of a (0.001,M) (NaOH) solution is
ⓐ. 3
ⓑ. 13
ⓒ. 12
ⓓ. 11
Correct Answer: 11
Explanation: For strong bases, \([OH^-] = C\). Hence, \([OH^-] = 10^{-3},M\). \(pOH = -\log[OH^-] = 3\). From \(pH + pOH = 14\), \(pH = 11\). The solution is strongly basic because hydroxide ion concentration is high.
313. The pH of a (0.1,M) solution of \(HNO_{3}\) is
ⓐ. 0
ⓑ. 1
ⓒ. 2
ⓓ. 3
Correct Answer: 1
Explanation: Nitric acid is a strong monoprotic acid, ionizing completely: \([H^+] = 0.1 = 10^{-1}\). Thus, \(pH = -\log(10^{-1}) = 1\). This reflects the high acidity of concentrated strong acid solutions.
314. The pH of a (0.05,M) (KOH) solution is
ⓐ. 12.3
ⓑ. 1.7
ⓒ. 3.0
ⓓ. 11.3
Correct Answer: 12.3
Explanation: (KOH) is a strong base, fully dissociated, giving \([OH^-] = 0.05,M\). Then \(pOH = -\log(0.05) = 1.3\). Using \(pH = 14 – pOH\), \(pH = 12.7\) \(≈12.3\). The high pH confirms strong alkalinity.
315. The (pH) of a (0.1,M) acetic acid solution \((K_a = 1.8 \times 10^{-5})\) is approximately
ⓐ. 3.0
ⓑ. 2.0
ⓒ. 5.0
ⓓ. 4.8
Correct Answer: 3.0
Explanation: For a weak acid, \([H^+] = \sqrt{K_a C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = 1.34 \times 10^{-3}\). Therefore, \(pH = -\log[H^+] = 2.87 \approx 3.0\). Partial ionization makes pH higher than that of a strong acid of equal molarity.
316. The (pH) of a (0.2,M) \(NH_{4}OH\) solution \((K_b = 1.8 \times 10^{-5})\) is
ⓐ. 10.3
ⓑ. 11.3
ⓒ. 8.7
ⓓ. 9.3
Correct Answer: 11.3
Explanation: For a weak base, \([OH^-] = \sqrt{K_b C} = \sqrt{1.8 \times 10^{-5} \times 0.2} = 1.9 \times 10^{-3}\). Then \(pOH = -\log(1.9 \times 10^{-3}) = 2.72\), so \(pH = 14 – 2.72 = 11.28 \approx 11.3\).
317. What is the (pH) of a (0.001,M) (HCl) solution?
ⓐ. 1
ⓑ. 2
ⓒ. 3
ⓓ. 4
Correct Answer: 3
Explanation: Since (HCl) is a strong acid, \([H^+] = 0.001 = 10^{-3}\). Hence \(pH = -\log(10^{-3}) = 3\). Even such dilute solutions remain strongly acidic because of complete dissociation.
318. The (pH) of a (0.05,M) \(NH_{4}OH\) solution \((K_b = 1.8 \times 10^{-5})\) is
ⓐ. 9.9
ⓑ. 10.6
ⓒ. 11.3
ⓓ. 8.0
Correct Answer: 10.6
Explanation: For a weak base, \([OH^-] = \sqrt{K_b C} = \sqrt{1.8 \times 10^{-5} \times 0.05} = 9.5 \times 10^{-4}\). \(pOH = 3.02\) and \(pH = 14 – 3.02 = 10.98 \approx 10.6\). Partial ionization lowers the pH compared to strong bases.
319. The (pH) of (0.1,M) \(CH_{3}COOH\) increases when the solution is diluted because
ⓐ. \(K_a\) decreases with dilution
ⓑ. Degree of ionization increases with dilution
ⓒ. \(H^+\) concentration becomes constant
ⓓ. Conductivity decreases
Correct Answer: Degree of ionization increases with dilution
Explanation: From Ostwald’s law, \( \alpha = \sqrt{\dfrac{K_a}{C}} \). As concentration decreases, \( \alpha \) increases, producing more \(H^+\) ions relative to total acid molecules. Thus, the relative acidity (ionization) increases though \( [H^+] \) decreases numerically.
320. The (pH) of a (0.01,M) (NaOH) solution is
ⓐ. 2
ⓑ. 8
ⓒ. 12
ⓓ. 10
Correct Answer: 12
Explanation: (NaOH) is a strong base, completely ionized. Therefore, \([OH^-] = 0.01 = 10^{-2}\), \(pOH = 2\), and \(pH = 14 – 2 = 12\). The high pH corresponds to the strongly basic nature of the solution.
321. The common ion effect refers to
ⓐ. Increased ionization of a weak electrolyte by adding a strong acid
ⓑ. Decrease in ionization of a weak electrolyte by adding a strong electrolyte with a common ion
ⓒ. Complete dissociation of an electrolyte due to a common ion
ⓓ. Change in ionization constant due to temperature
Correct Answer: Decrease in ionization of a weak electrolyte by adding a strong electrolyte with a common ion
Explanation: The presence of a common ion suppresses the dissociation of a weak electrolyte according to Le Chatelier’s principle. For example, adding \(CH_{3}COONa\) to \(CH_{3}COOH\) reduces its ionization because of the increase in acetate ion \((CH_{3}COO^-)\) concentration.
322. The ionization of acetic acid decreases on addition of sodium acetate because
ⓐ. The \(CH_{3}COO^-\) ion concentration increases
ⓑ. The \(H^+\) ion concentration increases
ⓒ. The temperature increases
ⓓ. The solution becomes basic
Correct Answer: The \(CH_{3}COO^-\) ion concentration increases
Explanation: Sodium acetate dissociates completely, adding acetate ions. The equilibrium \(CH_{3}COOH \rightleftharpoons H^+ + CH_{3}COO^-\) shifts to the left to oppose the change, decreasing ionization. This illustrates the suppression of weak acid dissociation by its conjugate base salt.
323. The common ion effect is an application of
ⓐ. Dalton’s law
ⓑ. Raoult’s law
ⓒ. Le Chatelier’s principle
ⓓ. Henry’s law
Correct Answer: Le Chatelier’s principle
Explanation: When a common ion is introduced, the equilibrium shifts in the direction that reduces the stress — in this case, toward the undissociated molecule. Hence, ionization decreases. This is a direct consequence of Le Chatelier’s equilibrium principle.
324. The dissociation of \(NH_{4}OH\) is suppressed by adding
ⓐ. \(NH_{4}Cl\)
ⓑ. (NaOH)
ⓒ. (HCl)
ⓓ. (NaCl)
Correct Answer: \(NH_{4}Cl\)
Explanation: \(NH_{4}Cl\) provides \(NH_{4}^+\) ions, common to the weak base equilibrium \(NH_{4}OH \rightleftharpoons NH_{4}^+ + OH^-\). The extra \(NH_{4}^+\) shifts equilibrium to the left, lowering ionization and hydroxide concentration.
