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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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451 Medium
451 of 680
Application +4 | -1
The pH\mathrm{pH} of mixture having 0.5 M NH4OH0.5 \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH} and 1.0 M NH4Cl1.0 \mathrm{~M} \mathrm{~NH}_4 \mathrm{Cl} is (Kb NH4OH=1×10−5)\left(\mathrm{K}_{\mathrm{b}} \mathrm{~NH}_4 \mathrm{OH}=1 \times 10^{-5}\right)
View answer and explanation
Correct option: A
8.78.7
Option A (8.7) is correct because the pH of the buffer solution is calculated using the Henderson-Hasselbalch equation for a basic buffer.
452 Easy
452 of 680
Remembering +4 | -1
The indicator used in the titration of sodium carbonate with sulphuric acid is
View answer and explanation
Correct option: C
Methyl orange
Option C (Methyl orange) is correct because methyl orange is a suitable indicator for titrations involving a strong acid and a weak base.
453 Medium
453 of 680
Application +4 | -1
What is the pH of a neutral solution at 37∘C,37^{\circ} \mathrm{C}, where KwK_{{w}} equals 2.5×10−14?2.5 \times 10^{-14}? (log⁡2=0.3)(\log 2=0.3)
View answer and explanation
Correct option: C
6.8
Option C (6.8) is correct because it accurately calculates the pH of a neutral solution at 37°C using the given Kw value.
454 Easy
454 of 680
Understanding +4 | -1
According to Le-Chatelier principle, adding heat a solid and liquid in equilibrium will cause the
View answer and explanation
Correct option: A
Amount of solid to decrease
Option A (Amount of solid to decrease) is correct because the fusion of a solid is an endothermic process, and adding heat favors the forward reaction (solid to liquid), thus decreasing the amount of solid.
455 Medium
455 of 680
Analyzing +4 | -1
Which statement is wrong about pH\mathrm{pH}?
View answer and explanation
Correct option: B
Adding 1 N1 \mathrm{~N}, solution of CH3COOH\mathrm{CH}_3 \mathrm{COOH} and 1 N NaOH1 \mathrm{~N} \mathrm{~NaOH} pH will be seven
Option B is wrong because a mixture of a weak acid (CH3COOH) and a strong base (NaOH) will result in a pH greater than 7 at equivalence point.
456 Medium
456 of 680
Analyzing +4 | -1
Statement (S): The [H+]\left[\mathrm{H}^{+}\right]in 10−8 NHCl10^{-8} \mathrm{~N} \mathrm{HCl} is 1.05×10−7M1.05 \times 10^{-7} \mathrm{M}.

Explanation(E): The H+\mathrm{H}^{+}is obtained from HCl\mathrm{HCl} and H2O\mathrm{H}_2 \mathrm{O}; the later furnishes less H+\mathrm{H}^{+}than 10−710^{-7} due to common ion effect.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation for S
The correct answer is C because both the statement and the explanation are correct, and the explanation correctly justifies why the [H+] in 10-8 N HCl is slightly higher than 10-7 M due to the contribution from water.
457 Medium
457 of 680
Application +4 | -1
If the solubility of Ag2CrO4A{g_2}Cr{O_4} is S   mol/litre.S\,\,\,{\text{mol/litre}}{\text{.}} Its solubility product is
View answer and explanation
Correct option: C
4S34{S^3}
Option C (4S34{S^3}) is correct because the solubility product for a salt of type A2BA_2B is given by 4S34S^3, where S is the solubility.
458 Medium
458 of 680
Application +4 | -1
The KspK_{\mathrm{sp}} of Ag2CrO4=1.2×10−11.\mathrm{Ag}_2 \mathrm{CrO}_4=1.2 \times 10^{-11}. What concentration of Ag+\mathrm{Ag}^{+} ion in aqueous solution will just fail to give a precipitate of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 with a solution in which [CrO4−2]=3×10−4M?\left[\mathrm{CrO}_4^{-2}\right]=3 \times 10^{-4} \mathrm{M}?
View answer and explanation
Correct option: D
2×10−4M2 \times 10^{-4} \mathrm{M}
Option D (2×10−4M2 \times 10^{-4} \mathrm{M}) is correct because it represents the maximum silver ion concentration that will not cause precipitation of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 given the chromate concentration and KspK_{\mathrm{sp}}.
459 Medium
459 of 680
Analyzing +4 | -1
Statement I : For the reaction, A+B⇌C;  KC=2A + B \rightleftharpoons C;\,\,{K_C} = 2
If we start with 1 mole each of A, B and C in 5 L flask, then at equilibrium, molar concentration of A and B increases and that of B decreases
Statement II : Reaction quotient Q>KCQ > {K_C} with given quantity of A, B and C, hence reaction is reversed.
