JEE Chemistry Equilibrium MCQs Set 17

Chemistry Objective Questions and Answers: Equilibrium

Explore reliable objective questions for Equilibrium tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

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Multiple Choice Questions

Question 1: The pH of blood does not appreciably change by a small addition of an acid or a base because blood
(a) contains serum protein which acts as a buffer
(b) contains iron as a part of the molecule
(c) can be easily coagulated
(d) is a body fluid
Show Answer & Explanation

Answer: (a) contains serum protein which acts as a buffer

Explanation:
1. Blood contains buffer systems such as proteins and the \( \ce{H2CO3/HCO3-} \) pair, which resist pH changes.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 2: The pH of a solution whose hydronium ion concentration is \( 6.2\times10^{-9} \) mol/L is
(a) 6.21
(b) 7.21
(c) 7.75
(d) 8.21
Show Answer & Explanation

Answer: (d) 8.21

Explanation:
1. pH = \( -\log(6.2\times10^{-9})=9-0.792=8.21 \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 3: What will be the sum of pH and pOH in an aqueous solution?
(a) 7
(b) \( pK_w \)
(c) zero
(d) 1
Show Answer & Explanation

Answer: (b) \( pK_w \)

Explanation:
1. \( [H^+][OH^-]=K_w \), so \( pH+pOH=pK_w \) (14 at 25 C).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 4: The hydrogen ion concentration in mol/L of a solution of pH = 5.4 will be
(a) \( 3.98\times10^{8} \)
(b) \( 3.88\times10^{6} \)
(c) \( 3.68\times10^{-6} \)
(d) \( 3.98\times10^{-6} \)
Show Answer & Explanation

Answer: (d) \( 3.98\times10^{-6} \)

Explanation:
1. \( [H^+]=10^{-5.4}=10^{0.6}\times10^{-6}=3.98\times10^{-6} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 5: When solid potassium cyanide is added to water then
(a) pH will increase
(b) pH will decrease
(c) pH will remain the same
(d) electrical conductivity will not change
Show Answer & Explanation

Answer: (a) pH will increase

Explanation:
1. \( \ce{CN-} \) hydrolyses (\( \ce{CN- + H2O <=> HCN + OH-} \)), producing hydroxide ions, so pH rises.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 6: The pH of a solution is 5. Its hydroxyl ion concentration is
(a) \( 5 \)
(b) \( 10 \)
(c) \( 10^{-5} \)
(d) \( 10^{-9} \)
Show Answer & Explanation

Answer: (d) \( 10^{-9} \)

Explanation:
1. pOH = 14 - 5 = 9, so \( [OH^-]=10^{-9} \) M.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 7: An aqueous solution of HCl has pH = 4. Its molarity would be
(a) 4 M
(b) 0.4 M
(c) 0.0001 M
(d) 10 M
Show Answer & Explanation

Answer: (c) 0.0001 M

Explanation:
1. HCl is a strong acid: \( [H^+]=10^{-4} \) M, so molarity = 0.0001 M.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 8: Which is a buffer solution?
(a) \( \ce{CH3COOH} \) + \( \ce{CH3COONa} \)
(b) \( \ce{CH3COOH} \) + \( \ce{CH3COONH4} \)
(c) \( \ce{CH3COOH} \) + \( \ce{NH4Cl} \)
(d) \( \ce{NaOH} \) + NaCl
Show Answer & Explanation

Answer: (a) \( \ce{CH3COOH} \) + \( \ce{CH3COONa} \)

Explanation:
1. A buffer is a weak acid with its salt of a strong base (or a weak base with its salt of a strong acid), here acetic acid with sodium acetate.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 9: The addition of solid sodium carbonate to pure water causes
(a) an increase in hydronium ion concentration
(b) an increase in alkalinity
(c) no change in acidity
(d) a decrease in hydroxide ion concentration
Show Answer & Explanation

Answer: (b) an increase in alkalinity

Explanation:
1. \( \ce{CO3^2-} \) hydrolyses to give \( \ce{OH-} \), so the solution becomes alkaline.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 10: The aqueous solution of which of the following salts has the lowest pH?
(a) \( \ce{NaClO} \)
(b) \( \ce{NaClO2} \)
(c) \( \ce{NaClO3} \)
(d) \( \ce{NaClO4} \)
Show Answer & Explanation

Answer: (d) \( \ce{NaClO4} \)

Explanation:
1. The conjugate base of the strongest acid (HClO4) is the weakest base, so \( \ce{ClO4-} \) does not hydrolyse and the solution is neutral, the lowest pH here.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 11: The pH of a \( 10^{-10} \) M NaOH solution is nearest to
(a) 10
(b) 7
(c) 4
(d) \( -10 \)
Show Answer & Explanation

Answer: (b) 7

Explanation:
1. Such a dilute base cannot raise the pH above that of water, so the pH is very close to 7 (about 7.0).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 12: Which will have the maximum pH?
(a) Distilled water
(b) 1 M \( \ce{NH3} \)
(c) 1 M NaOH
(d) water saturated with chlorine
Show Answer & Explanation

Answer: (c) 1 M NaOH

Explanation:
1. NaOH is a strong base, fully ionised, so 1 M NaOH has pH 14, higher than weak ammonia.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 13: The pH of a solution is 9.5. The solution is
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Show Answer & Explanation

Answer: (c) basic

Explanation:
1. pH greater than 7 means a basic solution.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 14: The pH of a \( 10^{-9} \) M solution of HCl in water is
(a) 8
(b) \( -8 \)
(c) between 7 and 8
(d) between 6 and 7
Show Answer & Explanation

