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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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331 Medium
331 of 401
Application +4 | -1
Equal volumes of two solutions of a strong acid having PH=3 and PH=4{P^H} = 3\,and\,{P^H} = 4 are mixed together. The PH{P^H} of the resulting solution is [log⁡0.5=−0.3]\left[ {\log 0.5 = - 0.3} \right]
View answer and explanation
Correct option: B
3.26
Option B (3.26) is correct because it's the calculated pH of the mixture of two strong acid solutions using the average hydrogen ion concentration.
332 Medium
332 of 401
Understanding +4 | -1
Which of the following salts when dissolved in water will hydrolyze?
View answer and explanation
Correct option: C
NH4Cl{NH}_4 {Cl}
Option C (NH4Cl{NH}_4 {Cl}) is correct because ammonium chloride is a salt of a weak base (ammonium hydroxide) and a strong acid (hydrochloric acid), and therefore undergoes hydrolysis in water.
333 Medium
333 of 401
Application +4 | -1
Some solid ammonium carbonate NH4COONH4\mathrm{NH}_4 \mathrm{COONH}_4 is take in a evacuated vessel and the vessel is sealed after a very long time a constant pressure of 0.6 atm0.6 \mathrm{~atm} is observed due to dissociation of the solid into NH3\mathrm{NH}_3 and CO2\mathrm{CO}_2 gas. Calculate the dissociation constant of solid.
View answer and explanation
Correct option: A
0.032 atm30.032 \mathrm{~atm}^3
Option A (0.032 atm30.032 \mathrm{~atm}^3) is correct because it is calculated by determining the partial pressures of the products and then using them to calculate the equilibrium constant Kp.
334 Medium
334 of 401
Analyzing +4 | -1
XeF6+H2O⇌XeOF4+2HF;\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{XeOF}_4+2 \mathrm{HF}; equilibrium constant =K1.=K_1.

XeO4+XeF6⇌XeOF4+XeO3F2;\mathrm{XeO}_4+\mathrm{XeF}_6 \rightleftharpoons \mathrm{XeOF}_4+\mathrm{XeO}_3 \mathrm{F}_2; equilibrium constant =K2.=K_2.

The equilibrium constant for the following reaction will be:

XeO4+2HF⇌XeO3F2+H2O\mathrm{XeO}_4+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_3 \mathrm{F}_2+\mathrm{H}_2 \mathrm{O}
View answer and explanation
Correct option: C
K2K1\frac{K_2}{K_1}
Option C is correct because the equilibrium constant for the overall reaction is obtained by dividing the equilibrium constant of the second reaction by that of the first.
335 Medium
335 of 401
Understanding +4 | -1
What is the best description of the change that occurs when Na2O(s)\mathrm{N}{\mathrm{a}_2}\mathrm{O}\left( s \right) is dissolved in water?
View answer and explanation
Correct option: B
Oxide ion accepts sharing in a pair of electrons
Option C (Oxide ion donates a pair of electrons) is correct because when Na₂O dissolves in water, the oxide ion (O²⁻) acts as a Lewis base and donates a pair of electrons to a water molecule, forming hydroxide ions (OH⁻).
336 Medium
336 of 401
Understanding +4 | -1
Favorable condition for product formation in the given reaction.
SO2+1/2O2→SO3(g)\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3(\mathrm{g})
View answer and explanation
Correct option: A
High pressure
Option A (High pressure) is correct because the forward reaction decreases the number of moles of gas, so high pressure favors product formation according to Le Chatelier's principle.
337 Medium
337 of 401
Analyzing +4 | -1
Assertion :- Solubility of sparingly soluble salt increases with increase in temperature.