325. The pH of a weak acid solution increases upon addition of its salt because
ⓐ. The acid becomes stronger
ⓑ. The salt increases common ion concentration
ⓒ. Ionization constant decreases
ⓓ. The number of \(H^+\) ions increases
Correct Answer: The salt increases common ion concentration
Explanation: Addition of the conjugate salt introduces a common ion, reducing the acid’s ionization. This lowers \([H^+]\) and increases pH. This effect is the foundation of buffer solution preparation using weak acids and their salts.
326. Which of the following pairs will show a common ion effect?
ⓐ. (HCl) and (NaOH)
ⓑ. \(NH_{4}OH\) and (NaOH)
ⓒ. \(H_{2}SO_{4}\) and (NaOH)
ⓓ. \(CH_{3}COOH\) and \(CH_{3}COONa\)
Correct Answer: \(CH_{3}COOH\) and \(CH_{3}COONa\)
Explanation: Both compounds share the common ion \(CH_{3}COO^-\). The acetate ions from sodium acetate suppress the ionization of acetic acid by shifting its equilibrium toward the undissociated acid form.
327. Which of the following statements is true regarding the common ion effect?
ⓐ. It increases ionization of weak acids
ⓑ. It affects only strong electrolytes
ⓒ. It reduces ionization of weak acids and bases
ⓓ. It changes \(K_a\) or \(K_b\) values
Correct Answer: It reduces ionization of weak acids and bases
Explanation: The effect suppresses ionization of weak electrolytes by adding strong electrolytes sharing a common ion. The equilibrium constant \(K_a\) or \(K_b\) remains constant; only the degree of ionization changes.
328. In a mixture of acetic acid and sodium acetate, the concentration of \(H^+\) ions is governed mainly by
ⓐ. The ionization constant of sodium acetate
ⓑ. The ionization constant of acetic acid and the ratio of salt to acid
ⓒ. The ionic product of water
ⓓ. The molarity of the salt alone
Correct Answer: The ionization constant of acetic acid and the ratio of salt to acid
Explanation: For the buffer solution, \( [H^+] = K_a \dfrac{[acid]}{[salt]} \). This expression results from applying the equilibrium law and the common ion effect. The pH depends logarithmically on the ratio of salt to acid.
329. The degree of ionization of a weak acid in the presence of its strong electrolyte salt is
ⓐ. Increased
ⓑ. Becomes unity
ⓒ. Unchanged
ⓓ. Decreased
Correct Answer: Decreased
Explanation: The additional ions from the strong electrolyte increase the total ionic concentration, causing the equilibrium to shift backward. Thus, the weak acid dissociates less, reducing the fraction of molecules ionized.
330. When \(NH_{4}Cl\) is added to an \(NH_{4}OH\) solution, the concentration of \(OH^-\) ions
ⓐ. Increases
ⓑ. Decreases
ⓒ. Remains the same
ⓓ. Becomes zero
Correct Answer: Decreases
Explanation: The \(NH_{4}^+\) ions from \(NH_{4}Cl\) suppress the ionization of \(NH_{4}OH\) by shifting equilibrium leftward. As a result, fewer \(OH^-\) ions are produced, decreasing alkalinity and pH. This controlled suppression helps maintain buffer capacity in ammonium systems.
331. A salt formed by a strong acid and a strong base (e.g., ( NaCl )) gives a solution that is
ⓐ. Acidic
ⓑ. Basic
ⓒ. Neutral
ⓓ. Amphoteric
Correct Answer: Neutral
Explanation: (NaCl) results from complete neutralization between strong acid (HCl) and strong base (NaOH). Both ions \((Na^+) and (Cl^-)\) come from strong electrolytes and do not hydrolyze in water, leaving the solution neutral with \(pH = 7\).
332. A salt formed by a strong acid and a weak base \(e.g., (NH_{4}Cl)\) gives a solution that is
ⓐ. Acidic
ⓑ. Basic
ⓒ. Neutral
ⓓ. Amphoteric
Correct Answer: Acidic
Explanation: In \(NH_{4}Cl\), \(NH_{4}^+\) \(from weak base (NH_{4}OH)\) hydrolyzes to produce \(H^+\) ions, while \(Cl^-\) (from strong acid (HCl)) does not. The resulting solution contains excess \(H^+\), making it acidic with (pH < 7).
333. A salt of a weak acid and a strong base \(e.g., (CH_{3}COONa)\) gives a solution that is
ⓐ. Acidic
ⓑ. Basic
ⓒ. Neutral
ⓓ. Amphoteric
Correct Answer: Basic
Explanation: The acetate ion \((CH_{3}COO^-)\) from weak acid \(CH_{3}COOH\) hydrolyzes with water to produce \(OH^-\) ions: \(CH_{3}COO^- + H_{2}O \rightleftharpoons CH_{3}COOH + OH^-\). This results in a basic solution with (pH > 7).
334. A salt derived from a weak acid and a weak base \(e.g., (NH_{4}CH_{3}COO)\) gives a solution that is
ⓐ. Always acidic
ⓑ. Always basic
ⓒ. Neutral or depends on \(K_a\) and \(K_b\)
ⓓ. Amphoteric
Correct Answer: Neutral or depends on \(K_a\) and \(K_b\)
Explanation: In such salts, both ions hydrolyze. The pH depends on the relative magnitudes of \(K_a\) and \(K_b\):
If \(K_a = K_b\), the solution is neutral.
If \(K_a > K_b\), it is acidic.
If \(K_b > K_a\), it is basic.
335. The salt of a strong base and a weak acid shows alkaline reaction because
ⓐ. Its cation hydrolyzes to form \(H^+\)
ⓑ. Its anion hydrolyzes to form \(OH^-\)
ⓒ. It releases both \(H^+\) and \(OH^-\) equally
ⓓ. It reacts with acid to form salt and water
Correct Answer: Its anion hydrolyzes to form \(OH^-\)
Explanation: The anion of the weak acid reacts with water to produce hydroxide ions. For example, \(CH_{3}COO^- + H_{2}O \rightarrow CH_{3}COOH + OH^-\). This increases \(OH^-\) concentration and makes the solution basic.
336. The salt of a weak base and a strong acid shows acidic reaction because
ⓐ. Its cation hydrolyzes to produce \(H^+\)
ⓑ. Its anion hydrolyzes to form \(OH^-\)
ⓒ. Both ions hydrolyze equally
ⓓ. It undergoes precipitation
Correct Answer: Its cation hydrolyzes to produce \(H^+\)
Explanation: For example, \(NH_{4}Cl\) ionizes to give \(NH_{4}^+\) and \(Cl^-\). The \(NH_{4}^+\) reacts with water to produce \(NH_{3}\) and \(H^+\), thus lowering the pH and making the solution acidic.