Answer: (d) between 6 and 7

Explanation:
1. Water supplies \( 10^{-7} \) M of \( \ce{H+} \), which far exceeds the acid's contribution, so the pH is just below 7 (between 6 and 7).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 15: Which of the following 0.1 M solutions will contain the largest concentration of hydronium ions?
(a) \( \ce{NaHCO3} \)
(b) \( \ce{NH4Cl} \)
(c) HCl
(d) \( \ce{NH3} \)
Show Answer & Explanation

Answer: (c) HCl

Explanation:
1. HCl is a strong acid, completely ionised, giving 0.1 M \( \ce{H3O+} \).
2. NH4Cl gives far less by hydrolysis; \( \ce{NaHCO3} \) and \( \ce{NH3} \) are basic.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 16: Which one has pH 12?
(a) 0.01 M KOH
(b) 1 N KOH, 1 mL
(c) 1 N NaOH, 1 mL
(d) 1 N \( \ce{Ca(OH)2} \), 1 mL
Show Answer & Explanation

Answer: (a) 0.01 M KOH

Explanation:
1. 0.01 M KOH: pOH = 2, so pH = 12.
2. The 1 N solutions have pH about 14.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 17: What is the correct relationship between the pH of isomolar solutions of sodium oxide (\( pH_1 \)), sodium sulphide (\( pH_2 \)), sodium selenide (\( pH_3 \)) and sodium telluride (\( pH_4 \))?
(a) \( pH_1>pH_2=pH_3>pH_4 \)
(b) \( pH_1(c) \( pH_1(d) \( pH_1>pH_2>pH_3>pH_4 \)
Show Answer & Explanation

Answer: (d) \( pH_1>pH_2>pH_3>pH_4 \)

Explanation:
1. \( \ce{Na2O} \) gives NaOH, the strongest base.
2. The anion basicity falls down the group: \( \ce{S^2->Se^2->Te^2-} \) (their conjugate acids \( \ce{H2S, H2Se, H2Te} \) get stronger).
3. So \( pH_1>pH_2>pH_3>pH_4 \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 18: The compound whose 0.1 M solution is basic is
(a) ammonium acetate
(b) calcium carbonate
(c) ammonium sulphate
(d) sodium acetate
Show Answer & Explanation

Answer: (d) sodium acetate

Explanation:
1. Sodium acetate is the salt of a weak acid and a strong base, so the acetate ion hydrolyses and the solution is basic.
2. Ammonium acetate is nearly neutral, ammonium sulphate is acidic and \( \ce{CaCO3} \) is insoluble.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 19: The following reaction occurs in the body: \( \ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-} \). If \( \ce{CO2} \) escapes from the system
(a) pH will decrease
(b) hydrogen ion concentration will decrease
(c) H2CO3 concentration will be unaltered
(d) the forward reaction will be promoted
Show Answer & Explanation

Answer: (b) hydrogen ion concentration will decrease

Explanation:
1. Loss of \( \ce{CO2} \) shifts the equilibrium to the left.
2. So \( \ce{H2CO3} \) and \( \ce{H+} \) decrease and the pH rises.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 20: For preparing a buffer solution of pH 6 by mixing sodium acetate and acetic acid, the ratio of the concentration of salt and acid should be (\( K_a=10^{-5} \))
(a) 1 : 10
(b) 10 : 1
(c) 100 : 1
(d) 1 : 100
Show Answer & Explanation

Answer: (b) 10 : 1

Explanation:
1. \( pH=pK_a+\log\frac{[salt]}{[acid]} \): 6 = 5 + log(ratio).
2. So the ratio is 10 : 1.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 21: Which is incorrect for a buffer solution?
(a) It contains a weak acid and its conjugate base
(b) It contains a weak base and its conjugate acid
(c) There is very little change in pH when a small amount of acid or base is added
(d) None of the above
Show Answer & Explanation

Answer: (d) None of the above

Explanation:
1. All three statements correctly describe a buffer, so none of them is incorrect.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 22: The pH values of HCl and NaOH solutions each of strength N/100 are respectively
(a) 2 and 2
(b) 2 and 12
(c) 12 and 2
(d) 12 and 10
Show Answer & Explanation

Answer: (b) 2 and 12

Explanation:
1. 0.01 N HCl: pH = 2.
2. 0.01 N NaOH: pOH = 2, so pH = 12.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 23: When rain is accompanied by a thunderstorm, the collected rain water will have a pH value
(a) slightly lower than that of rain water without thunderstorm
(b) slightly higher than that when the thunderstorm is not there
(c) uninfluenced by occurrence of thunderstorm
(d) which depends on the amount of dust in air
Show Answer & Explanation

Answer: (a) slightly lower than that of rain water without thunderstorm

Explanation:
1. Lightning forms nitrogen oxides which dissolve to give nitric acid, so the rain is slightly more acidic (lower pH).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 24: The dissociation constant of an acid HA is \( 1\times10^{-5} \). The pH of a 0.1 molar solution of the acid will be
(a) five
(b) four
(c) three
(d) one
Show Answer & Explanation

Answer: (c) three

Explanation:
1. \( [H^+]=\sqrt{K_aC}=\sqrt{10^{-5}\times0.1}=10^{-3} \), so pH = 3.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 25: The pH value of \( 1.0\times10^{-8} \) M HCl is less than 8 because
(a) HCl is completely ionised at this concentration
(b) the ionisation of water is negligible
(c) the ionisation of water cannot be assumed to be negligible in comparison with this low concentration of HCl
(d) the pH cannot be calculated at such a low concentration of HCl
Show Answer & Explanation

Answer: (c) the ionisation of water cannot be assumed to be negligible in comparison with this low concentration of HCl

Explanation:
1. At such low concentration the \( 10^{-7} \) M of \( \ce{H+} \) from water is larger than that from the acid and must be included, giving pH about 6.98.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

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Equilibrium Objective Questions & Solutions for JEE Chemistry

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