Reason :- Enthalpy of solution is negative for all salts.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the assertion is true (solubility of sparingly soluble salts generally increases with temperature), but the reason is false (enthalpy of solution can be positive or negative).
338 Medium
338 of 401
Application +4 | -1
A 0.1 M0.1 \, \mathrm{M} solution of ammonium acetate is diluted by adding 100 ml100 \, \mathrm{ml} of water. The pH{pH} of the resulting solution will be (pKa\left({pK}_{{a}}\right. of acetic acid is equal to pKb{pK}_{{b}} of NH4OH{NH}_4{OH})
View answer and explanation
Correct option: D
7.07.0
Option D (7.07.0) is correct because ammonium acetate is a salt of a weak acid and a weak base, and since pKa = pKb, the pH is 7.
339 Medium
339 of 401
Application +4 | -1
A volume of 20 ml20 \mathrm{~ml} of 0.8 M−HCN0.8 \mathrm{~M}-\mathrm{HCN} solution is mixed with 80 ml80 \mathrm{~ml} of 0.4 M−NaCN0.4 \mathrm{~M}-\mathrm{NaCN} solution. Calculate the pH\mathrm{pH} of the resulting solution. The value of KaK_{{a}} of HCN=2.5×10−10(log⁡2=0.3).\mathrm{HCN}=2.5 \times 10^{-10}(\log 2=0.3).
View answer and explanation
Correct option: A
9.9
Option A (9.9) is correct because the pH of the buffer solution is calculated using the Henderson-Hasselbalch equation.
340 Medium
340 of 401
Application +4 | -1
150 mL150 \mathrm{~mL} of 0.0008 M0.0008 \, \mathrm{M} ammonium sulphate is mixed with 150 mL150 \mathrm{~mL} of 0.04 M0.04 \, \mathrm{M} calcium nitrate. The ionic product of CaSO4{CaSO}_4 will be: (KSP=2.4×10−5\left({K}_{{SP}}=2.4 \times 10^{-5}\right. for CaSO4)\left.{CaSO}_4\right)
View answer and explanation
Correct option: A
<KSP< {K}_{{SP}}
Option A (<KSP< {K}_{{SP}}) is correct because the calculated ionic product is less than the given solubility product, indicating that a precipitate will not form.
341 Medium
341 of 401
Application +4 | -1
What should be the pHpH of solution to dissolve the Cr(OH)3  precipitate\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3}\mathrm{\;precipitate}
[Given, [Cr3+]=1.0  mol/L,  Ksp=6×10−31\left[ {\mathrm{C}{\mathrm{r}^{3 + }}} \right] = 1.0\;\mathrm{mol/L},\;{K_{\mathrm{sp}}} = 6 \times {10^{ - 31}}
View answer and explanation
Correct option: C
5
Option C (5) is correct because the pH must be greater than or equal to 4.253 to dissolve the precipitate, making pH 5 the appropriate choice.
342 Medium
342 of 401
Application +4 | -1
18  mL18\;\mathrm{mL} of mixture of acetic acid and sodium acetate required 6  mL6\;mL of 0.1  M  NaOH0.1\;M\;\mathrm{NaOH} for neutralization of the acid and 12  mL\mathrm{12}\;\mathrm{mL} of pKa\mathrm{p}{K_a} of the acid is 4.75,4.75, what is the pHpH of the mixture?
View answer and explanation
Correct option: A
5.055.05
Option A (5.05) is correct because it accurately calculates the pH of the buffer solution using the Henderson-Hasselbalch equation.
343 Medium
343 of 401
Application +4 | -1
0.04  g0.04\;\mathrm{g} of pure NaOH\mathrm{NaOH} is dissolved in 10  litre10\;\mathrm{litre} of distilled water. The pH\mathrm{pH} of the solution is:
View answer and explanation
Correct option: B
10
Option B (10) is correct because it is calculated by first finding the molarity of the NaOH solution, then calculating the pOH, and finally determining the pH.
344 Medium
344 of 401
Analyzing +4 | -1
Assertion (A): pH value of HClHCl solution is less than that of acetic acid of the some concentration.
Reason (R): Inequimolar solution, the number of titrable protons present in HClHCl is less than that present in acetic acid.
View answer and explanation
Correct option: C
Assertion is true but Reason is false
Option C is correct. HCl is a strong acid and dissociates completely, resulting in a lower pH than acetic acid, which is a weak acid. The reason is incorrect because HCl has more titrable protons.
345 Medium
345 of 401
Understanding +4 | -1
From a weak acid of pKa=6\mathrm{pK}_{\mathrm{a}}=6, the pH\mathrm{pH} range of practically possible buffer solution is
View answer and explanation
Correct option: A
5 to 7
Option A is correct because the pH range of a buffer solution is typically pKa±1\mathrm{pK}_{\mathrm{a}} \pm 1, so for a weak acid with pKa=6\mathrm{pK}_{\mathrm{a}}=6, the pH range is 5 to 7.
346 Medium
346 of 401
Application +4 | -1
The solubility of AgCl\mathrm{AgCl} in 0.1 M CaCl20.1 \mathrm{~M} \mathrm{~CaCl} 2 is (Ksp\left(\mathrm{K}_{\mathrm{sp}}\right. of AgCl=1×10−10\mathrm{AgCl}=1 \times 10^{-10} )
View answer and explanation
Correct option: B
5×10−10M5 \times 10^{-10} \mathrm{M}
Option B (5×10−10M5 \times 10^{-10} \mathrm{M}) is correct because the solubility product expression is used to calculate the solubility of AgCl in the presence of a common ion.
347 Medium
347 of 401
Understanding +4 | -1
The correct statement about buffer solution is:
View answer and explanation
Correct option: C
It shows little change in pHpH on adding small amount of an acid or base
Option C is correct because a buffer solution resists changes in pH upon the addition of small amounts of acid or base.
348 Medium
348 of 401
Understanding +4 | -1
According to Bronsted-Lowry concept, the correct order of relative strength of bases follows the order:
View answer and explanation
Correct option: C
OH−>CH3COO−>Cl−\mathrm{OH}^{-} > \mathrm{CH}_3 \mathrm{COO}^{-} > \mathrm{Cl}^{-}
Option C is correct. The order of relative base strength is determined by the strength of their conjugate acids, with weaker acids corresponding to stronger bases. OH- is the strongest base, followed by CH3COO-, and then Cl-.
349 Medium
349 of 401
Application +4 | -1
5 moles of XX are mixed with 3 moles of YY. At equilibrium for the reaction, X+Y  ⇋  Z,X + Y\;\leftrightharpoons\;Z, 2 moles of ZZ are formed. The equilibrium constant for the reaction will be
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (23\frac{2}{3}) is correct because the equilibrium constant is calculated as the ratio of the concentration of products to reactants at equilibrium, which in this case is 23\frac{2}{3}.
350 Medium
350 of 401
Remembering +4 | -1
If KcK_c is in the range of _____ appreciable concentrations of both reactants and products are present.
View answer and explanation
Correct option: B
10−310^{-3} to 10310^3
Option B (10−310^{-3} to 10310^3) is correct because when the equilibrium constant KcK_c is within this range, the reaction mixture contains appreciable amounts of both reactants and products.
351 Medium
351 of 401
Analyzing +4 | -1
(A): If solubility of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 is S\mathrm{S} mole/litre, its solubility product will be S2S^2