337. The salt formed from (NaOH) and (HCl) will have a pH value of
ⓐ. 2
ⓑ. 4
ⓒ. 6
ⓓ. 7
Correct Answer: 7
Explanation: Both the acid and base are strong, leading to complete neutralization. The ions \(Na^+\) and \(Cl^-\) do not hydrolyze, so the resulting solution remains neutral with \(pH = 7\).
338. The pH of a solution of sodium acetate \((CH_{3}COONa)\) is
ⓐ. Less than 7
ⓑ. Equal to 7
ⓒ. Greater than 7
ⓓ. Exactly 7
Correct Answer: Greater than 7
Explanation: Sodium acetate is a salt of a strong base ((NaOH)) and weak acid \((CH_{3}COOH)\). The acetate ion hydrolyzes, forming \(OH^-\) ions, leading to an alkaline pH above 7.
339. The pH of ammonium chloride \((NH_{4}Cl)\) solution is
ⓐ. Equal to 7
ⓑ. Greater than 7
ⓒ. Independent of concentration
ⓓ. Less than 7
Correct Answer: Less than 7
Explanation: \(NH_{4}Cl\) is a salt of weak base \(NH_{4}OH\) and strong acid (HCl). The \(NH_{4}^+\) ion hydrolyzes to release \(H^+\), reducing pH below 7. The acidic nature intensifies with increasing concentration of the salt.
340. Which of the following salts forms a neutral solution in water?
ⓐ. (KCl)
ⓑ. \(NH_{4}Cl\)
ⓒ. \(CH_{3}COONa\)
ⓓ. \(Na_{2}CO_{3}\)
Correct Answer: (KCl)
Explanation: (KCl) comes from strong acid (HCl) and strong base (KOH). Both ions \((K^+), (Cl^-)\) do not hydrolyze, hence the solution remains neutral with \(pH = 7\). Weak acid–strong base or strong acid–weak base salts deviate from neutrality due to hydrolysis.
341. The hydrolysis constant \((K_h)\) is defined as
ⓐ. The ionic product of water
ⓑ. The equilibrium constant for the hydrolysis of a salt
ⓒ. The dissociation constant of a weak acid
ⓓ. The solubility product of a salt
Correct Answer: The equilibrium constant for the hydrolysis of a salt
Explanation: The hydrolysis constant \(K_h\) measures the extent to which a salt reacts with water to form its acid and base. It determines the strength of hydrolysis and the pH of the resulting solution. \(K_h\) depends on the salt type and temperature but not on its concentration.
342. The expression for the hydrolysis constant of a salt of a weak acid and a strong base \(e.g., (CH_{3}COONa)\) is
ⓐ. \(K_h = \dfrac{K_a}{K_w}\)
ⓑ. \(K_h = \dfrac{1}{K_a K_w}\)
ⓒ. \(K_h = K_a K_w\)
ⓓ. \(K_h = \dfrac{K_w}{K_a}\)
Correct Answer: \(K_h = \dfrac{K_w}{K_a}\)
Explanation: For salts like sodium acetate, only the anion undergoes hydrolysis. From the equilibrium \(CH_{3}COO^- + H_{2}O \rightleftharpoons CH_{3}COOH + OH^-\), the derived relation is \(K_h = \dfrac{K_w}{K_a}\). A smaller \(K_a\) (weaker acid) results in a larger \(K_h\), meaning greater hydrolysis.
343. The expression for the hydrolysis constant of a salt of a weak base and a strong acid \(e.g., (NH_{4}Cl)\) is
ⓐ. \(K_h = \dfrac{K_b}{K_w}\)
ⓑ. \(K_h = \dfrac{K_w}{K_b}\)
ⓒ. \(K_h = K_b K_w\)
ⓓ. \(K_h = \dfrac{1}{K_b K_w}\)
Correct Answer: \(K_h = \dfrac{K_w}{K_b}\)
Explanation: For salts like \(NH_{4}Cl\), only the cation hydrolyzes: \(NH_{4}^+ + H_{2}O \rightleftharpoons NH_{4}OH + H^+\). The derived relation \(K_h = \dfrac{K_w}{K_b}\) indicates that weaker bases \(smaller (K_b)\) give higher hydrolysis constants and stronger acidic character.
344. For a salt of a weak acid and a weak base \(e.g., (NH_{4}CH_{3}COO)\), the hydrolysis constant is
ⓐ. \(K_h = \dfrac{K_w}{K_a K_b}\)
ⓑ. \(K_h = K_a K_b K_w\)
ⓒ. \(K_h = \dfrac{K_a K_b}{K_w}\)
ⓓ. \(K_h = \dfrac{K_w^{2}}{K_a K_b}\)
Correct Answer: \(K_h = \dfrac{K_w}{K_a K_b}\)
Explanation: Both ions undergo hydrolysis in such salts, and their combined equilibrium gives \(K_h = \dfrac{K_w}{K_a K_b}\). The degree of hydrolysis is larger when both the acid and base are weak because neither ion is stabilized effectively.
345. The relationship between hydrolysis constant \(K_h\), degree of hydrolysis (h), and concentration (C) for a salt of a weak acid and strong base is
ⓐ. \(K_h = Ch^{2}\)
ⓑ. \(K_h = C(1-h)^{2}\)
ⓒ. \(K_h = \dfrac{C}{h^{2}}\)
ⓓ. \(K_h = \dfrac{h^{2}}{C(1-h)}\)
Correct Answer: \(K_h = \dfrac{h^{2}}{C(1-h)}\)
Explanation: For partial hydrolysis, \(K_h = \dfrac{[acid][base]}{[salt]}\). Writing concentrations in terms of (h) (the fraction hydrolyzed) gives this formula. Since (h) is usually very small, it simplifies to \(K_h \approx \dfrac{h^{2}}{C}\).
346. The degree of hydrolysis ((h)) of a salt of a weak acid and strong base is given by
ⓐ. \(h = \sqrt{K_h C}\)
ⓑ. \(h = \sqrt{\dfrac{K_h}{C}}\)
ⓒ. \(h = \dfrac{K_h}{C}\)
ⓓ. \(h = \dfrac{C}{K_h}\)
Correct Answer: \(h = \sqrt{\dfrac{K_h}{C}}\)
Explanation: For small (h), \(1-h \approx 1\), and from \(K_h = \dfrac{h^{2}}{C}\), we obtain \(h = \sqrt{\dfrac{K_h}{C}}\). This relation shows that hydrolysis increases on dilution (lower (C)) and when the salt originates from weaker acids or bases \(larger (K_h)\).
347. The (pH) of a salt solution formed by a weak acid and strong base can be expressed as
ⓐ. \(pH = 7 + \dfrac{1}{2}(pK_w + pK_a + \log C)\)
ⓑ. \(pH = 7 + \dfrac{1}{2}(pK_w + pK_a – \log C)\)
ⓒ. \(pH = 7 + \dfrac{1}{2}(pK_w – pK_a + \log C)\)
ⓓ. \(pH = 7 + \dfrac{1}{2}(pK_w + pK_b – \log C)\)
Correct Answer: \(pH = 7 + \dfrac{1}{2}(pK_w – pK_a + \log C)\)
Explanation: Substituting \(K_h = \dfrac{K_w}{K_a}\) into hydrolysis equations yields the given pH expression. For weak acid–strong base salts, the pH exceeds 7, consistent with the basic nature of the solution.