View answer and explanation
Correct option: C
Both Statements are true
Both statements are true. Statement I is true because the reaction will shift to reach equilibrium. Statement II is true because Q > Kc, indicating the reaction will proceed in reverse.
460 Hard
460 of 680
Analyzing +4 | -1
The correct increasing order of solubility of the following substances in g/100 ml\mathrm{g} / 100 \mathrm{~ml} is PbSO4(Ksp=2×10−9),ZnS⁡(Ksp=1×10−22),AgBr⁡(Ksp=4×10−13),\mathrm{PbSO}_4\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right), \operatorname{ZnS}\left(K_{\mathrm{sp}}=1 \times 10^{-22}\right), \operatorname{AgBr}\left(K_{\mathrm{sp}}=4 \times 10^{-13}\right), CuCO3(Ksp=1×10−8).\mathrm{CuCO}_3\left(K_{\mathrm{sp}}=1 \times 10^{-8}\right).

(Atomic masses: Pb=208,Zn=65,Ag=108,Br=80,Cu=63)\mathrm{Pb}=208, \mathrm{Zn}=65, \mathrm{Ag}=108, \mathrm{Br}=80, \mathrm{Cu}=63)
View answer and explanation
Correct option: C
ZnS<AgBr<CuCO3<PbSO4\mathrm{ZnS} < \mathrm{AgBr} < \mathrm{CuCO}_3 < \mathrm{PbSO}_4
Option C (ZnS<AgBr<CuCO3<PbSO4\mathrm{ZnS} < \mathrm{AgBr} < \mathrm{CuCO}_3 < \mathrm{PbSO}_4) is correct because it represents the increasing order of solubility based on the calculated solubility from the given Ksp values.
461 Easy
461 of 680
Application +4 | -1
For the reaction, N2(g)+3H2(g) ⇋ 2NH3(g), {N_2}(g) + 3{H_2}(g)\,\leftrightharpoons\,2N{H_3}(g),\, If the value of equilibrium constant is K,K, then the equilibrium constant for the reaction, NH3(g) ⇋ 12N2(g)+32H2(g) N{H_3}(g)\,\leftrightharpoons\,\frac{1}{2}{N_2}(g) + \frac{3}{2}{H_2}(g)\, will be
View answer and explanation
Correct option: B
1K\frac{1}{{\sqrt K }}
Option B (1K\frac{1}{{\sqrt K }}) is correct because reversing the reaction and halving the stoichiometric coefficients results in taking the reciprocal of the square root of the original equilibrium constant.
462 Medium
462 of 680
Application +4 | -1
A saturated solution prepared by dissolved CaF2(s)\mathrm{Ca}{\mathrm{F}_2}\left( s \right) in water, has [Ca2+]=3.3×10−4\left[ {\mathrm{C}{\mathrm{a}^{2 + }}} \right] = 3.3 \times {10^{ - 4}} M. What is the Ksp{K_{\mathrm{sp}}} of CaF2\mathrm{Ca}{\mathrm{F}_2}?
View answer and explanation
Correct option: A
1.44×10−101.44 \times {10^{ - 10}}
Option A (1.44×10−101.44 \times {10^{ - 10}}) is correct because it is calculated using the solubility product expression for CaF2\mathrm{Ca}{\mathrm{F}_2}, where Ksp=[Ca2+][F−]2K_{sp} = [Ca^{2+}][F^-]^2.
463 Medium
463 of 680
Understanding +4 | -1
The total number of different kind of buffers obtained during the titration of H3PO4{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4} with NaOH\mathrm{NaOH} are:
View answer and explanation
Correct option: A
3
Option A (3) is correct because during the titration of H3PO4H_3PO_4 with NaOH, three buffer regions are formed corresponding to the three acidic protons.
464 Hard
464 of 680
Analyzing +4 | -1
The pH\mathrm{pH} value of 0.1 M0.1 \mathrm{~M} solutions of CH3COONa(I),CH3COOH(II),CH3COONH4(III),NaOH(IV)\mathrm{CH}_3 \mathrm{COONa} \text{(I)}, \mathrm{CH}_3 \mathrm{COOH} \text{(II)}, \mathrm{CH}_3 \mathrm{COONH}_4 \text{(III)}, \mathrm{NaOH} \text{(IV)} and HCl(V){HCl}(\mathrm{V}) is in the order
View answer and explanation
Correct option: C
V << II << III << I << IV
Option C is correct because it correctly orders the pH values of the given solutions from most acidic to most basic.