(R): Solubility product of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 is given by expression Ksp=[Ag+][CrO42−]K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CrO}_4^{2-}\right]
View answer and explanation
Correct option: D
Both (A) and (R) is false
Option D is correct because both the assertion and the reason are false. The solubility product of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 is 4S34S^3, and the expression for KspK_{sp} is Ksp=[Ag+]2[CrO42−]\mathrm{K_{sp}} = [\mathrm{Ag}^+]^2[\mathrm{CrO}_4^{2-}].
352 Medium
352 of 401
Application +4 | -1
What is the H+{H}^{+}ion concentration of a solution prepared by dissolving 4 g4 {~g} of NaOH{NaOH} (Atomic weight of Na=23 amu{Na}=23 \,{amu} ) in 1000 ml1000\, {ml} ?
View answer and explanation
Correct option: D
10−13M10^{-13} {M}
Option D (10−13M10^{-13} {M}) is correct because the hydrogen ion concentration is calculated using the ion product of water (Kw=10−14K_w = 10^{-14}).
353 Medium
353 of 401
Application +4 | -1
A buffer solution is prepared by mixing 0.1 M ammonia and 1.0 M ammonium chloride. At 298 K, the pKb  of  NH4OH  is  5.0.\mathrm{p}{K_b}\;\mathrm{of}\;\mathrm{ N}{\mathrm{H}_4}\mathrm{OH }\;\mathrm{is}\;5.0. The pH of the buffer is
View answer and explanation
Correct option: D
8
Option D (8) is correct because it is calculated using the Henderson-Hasselbalch equation for a buffer solution, considering the pKb of ammonia and the concentrations of ammonia and ammonium chloride.
354 Medium
354 of 401
Application +4 | -1
The equilibrium PCl5(g)    ⇋PCl3(g)+Cl2(g)\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\;\leftrightharpoons\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) shows that Kp{K_p} is 24.6324.63 times the value of Kc{K_c} at a particular temperature TT. Then TT (in KK) is:
View answer and explanation
Correct option: C
300
Option A (80%80 \% ) is correct because it's the degree of association calculated using the van't Hoff factor and the freezing point depression.
355 Medium
355 of 401
Application +4 | -1
The number of H+{H^ + } ions present in 1 ml of a solution having PH=13{P^H} = 13 is
View answer and explanation
Correct option: C
6.02×1076.02 \times {10^{7}}
Option C (6.02×1076.02 \times {10^{7}}) is correct because it calculates the number of H+ ions from pH and converts concentration to number of ions in 1 ml.
356 Medium
356 of 401
Analyzing +4 | -1
At 30∘C30^{\circ} \mathrm{C}, the solubility of Ag2CO3( Ksp=8×10−2){Ag}_2{CO}_3\left({~K}_{\mathrm{sp}}=8 \times 10^{-2}\right) will be maximum in -
View answer and explanation
Correct option: D
0.05 M NH30.05\, \mathrm{M~} {NH}_3
Option D (0.05 M NH30.05\, \mathrm{M~} {NH}_3) is correct because the solubility of Ag2CO3{Ag}_2{CO}_3 increases in NH3{NH}_3 due to the formation of a complex ion, while it decreases in Na2CO3{Na}_2{CO}_3 and AgNO3{AgNO}_3 due to the common ion effect.
357 Medium
357 of 401
Analyzing +4 | -1
Assertion (A): The degree of dissociation of CH3COOH\mathrm{CH}_3 \mathrm{COOH} is more in a solution which is basic than in water.