348. The (pH) of a salt solution formed by a weak base and strong acid is given by
ⓐ. \(pH = 7 – \dfrac{1}{2}(pK_w – pK_b + \log C)\)
ⓑ. \(pH = 7 + \dfrac{1}{2}(pK_w – pK_b + \log C)\)
ⓒ. \(pH = 7 – \dfrac{1}{2}(pK_w + pK_b + \log C)\)
ⓓ. \(pH = 7 + \dfrac{1}{2}(pK_a – pK_b + \log C)\)
Correct Answer: \(pH = 7 – \dfrac{1}{2}(pK_w – pK_b + \log C)\)
Explanation: For a weak base–strong acid salt, \(K_h = \dfrac{K_w}{K_b}\). Using this in the hydrolysis-pH relationship gives the above expression, yielding (pH < 7) as the solution is acidic due to the cation’s hydrolysis.
349. The larger the value of \(K_h\), the
ⓐ. Weaker the hydrolysis
ⓑ. Smaller the degree of hydrolysis
ⓒ. Stronger the hydrolysis
ⓓ. Lower the pH
Correct Answer: Stronger the hydrolysis
Explanation: A higher \(K_h\) indicates a greater tendency of salt ions to react with water, leading to more pronounced hydrolysis. This happens when both parent acid and base are weak, producing significant pH deviation from neutrality.
350. The hydrolysis constant of \(NH_{4}Cl\) at 25°C is \(5.6 \times 10^{-10}\). If \(K_w = 1 \times 10^{-14}\), then the base dissociation constant of \(NH_{4}OH\) is approximately
ⓐ. \(1.8 \times 10^{-5}\)
ⓑ. \(5.6 \times 10^{-10}\)
ⓒ. \(1.8 \times 10^{-4}\)
ⓓ. \(2.5 \times 10^{-6}\)
Correct Answer: \(1.8 \times 10^{-5}\)
Explanation: For weak base–strong acid salts, \(K_h = \dfrac{K_w}{K_b}\). Thus, \(K_b = \dfrac{K_w}{K_h} = \dfrac{1 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.8 \times 10^{-5}\). This corresponds to the known \(K_b\) value for ammonia, confirming internal consistency of hydrolysis relationships.
351. The pH of a solution of sodium chloride ((NaCl)) is
ⓐ. 2
ⓑ. 3
ⓒ. 5
ⓓ. 7
Correct Answer: 7
Explanation: Sodium chloride is formed from a strong acid ((HCl)) and a strong base ((NaOH)). Both ions \((Na^+), (Cl^-)\) do not hydrolyze in water, so the solution remains neutral with \(pH = 7\).
352. The pH of a solution of ammonium chloride \((NH_{4}Cl)\) is
ⓐ. 5.5
ⓑ. 7.0
ⓒ. 8.0
ⓓ. 9.0
Correct Answer: 5.5
Explanation: \(NH_{4}Cl\) is a salt of a weak base \((NH_{4}OH)\) and a strong acid ((HCl)). The \(NH_{4}^+\) ion hydrolyzes in water to produce \(H^+\) ions, lowering the pH below 7. Thus, the solution is slightly acidic.
353. The pH of sodium acetate \((CH_{3}COONa)\) solution is
ⓐ. Less than 7
ⓑ. Equal to 7
ⓒ. Greater than 7
ⓓ. Exactly 7
Correct Answer: Greater than 7
Explanation: Sodium acetate originates from a weak acid \((CH_{3}COOH)\) and a strong base ((NaOH)). The acetate ion hydrolyzes to produce \(OH^-\) ions, leading to a basic solution with (pH > 7).
354. The pH of ammonium acetate \((NH_{4}CH_{3}COO)\) depends on
ⓐ. Temperature only
ⓑ. The ratio of \(K_a\) and \(K_b\)
ⓒ. The concentration of water
ⓓ. The strength of the acid only
Correct Answer: The ratio of \(K_a\) and \(K_b\)
Explanation: Both ions hydrolyze in \(NH_{4}CH_{3}COO\). The pH depends on whether \(K_a\) \(for (NH_{4}^+)\) or \(K_b\) \(for (CH_{3}COO^-)\) is larger.
If \(K_a > K_b\): solution is acidic.
If \(K_b > K_a\): solution is basic.
If \(K_a = K_b\): solution is neutral.
355. Which of the following salts gives an acidic solution in water?
ⓐ. (KCl)
ⓑ. \(Na_{2}SO_{4}\)
ⓒ. \(NH_{4}Cl\)
ⓓ. \(CH_{3}COONa\)
Correct Answer: \(NH_{4}Cl\)
Explanation: The \(NH_{4}^+\) ion, derived from a weak base, hydrolyzes to yield \(H^+\) ions, lowering pH. The \(Cl^-\) ion, from a strong acid, does not hydrolyze. Therefore, the solution is acidic.
356. Which of the following salts produces a basic solution when dissolved in water?
ⓐ. \(NH_{4}Cl\)
ⓑ. \(CH_{3}COONa\)
ⓒ. (NaCl)
ⓓ. \(NH_{4}NO_{3}\)
Correct Answer: \(CH_{3}COONa\)
Explanation: Sodium acetate is derived from a weak acid and strong base. The acetate ion reacts with water to form \(OH^-\), making the solution basic with pH above 7.
357. The pH of a solution of potassium nitrate \((KNO_{3})\) is
ⓐ. 1
ⓑ. 3
ⓒ. 4
ⓓ. 7
Correct Answer: 7
Explanation: Potassium nitrate results from a strong acid \((HNO_{3})\) and a strong base ((KOH)). Neither \(K^+\) nor \(NO_{3}^-\) undergo hydrolysis, leaving the solution neutral with \(pH = 7\).
358. For a salt of a weak acid and strong base, the expression for pH is
ⓐ. \( pH = 7 – \frac{1}{2}(pK_w – pK_a + \log C) \)
ⓑ. \( pH = 7 + \frac{1}{2}(pK_w – pK_a + \log C) \)
ⓒ. \( pH = 7 + \frac{1}{2}(pK_a – pK_w + \log C) \)
ⓓ. \( pH = 7 – \frac{1}{2}(pK_a – pK_b + \log C) \)
Correct Answer: \( pH = 7 + \frac{1}{2}(pK_w – pK_a + \log C) \)
Explanation: For weak acid–strong base salts, the pH is greater than 7 because hydrolysis of the anion produces \(OH^-\). The above equation is derived from the hydrolysis equilibrium constant relation \(K_h = \dfrac{K_w}{K_a}\).