465 Easy
465 of 680
Application +4 | -1
In the reaction, N2O4  ⇋  2NO2{\mathrm{N}_2}{\mathrm{O}_4}\;\leftrightharpoons\;2\mathrm{N}{\mathrm{O}_2}, α\mathrm{\alpha } is that part of N2O4{\mathrm{N}_2}{\mathrm{O}_4} which dissociates, then the number of moles at equilibrium will be
View answer and explanation
Correct option: C
(1+α)\left( {1 + \mathrm{\alpha }} \right)
The correct answer is C ((1+α)\left( {1 + \mathrm{\alpha }} \right)) because it accurately calculates the total number of moles at equilibrium based on the dissociation of N2O4.
466 Medium
466 of 680
Understanding +4 | -1
The total number of different kind of buffers obtained during the titration of H3PO4{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4} with NaOH\mathrm{NaOH} are:
View answer and explanation
Correct option: A
3
Option A (3) is correct because three different buffer solutions can be formed during the titration of H3PO4 with NaOH.
467 Medium
467 of 680
Analyzing +4 | -1
STATEMENT I : pH\mathrm{pH} of a aqueous solution of acetic acid remains unchanged on the addition of sodium acetate.

STATEMENT II : Dissociation of acetic acid is increased by the addition of sodium acetate due to common ion effect.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B (Statement I, II are incorrect) is correct because the pH changes due to the common ion effect, and the dissociation of acetic acid decreases.
468 Hard
468 of 680
Application +4 | -1
A gaseous phase reaction taking place in 1 L1 \mathrm{~L} flask at 400 K400 \mathrm{~K} is given, N2(g)+3H2(g)⇌2NH3(g)\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g).
Starting with 1 mole N2\mathrm{N}_2 and 3 moles H2\mathrm{H}_2, equilibrium mixture required 250 mL250 \mathrm{~mL} of 1 M H2SO41 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4. Thus, KcK_c is
View answer and explanation
Correct option: C
0.029
Option C (0.029) is correct because the equilibrium constant Kc is calculated based on the stoichiometry of the reaction and the amount of ammonia produced, which is determined by the titration with sulfuric acid.
469 Medium
469 of 680
Application +4 | -1
An aqueous solution contains Ni2+,Co2+  and  Pb2+\mathrm{N}{\mathrm{i}^{2 + }},\mathrm{C}{\mathrm{o}^{2 + }}\;\mathrm{and}\;\mathrm{ P}{\mathrm{b}^{2 + }} ions at equal concentrations. The solubility product of NiS,PbS  and  CoS\mathrm{NiS},\mathrm{PbS}\;\mathrm{ and }\;\mathrm{CoS} in water at 25∘C25^\circ C are 1.4×10−24,  3.4×10−28  and  3×10−261.4 \times {10^{ - 24}},\;3.4 \times {10^{ - 28}}\;\mathrm{and}\;3 \times {10^{ - 26}}, respectively. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero?
View answer and explanation
Correct option: D
PbS  and  NiS\mathrm{PbS }\;\mathrm{and }\;\mathrm{NiS}
Option D (PbS  and  NiS\mathrm{PbS }\;\mathrm{and }\;\mathrm{NiS}) is correct because PbS, having the lowest solubility product, will precipitate first, and NiS, with the highest solubility product, will precipitate last.
470 Medium
470 of 680
Analyzing +4 | -1
Assertion : Mixture of CH3COOH{CH}_3 {COOH} and CH3COONH4{CH}_3 {COONH}_4 is an example of acidic buffer.

Reason: Acidic buffer contains equimolar mixture of a weak acid and its salt with weak base.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because an acidic buffer requires a weak acid and a salt of that weak acid with a strong base, not a weak base.
471 Medium
471 of 680
Analyzing +4 | -1
The solubility of CaCO3\mathrm{CaCO}_3 is 7 mg/litre.7 \mathrm{~mg/litre}. Calculate the solubility product of BaCO3\mathrm{BaCO}_3 from this information and from the fact that when Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is added slowly to a solution containing equimolar concentration of Ca2+\mathrm{Ca}^{2+} and Ba2+,\mathrm{Ba}^{2+}, no precipitate of CaCO3\mathrm{CaCO}_3 is formed until 90%90 \% of Ba2+\mathrm{Ba}^{2+} has been precipitated as BaCO3.\mathrm{BaCO}_3.
View answer and explanation
Correct option: C
4.9×10−104.9 \times 10^{-10}
Option C (4.9×10−104.9 \times 10^{-10}) is correct because it is calculated using the solubility of CaCO3 and the relative precipitation of BaCO3 compared to CaCO3.
472 Medium
472 of 680
Application +4 | -1
pH\mathrm{pH} of an acidic buffer is 7 and pKa\mathrm{pKa} is 5 then the ratio of [ salt ][ Acid ]\frac{[\text { salt }]}{[\text { Acid }]} is:-
View answer and explanation
Correct option: B
100
Option B (100) is correct because it is the ratio of [salt]/[acid] calculated using the Henderson-Hasselbalch equation for an acidic buffer.