Reason (R): Ka\mathrm{K}_{\mathrm{a}} of CH3COOH\mathrm{CH}_3 \mathrm{COOH} increases in basic solution.
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C ((A) is true but (R) is false) is correct because the degree of dissociation increases in a basic solution, but the Ka value remains constant as it's a characteristic of the acid.
358 Medium
358 of 401
Application +4 | -1
When 0.5 mole of N2O4\mathrm{N}_2 \mathrm{O}_4 is placed in a 4.00 L4.00 \mathrm{~L} reaction vessel and heated at 405 K,79.3%405 \mathrm{~K}, 79.3 \% of the N2O4\mathrm{N}_2 \mathrm{O}_4 decomposes to NO2\mathrm{NO}_2. Calculate value of KP\mathrm{K}_{\mathrm{P}} at 405 K405 \mathrm{~K}, for the reaction : N2O4( g)⇌2NO2( g)\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}).
View answer and explanation
Correct option: A
50
The correct answer is A (50) because Kp is calculated using the equilibrium constant Kc and the ideal gas constant R, along with the temperature and change in moles.
359 Medium
359 of 401
Analyzing +4 | -1
What are the product formed when Deuterium is added equilibrium reaction of H2{H}_2 and I2{I}_2 ?

(i) HD{HD}

(ii) DI{DI}

(iii) D2{D}_2

(iv) HI{HI}
View answer and explanation
Correct option: D
All of these
The correct answer is D (All of these) because in a dynamic equilibrium, all possible isotopic combinations can form.
360 Medium
360 of 401
Analyzing +4 | -1
Which of the following statements are correct ?

(i) The extent of dissociation of an acid depends on the strength and polarity of the H−A{H}-{A} bond (where A{A} is an electronegative element.)

(ii) As the strength of H−A{H}-{A} bond increases, the energy required to break the bond decreases.

(iii) As the electronegativity difference between the atoms H{H} and A{A} increases, acidity increases
View answer and explanation
Correct option: C
(i) and (iii)
Option C ((i) and (iii)) is correct because the extent of dissociation depends on the H-A bond strength and polarity, and acidity increases with increasing electronegativity difference.
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