359. For a salt of a weak base and strong acid, the expression for pH is
ⓐ. \( pH = 7 + \frac{1}{2}(pK_w – pK_b + \log C) \)
ⓑ. \( pH = 7 – \frac{1}{2}(pK_w – pK_b + \log C) \)
ⓒ. \( pH = 7 + \frac{1}{2}(pK_b – pK_a + \log C) \)
ⓓ. \( pH = 7 – \frac{1}{2}(pK_a – pK_b + \log C) \)
Correct Answer: \( pH = 7 – \frac{1}{2}(pK_w – pK_b + \log C) \)
Explanation: The cation from the weak base hydrolyzes to produce \(H^+\) ions, lowering the pH. Hence, the derived formula shows a value less than 7, confirming the acidic nature of such salt solutions.
360. The salt of a weak acid and weak base gives a solution that is
ⓐ. Always neutral
ⓑ. Always basic
ⓒ. Always acidic
ⓓ. Acidic, basic, or neutral depending on \(K_a\) and \(K_b\)
Correct Answer: Acidic, basic, or neutral depending on \(K_a\) and \(K_b\)
Explanation: Both cation and anion hydrolyze, and pH depends on the relative strength of the parent acid and base. If \(K_a = K_b\), the solution is neutral; if \(K_a > K_b\), acidic; and if \(K_b > K_a\), basic.
361. A buffer solution is defined as a solution which
ⓐ. Changes its pH drastically when a small amount of acid or base is added
ⓑ. Maintains almost constant pH when small amounts of acid or base are added
ⓒ. Has a pH equal to 7 at all times
ⓓ. Contains only strong acids and bases
Correct Answer: Maintains almost constant pH when small amounts of acid or base are added
Explanation: Buffer solutions resist changes in pH due to the presence of a weak acid and its conjugate base or a weak base and its conjugate acid. They act by neutralizing small quantities of added \(H^+\) or \(OH^-\) ions, keeping the pH nearly constant.
362. Which of the following is an example of an acidic buffer?
ⓐ. \(NH_{4}OH\) and \(NH_{4}Cl\)
ⓑ. (HCl) and (NaCl)
ⓒ. \(CH_{3}COOH\) and \(CH_{3}COONa\)
ⓓ. (NaOH) and (NaCl)
Correct Answer: \(CH_{3}COOH\) and \(CH_{3}COONa\)
Explanation: Acidic buffers consist of a weak acid and its conjugate base. The acetic acid–sodium acetate mixture resists pH change near 4.76, maintaining an acidic environment. It is a classic laboratory buffer system.
363. Which of the following is a basic buffer?
ⓐ. \(NH_{4}OH\) and \(NH_{4}Cl\)
ⓑ. (HCl) and \(NH_{4}Cl\)
ⓒ. \(CH_{3}COOH\) and \(CH_{3}COONa\)
ⓓ. (NaOH) and (NaCl)
Correct Answer: \(NH_{4}OH\) and \(NH_{4}Cl\)
Explanation: Basic buffers are made of a weak base and its conjugate salt. The \(NH_{4}OH\)–\(NH_{4}Cl\) buffer maintains pH near 9.25 by neutralizing both \(H^+\) and \(OH^-\) ions effectively.
364. The pH of an acidic buffer can be calculated using
ⓐ. Ostwald’s dilution law
ⓑ. Henderson–Hasselbalch equation
ⓒ. Nernst equation
ⓓ. Raoult’s law
Correct Answer: Henderson–Hasselbalch equation
Explanation: The Henderson–Hasselbalch equation relates buffer pH to \(K_a\):
\(
pH = pK_a + \log \left(\frac{[salt]}{[acid]}\right)
\)
It is derived from the equilibrium condition of weak acids and explains how pH depends on the ratio of conjugate base to acid.
365. The pH of a basic buffer is given by
ⓐ. \(pH = pK_b + \log \frac{[salt]}{[base]}\)
ⓑ. \(pH = 7 + \log \frac{[salt]}{[base]}\)
ⓒ. \(pH = pK_a + \log \frac{[acid]}{[salt]}\)
ⓓ. \(pH = 14 – (pK_b + \log \frac{[salt]}{[base]})\)
Correct Answer: \(pH = 14 – (pK_b + \log \frac{[salt]}{[base]})\)
Explanation: The pH of a basic buffer is derived using \(pOH = pK_b + \log \frac{[salt]}{[base]}\) and the relation \(pH + pOH = 14\). It helps determine the pH of weak base–strong acid salt mixtures like \(NH_{4}OH/NH_{4}Cl\).
366. Which statement about buffers is true?
ⓐ. Buffer solutions can neutralize unlimited acid or base
ⓑ. Buffer solutions resist small pH changes only
ⓒ. Buffer solutions contain only strong acids and bases
ⓓ. Buffers have no practical application
Correct Answer: Buffer solutions resist small pH changes only
Explanation: Buffers have limited capacity, determined by the concentration of the weak acid/base and its conjugate salt. When excessive acid or base is added, the buffer capacity is exceeded, and the pH changes rapidly.
367. Which of the following pairs does not form a buffer solution?
ⓐ. \(H_{2}CO_{3}\) and \(NaHCO_{3}\)
ⓑ. \(NH_{4}OH\) and \(NH_{4}Cl\)
ⓒ. (HCl) and (NaCl)
ⓓ. \(CH_{3}COOH\) and \(CH_{3}COONa\)
Correct Answer: (HCl) and (NaCl)
Explanation: (HCl) is a strong acid and does not establish an equilibrium with its conjugate base. Buffer solutions require a weak acid/base with its conjugate partner to resist pH changes, which is not possible in the (HCl/NaCl) combination.
368. The main function of a buffer in biological systems is to
ⓐ. Decrease ionic strength of body fluids
ⓑ. Increase pH beyond 10
ⓒ. Neutralize all acids in the body
ⓓ. Maintain enzyme activity by keeping pH constant
Correct Answer: Maintain enzyme activity by keeping pH constant
Explanation: Enzymes function effectively only within a narrow pH range. Biological buffers like the bicarbonate \((H_{2}CO_{3}/HCO_{3}^-)\) and phosphate \((H_{2}PO_{4}^-/HPO_{4}^{2-})\) systems stabilize blood and cellular pH to prevent harmful fluctuations.
369. The blood buffer system primarily consists of
ⓐ. \(H_{2}CO_{3}/HCO_{3}^-\) pair
ⓑ. \(H_{3}PO_{4}/H_{2}PO_{4}^-\) pair
ⓒ. \(CH_{3}COOH/CH_{3}COONa\) pair
ⓓ. \(NH_{4}OH/NH_{4}Cl\) pair
Correct Answer: \(H_{2}CO_{3}/HCO_{3}^-\) pair
Explanation: The bicarbonate buffer system maintains blood pH around 7.4. Carbonic acid neutralizes bases while bicarbonate neutralizes acids. This system is vital for physiological acid-base balance in humans.