473 Medium
473 of 680
Application +4 | -1
A 0.2 molar solution of formic acid is 3.2%3.2 \% ionised, its ionisation constant is
View answer and explanation
Correct option: B
2.1×10−42.1 \times 10^{-4}
The correct answer is B (2.1×10−42.1 \times 10^{-4}) because the ionization constant is calculated using the degree of ionization and the initial concentration of the formic acid.
474 Medium
474 of 680
Analyzing +4 | -1
Assertion :- For the reaction A⇌B+C+DA \rightleftharpoons B + C + D at equilibrium state [A]=[B]=[C]=3×10−4M[A]=[B]=[C]=3 \times 10^{-4} \mathrm{M} and equilibrium constant is 3×10−4.3 \times 10^{-4}.
Reason :- Given reaction is exothermic.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because the assertion is true (equilibrium state is given), but the reason is false (the reaction's equilibrium constant doesn't imply it's exothermic).
475 Medium
475 of 680
Analyzing +4 | -1
Statement I : For the equilibrium H2(g)+I2(g)⇌2HI(g),KCH_{2(g)}+I_{2(g)} \rightleftharpoons 2 H I_{(g)}, K_C is doubled if volume of the vessel is reduced to half its initial volume.
Statement II : Decrease in volume proportionately decrease the concentrations of both reactants & products.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B is correct because the equilibrium constant KcK_c depends only on temperature and is independent of volume changes for reactions with Δng=0\Delta n_g = 0.
476 Medium
476 of 680
Application +4 | -1
The values of Ksp\mathrm{K}_{\mathrm{sp}} of CaCO3\mathrm{CaCO}_3 and CaC2O4\mathrm{CaC}_2 \mathrm{O}_4 are 4.7×10−94.7 \times 10^{-9} and 1.3×10−91.3 \times 10^{-9} respectively at 25∘C25^{\circ} \mathrm{C}. If the mixture of these two is washed with water, what is the concentration of Ca2+\mathrm{Ca}^{2+} ions in water?
View answer and explanation
Correct option: A
7.746×10−5 M7.746 \times 10^{-5} \mathrm{~M}
Option A (7.746×10−5 M7.746 \times 10^{-5} \mathrm{~M}) is correct because it involves calculating the total calcium ion concentration by considering the solubility product constants of both calcium carbonate and calcium oxalate.
477 Medium
477 of 680
Application +4 | -1
The pH of an aqueous solution having hydroxide ion concentration as 1×10−51 \times {10^{ - 5}} is
View answer and explanation
Correct option: B
9
Option B (9) is correct because pH is calculated from pOH using the relationship pH + pOH = 14, where pOH is derived from the hydroxide ion concentration.
478 Medium
478 of 680
Application +4 | -1
Find change in pH\mathrm{pH} of water when 1 mL1 \mathrm{~mL} of 10−3M10^{-3} \mathrm{M} KOH\mathrm{KOH} is added to 99 mL99 \mathrm{~mL} of H2O\mathrm{H}_2 \mathrm{O} at 25∘C25^{\circ} \mathrm{C}.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the addition of KOH to water increases the pH from 7 to 9, resulting in a change of 2 pH units.
479 Hard
479 of 680
Application +4 | -1
Hydrolysis of sucrose gives

sucrose + H2O⇋{H_2}O\underset{{}}{\overset{{}}{\leftrightharpoons}} Glucose + Fructose

Equilibrium constant KC{K_C} for the reaction is 2×10132 \times {10^{13}} at 300 K. calculate ΔGo\Delta {G^\mathrm{o}} at 300 K
View answer and explanation
Correct option: A
−7.64×104 J mol−1- 7.64 \times {10^4}\,J\,mo{l^{ - 1}}
Option A (−7.64×104 J mol−1-7.64 \times 10^4 \,J\,mol^{-1}) is correct because it is calculated using the formula ΔG∘=−2.303RTlog⁡Kc\Delta G^{\circ} = -2.303RT\log K_c.
480 Medium
480 of 680
Analyzing +4 | -1
Which statement is false? (Assume complete dissociation in each case)
View answer and explanation
Correct option: D
In a 0.001 M solution of NaOH\mathrm{NaOH} the concentration of H+ is 10−3{\mathrm{H}^ + }\,\mathrm{is}\,{10^{ - 3}} mol/L
Option D (In a 0.001 M solution of NaOH\mathrm{NaOH} the concentration of H+ is 10−3{\mathrm{H}^ + }\,\mathrm{is}\,{10^{ - 3}} mol/L) is false because in a 0.001 M NaOH solution, the [H+][H^+] is 10−1110^{-11} M, not 10−310^{-3} M.
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