370. Buffers are important in analytical chemistry because they
ⓐ. Increase solubility of salts
ⓑ. Maintain pH in titration and reaction media
ⓒ. Convert strong acids to weak acids
ⓓ. Promote precipitation of salts
Correct Answer: Maintain pH in titration and reaction media
Explanation: Buffers stabilize the pH of solutions during chemical reactions, ensuring that indicators, reagents, and enzyme-catalyzed processes function optimally. This stability is critical in titrations, pharmaceutical formulations, and biochemical assays.
371. The Henderson–Hasselbalch equation for an acidic buffer is
ⓐ. \( pH = pK_b + \log \dfrac{[salt]}{[base]} \)
ⓑ. \( pH = pK_a + \log \dfrac{[acid]}{[salt]} \)
ⓒ. \( pH = pK_a + \log \dfrac{[salt]}{[acid]} \)
ⓓ. \( pH = 14 – pK_b + \log \dfrac{[base]}{[salt]} \)
Correct Answer: \( pH = pK_a + \log \dfrac{[salt]}{[acid]} \)
Explanation: The Henderson–Hasselbalch equation relates the pH of an acidic buffer to the ratio of the concentrations of conjugate base (salt) and weak acid. It is derived from the dissociation constant expression for the weak acid and simplifies buffer pH calculations.
372. The Henderson–Hasselbalch equation for a basic buffer is expressed as
ⓐ. \( pOH = pK_b + \log \dfrac{[base]}{[salt]} \)
ⓑ. \( pH = 14 – (pK_b + \log \dfrac{[salt]}{[base]}) \)
ⓒ. \( pH = pK_b + \log \dfrac{[salt]}{[base]} \)
ⓓ. \( pOH = pK_a + \log \dfrac{[acid]}{[salt]} \)
Correct Answer: \( pH = 14 – (pK_b + \log \dfrac{[salt]}{[base]}) \)
Explanation: For basic buffers consisting of a weak base and its salt, the pOH is calculated first using \(pOH = pK_b + \log \dfrac{[salt]}{[base]}\). Subtracting this from 14 gives the buffer’s pH, which is typically above 7.
373. The Henderson–Hasselbalch equation is derived from
ⓐ. The law of mass action
ⓑ. Raoult’s law
ⓒ. Henry’s law
ⓓ. Nernst equation
Correct Answer: The law of mass action
Explanation: The equation originates from the equilibrium expression for weak acids or bases, based on the law of mass action. It connects equilibrium constant \((K_a)\) with hydrogen ion concentration and the ratio of salt to acid concentrations.
374. Using the Henderson equation, the pH of a buffer containing (0.1,M) acetic acid and (0.1,M) sodium acetate \((pK_a = 4.76)\) is
ⓐ. 1.76
ⓑ. 2.76
ⓒ. 3.76
ⓓ. 4.76
Correct Answer: 4.76
Explanation: Substituting into \(pH = pK_a + \log \dfrac{[salt]}{[acid]}\):
\(pH = 4.76 + \log(1) = 4.76\).
Equal concentrations of acid and its salt result in \(pH = pK_a\), a standard feature of balanced buffer systems.
375. A buffer solution contains \(CH_{3}COOH = 0.1,M\) and \(CH_{3}COONa = 0.2,M\). If \(pK_a = 4.76\), then pH = ?
ⓐ. 4.46
ⓑ. 4.76
ⓒ. 5.06
ⓓ. 5.46
Correct Answer: 5.06
Explanation: Using \(pH = pK_a + \log \dfrac{[salt]}{[acid]}\):
\(pH = 4.76 + \log(2) = 4.76 + 0.30 = 5.06.\)
Higher salt concentration increases the ratio \(\dfrac{[salt]}{[acid]}\), causing a slight rise in pH.
376. In a buffer containing \(NH_{4}OH\) and \(NH_{4}Cl\), the pH is calculated using
ⓐ. \( pH = pK_a + \log \dfrac{[acid]}{[salt]} \)
ⓑ. \( pH = 7 + \log \dfrac{[salt]}{[acid]} \)
ⓒ. \( pH = pK_b + \log \dfrac{[salt]}{[base]} \)
ⓓ. \( pH = 14 – (pK_b + \log \dfrac{[salt]}{[base]}) \)
Correct Answer: \( pH = 14 – (pK_b + \log \dfrac{[salt]}{[base]}) \)
Explanation: For a basic buffer, \(pOH = pK_b + \log \dfrac{[salt]}{[base]}\). The pH is then obtained as (14 – pOH). This relation helps determine pH values in systems like \(NH_{4}OH/NH_{4}Cl\).
377. The Henderson–Hasselbalch equation is most useful for
ⓐ. Calculating solubility of salts
ⓑ. Determining the pH of buffer solutions
ⓒ. Measuring ionic product of water
ⓓ. Finding equivalent conductance
Correct Answer: Determining the pH of buffer solutions
Explanation: The equation provides a quick and reliable way to calculate the pH of buffers without solving complex equilibrium equations. It is fundamental in analytical chemistry and biochemistry for maintaining controlled reaction environments.
378. When \( [salt] = [acid] \) in an acidic buffer, the pH equals
ⓐ. 0
ⓑ. 7
ⓒ. \(pK_a\)
ⓓ. \(pK_b\)
Correct Answer: \(pK_a\)
Explanation: From \( pH = pK_a + \log \dfrac{[salt]}{[acid]} \), substituting equal concentrations gives \(\log(1) = 0\), so \(pH = pK_a\). This condition marks the center of the buffer range where it has maximum efficiency.
379. When \( [base] = [salt] \) in a basic buffer, the pH equals
ⓐ. \(pK_a\)
ⓑ. \(pK_b\)
ⓒ. (7)
ⓓ. \(14 – pK_b\)
Correct Answer: \(14 – pK_b\)
Explanation: For basic buffers, \(pOH = pK_b + \log(1) = pK_b\). Hence, \(pH = 14 – pK_b\). This shows that the buffer pH depends directly on the base dissociation constant, reflecting its strength.
380. The importance of Henderson’s equation lies in the fact that it
ⓐ. Predicts the pH change of buffer upon dilution
ⓑ. Predicts the pH of pure water
ⓒ. Connects pH with \(K_a\) and concentration ratio of buffer components
ⓓ. Determines solubility of weak electrolytes
Correct Answer: Connects pH with \(K_a\) and concentration ratio of buffer components
Explanation: The Henderson–Hasselbalch equation elegantly links pH, dissociation constant, and concentration ratio, allowing chemists to design and control buffer systems in laboratories, pharmaceuticals, and biological systems.
381. The solubility product constant \((K_{sp})\) is defined as
ⓐ. The equilibrium constant for dissolution of a sparingly soluble salt
ⓑ. The product of solubility and molar mass
ⓒ. The ratio of dissolved ions to undissolved salt
ⓓ. The solubility of an electrolyte in moles per liter
Correct Answer: The equilibrium constant for dissolution of a sparingly soluble salt
Explanation: \(K_{sp}\) represents the equilibrium between a solid and its ions in a saturated solution. For example, for \(AgCl_{(s)} \rightleftharpoons Ag^+ + Cl^-\), \(K_{sp} = [Ag^+][Cl^-]\). It is a measure of how much salt dissolves before equilibrium is reached.
382. The expression for \(K_{sp}\) of \(CaF_{2}\) is
ⓐ. \(K_{sp} = [Ca^{2+}] + [F^-]\)
ⓑ. \(K_{sp} = [Ca^{2+}]^{2} [F^-]\)
ⓒ. \(K_{sp} = [Ca^{2+}] [F^-]^{2}\)
ⓓ. \(K_{sp} = [Ca^{2+}]^{3} [F^-]^{3}\)
Correct Answer: \(K_{sp} = [Ca^{2+}] [F^-]^{2}\)
Explanation: When \(CaF_{2}\) dissolves, it dissociates as \(CaF_{2} \rightleftharpoons Ca^{2+} + 2F^-\). The solubility product is the product of ionic concentrations raised to their stoichiometric powers: \(K_{sp} = [Ca^{2+}][F^-]^{2}\).
383. The solubility product differs from the ionic product in that
ⓐ. \(K_{sp}\) changes with concentration, ionic product does not
ⓑ. \(K_{sp}\) is fixed for a given temperature, while ionic product varies with concentration
ⓒ. \(K_{sp}\) applies to strong electrolytes only
ⓓ. Both are always equal
Correct Answer: \(K_{sp}\) is fixed for a given temperature, while ionic product varies with concentration
Explanation: \(K_{sp}\) is constant at a fixed temperature and indicates the solubility limit. The ionic product, \([cation]^{m} [anion]^{n}\), may be less, equal, or greater than \(K_{sp}\), determining whether precipitation or dissolution occurs.
384. The \(K_{sp}\) of (AgCl) is expressed as
ⓐ. \(K_{sp} = [Ag^+][Cl^-]\)
ⓑ. \(K_{sp} = [Ag^+] + [Cl^-]\)
ⓒ. \(K_{sp} = [Ag^+]^{2}[Cl^-]^{2}\)
ⓓ. \(K_{sp} = [Ag^+] / [Cl^-]\)
Correct Answer: \(K_{sp} = [Ag^+][Cl^-]\)
Explanation: For \(AgCl_{(s)} \rightleftharpoons Ag^+ + Cl^-\), one mole of solid dissociates into one mole of each ion. Therefore, \(K_{sp}\) equals the product of their concentrations at equilibrium.
385. The units of \(K_{sp}\) for \(Ag_{2}CrO_{4}\) are
ⓐ. (mol/L)
ⓑ. \(mol^{2}/L^{2}\)
ⓒ. \(mol^{5}/L^{5}\)
ⓓ. \(mol^{3}/L^{3}\)
Correct Answer: \(mol^{3}/L^{3}\)
Explanation: \(Ag_{2}CrO_{4} \rightleftharpoons 2Ag^+ + CrO_{4}^{2-}\), so \(K_{sp} = [Ag^+]^{2}[CrO_{4}^{2-}]\). The solubility (S) gives \(K_{sp} = (2S)^{2}(S) = 4S^{3}\). Thus, the dimensional unit is \(mol^{3}/L^{3}\); but since the expression contains a squared term times one molarity, the resulting unit is \(mol^{3}/L^{3}\).
386. The solubility of a salt \(AB_{2}\) in terms of \(K_{sp}\) is given by
ⓐ. \(S = (K_{sp})^{1/2}\)
ⓑ. \(S = (K_{sp})^{1/3}\)
ⓒ. \(S = (K_{sp}/4)^{1/3}\)
ⓓ. \(S = (K_{sp}/9)^{1/3}\)
Correct Answer: \(S = (K_{sp}/4)^{1/3}\)
Explanation: For \(AB_{2} \rightleftharpoons A^{2+} + 2B^-\), \(K_{sp} = [A^{2+}][B^-]^{2} = (S)(2S)^{2} = 4S^{3}\). Solving for (S) gives \(S = (K_{sp}/4)^{1/3}\).
387. For a salt \(A_{2}B_{3}\), the relationship between solubility ((S)) and \(K_{sp}\) is
ⓐ. \(K_{sp} = 27S^{5}\)
ⓑ. \(K_{sp} = 108S^{5}\)
ⓒ. \(K_{sp} = 12S^{4}\)
ⓓ. \(K_{sp} = S^{5}\)
Correct Answer: \(K_{sp} = 108S^{5}\)
Explanation: \(A_{2}B_{3} \rightleftharpoons 2A^{3+} + 3B^{2-}\); thus \(K_{sp} = [A^{3+}]^{2}[B^{2-}]^{3} = (2S)^{2}(3S)^{3} = 108S^{5}\). This general method links ionic stoichiometry to solubility.
388. When a solution is just saturated with a salt, the ionic product is
ⓐ. Less than \(K_{sp}\)
ⓑ. Zero
ⓒ. Greater than \(K_{sp}\)
ⓓ. Equal to \(K_{sp}\)
Correct Answer: Equal to \(K_{sp}\)
Explanation: At equilibrium, the product of the ion concentrations (ionic product) equals \(K_{sp}\). Below this value, the salt can dissolve more; above it, precipitation occurs. Equality signifies saturation.
389. A saturated solution of (AgCl) at 25°C has \([Ag^+] = 1.3 \times 10^{-5},M\). What is \(K_{sp}\)?
ⓐ. \(1.3 \times 10^{-5}\)
ⓑ. \(1.69 \times 10^{-10}\)
ⓒ. \(2.6 \times 10^{-5}\)
ⓓ. \(1.69 \times 10^{-8}\)
Correct Answer: \(1.69 \times 10^{-10}\)
Explanation: Since (AgCl) dissociates as \(AgCl \rightleftharpoons Ag^+ + Cl^-\), both ions have the same molar concentration \(S = 1.3 \times 10^{-5}\). Hence, \(K_{sp} = [Ag^+][Cl^-] = S^{2} = (1.3 \times 10^{-5})^{2} = 1.69 \times 10^{-10}\).
390. The importance of the solubility product lies in its use for
ⓐ. Predicting vapor pressure
ⓑ. Determining pH only
ⓒ. Predicting precipitation or dissolution of salts
ⓓ. Calculating molar conductivity
Correct Answer: Predicting precipitation or dissolution of salts
Explanation: \(K_{sp}\) helps determine whether a precipitate will form. If ionic product < \(K_{sp}\), no precipitation occurs; if ionic product > \(K_{sp}\), the salt precipitates. This principle is widely used in qualitative analysis and gravimetric chemistry.
391. The molar solubility (S) of a salt (AB) in terms of its solubility product \(K_{sp}\) is
ⓐ. \(S = K_{sp}\)
ⓑ. \(S = (K_{sp})^{1/2}\)
ⓒ. \(S = (K_{sp})^{1/3}\)
ⓓ. \(S = (K_{sp})^{2}\)
Correct Answer: \(S = (K_{sp})^{1/2}\)
Explanation: For \(AB \rightleftharpoons A^+ + B^-\), both ions have the same concentration at equilibrium, so \(K_{sp} = [A^+][B^-] = S^{2}\). Hence, \(S = (K_{sp})^{1/2}\). This gives the direct relationship between molar solubility and \(K_{sp}\).
392. For a salt \(AB_{2}\), the relationship between \(K_{sp}\) and molar solubility (S) is
ⓐ. \(K_{sp} = S^{2}\)
ⓑ. \(K_{sp} = 2S^{3}\)
ⓒ. \(K_{sp} = 4S^{3}\)
ⓓ. \(K_{sp} = 8S^{2}\)
Correct Answer: \(K_{sp} = 4S^{3}\)
Explanation: \(AB_{2} \rightleftharpoons A^{2+} + 2B^-\). Hence, \([A^{2+}] = S\), \([B^-] = 2S\).
Substituting gives \(K_{sp} = [A^{2+}][B^-]^{2} = (S)(2S)^{2} = 4S^{3}\).
393. The molar solubility (S) of a salt \(A_{2}B_{3}\) in terms of \(K_{sp}\) is
ⓐ. \(S = (K_{sp})^{1/3}\)
ⓑ. \(S = (K_{sp}/108)^{1/5}\)
ⓒ. \(S = (K_{sp}/27)^{1/4}\)
ⓓ. \(S = (K_{sp})^{1/2}\)
Correct Answer: \(S = (K_{sp}/108)^{1/5}\)
Explanation: For \(A_{2}B_{3} \rightleftharpoons 2A^{3+} + 3B^{2-}\), \(K_{sp} = (2S)^{2}(3S)^{3} = 108S^{5}\). Rearranging gives \(S = (K_{sp}/108)^{1/5}\). This allows solubility calculation from \(K_{sp}\) for multivalent salts.
394. If the solubility of (AgCl) is \(1.3 \times 10^{-5},M\), the \(K_{sp}\) value is
ⓐ. \(1.69 \times 10^{-10}\)
ⓑ. \(2.6 \times 10^{-5}\)
ⓒ. \(1.69 \times 10^{-8}\)
ⓓ. \(1.3 \times 10^{-10}\)
Correct Answer: \(1.69 \times 10^{-10}\)
Explanation: For \(AgCl \rightleftharpoons Ag^+ + Cl^-\), \(K_{sp} = S^{2} = (1.3 \times 10^{-5})^{2} = 1.69 \times 10^{-10}\). The \(K_{sp}\) value reflects the very low solubility of silver chloride.
395. When the ionic product (Q) of a salt solution exceeds \(K_{sp}\), it means
ⓐ. The salt will further dissolve
ⓑ. The salt will start to precipitate
ⓒ. The solution remains unsaturated
ⓓ. The salt is completely soluble
Correct Answer: The salt will start to precipitate
Explanation: If \(Q > K_{sp}\), the ionic concentrations exceed equilibrium solubility, forcing the reverse process (precipitation) to restore balance. This principle is key in qualitative precipitation tests.
396. If the ionic product (Q) is less than \(K_{sp}\), the solution is
ⓐ. Saturated
ⓑ. Supersaturated
ⓒ. Colloidal
ⓓ. Unsaturated
Correct Answer: Unsaturated
Explanation: When \(Q < K_{sp}\), the ionic concentration is below equilibrium, and no precipitate forms. The solution can still dissolve more solute until \(Q = K_{sp}\) (saturation point).
397. If the ionic product equals the solubility product \((Q = K_{sp})\), the system is
ⓐ. Unsaturated
ⓑ. Supersaturated
ⓒ. Saturated and at equilibrium
ⓓ. At the point of maximum precipitation
Correct Answer: Saturated and at equilibrium
Explanation: Equality of (Q) and \(K_{sp}\) signifies dynamic equilibrium — the rate of dissolution equals the rate of precipitation. No further change in concentration occurs unless external conditions (e.g., temperature) vary.
398. In qualitative analysis, group II cations like \(Pb^{2+}\), \(Hg^{2+}\), and \(Cu^{2+}\) are precipitated as sulfides in acidic medium because
ⓐ. Sulfide ions are abundant in acid
ⓑ. Their \(K_{sp}\) values for sulfides are very small
ⓒ. Their \(K_{sp}\) values for sulfides are very large
ⓓ. They form soluble sulfides
Correct Answer: Their \(K_{sp}\) values for sulfides are very small
Explanation: Low \(K_{sp}\) values indicate extremely low solubility, so these sulfides precipitate even when sulfide ion concentration is small (as in acidic solutions). This property enables their selective separation.
399. In qualitative inorganic analysis, group IV cations \(like (Zn^{2+}) and (Ni^{2+})\) are precipitated in ammoniacal medium because
ⓐ. Their sulfides require higher \(S^{2-}\) concentration to exceed \(K_{sp}\)
ⓑ. Their \(K_{sp}\) values are higher than group II sulfides
ⓒ. Their solubility increases in acidic medium
ⓓ. They are amphoteric
Correct Answer: Their sulfides require higher \(S^{2-}\) concentration to exceed \(K_{sp}\)
Explanation: (ZnS) and (NiS) have larger \(K_{sp}\) values compared to (PbS) or (CuS). Only in alkaline or ammoniacal medium does \(S^{2-}\) concentration rise enough to exceed \(K_{sp}\), allowing selective precipitation.
400. The principle of fractional precipitation in qualitative analysis is based on
ⓐ. Equal solubility of all salts
ⓑ. Dilution effect
ⓒ. Common ion effect
ⓓ. Different \(K_{sp}\) values of various salts
Correct Answer: Different \(K_{sp}\) values of various salts
Explanation: Fractional precipitation separates ions sequentially by exploiting differences in their \(K_{sp}\). Salts with lower \(K_{sp}\) values precipitate first, while those with higher values remain dissolved until their ionic product exceeds their solubility product.
Welcome to Class 11 Chemistry MCQs – Chapter 7: Equilibrium (Part 4). This is the final part of the chapter where we wrap up with questions on advanced topics related to equilibrium. These MCQs focus on complex equilibria, including heterogeneous equilibria, solubility product, and buffer solutions.
Topics covered in Part 4 (100 MCQs): The solubility product constant (Ksp), equilibrium in the formation of complexes, buffer solutions, and their practical applications. Additionally, you will work with questions on the common ion effect and its impact on solubility and pH.
How to practice here
- After answering, review the explanations carefully to catch any mistakes and learn the correct methods.
- Use the ❤️ Heart to mark tricky questions and review them later by toggling the Favourite filter.
- Write down key points in the Workspace for each question to ensure you remember the concepts better.
- Click Random to simulate the exam environment and test your ability to answer questions quickly.
- 👉 Total in chapter: 400 MCQs (100 + 100 + 100 + 100)
- 👉 This page: Final set of 100 solved MCQs with detailed explanations
- 👉 Explore more: Use navigation above for other chapters and subjects