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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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271 Medium
271 of 401
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}}.
272 Medium
272 of 401
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.
273 Medium
273 of 401
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.
274 Medium
274 of 401
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.
275 Medium
275 of 401
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.
276 Medium
276 of 401
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.
277 Medium
277 of 401
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.
278 Medium
278 of 401
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.
279 Medium
279 of 401
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.
280 Medium
280 of 401
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.
281 Medium
281 of 401
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.
282 Medium
282 of 401
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).
283 Medium
283 of 401
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.
284 Medium
284 of 401
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.
285 Medium
285 of 401
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.
286 Medium
286 of 401
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.
287 Medium
287 of 401
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.
288 Medium
288 of 401
Analyzing +4 | -1
Assertion (A): At an equilibrium A(g)+2B(g)⇌C(g)\mathrm{A}(\mathrm{g})+2 \mathrm{B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g}) if substantial amount of water is added to the mixture and stated that only A(g)\mathrm{A}(\mathrm{g}) gets dissolved to a certain extent in water then equilibrium shifts towards forward direction.

Reason (R): On decreasing the volume of reaction mixture and keeping rest of things same the equilibrium shifts to a direction having more number of gaseous molecules.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because both the assertion and the reason are false. Adding water and dissolving A(g) doesn't necessarily shift the equilibrium forward, and decreasing volume shifts the equilibrium towards fewer gaseous molecules, not more.
289 Medium
289 of 401
Application +4 | -1
The pH\mathrm{pH} of a buffer solution containing 2.0 mol2.0 \mathrm{~mol} per litre CH3COONa\mathrm{CH}_3 \mathrm{COONa} and 1.5 mol1.5 \mathrm{~mol} per litre CH3COOH\mathrm{CH}_3 \mathrm{COOH} is: (Ka\left(K_a\right. for acetic acid is 1.8×10−51.8 \times 10^{-5} )
View answer and explanation
Correct option: A
4.874.87
Option A (4.87) is correct because the pH of the buffer solution is calculated using the Henderson-Hasselbalch equation.
290 Medium
290 of 401
Application +4 | -1
The pKb{pK}_{{b}} value of ammonium hydroxide is 4.75. An aqueous solution of ammonium hydroxide is titrated with HCl{HCl}. The pH{pH} of the solution at the point where half of ammonium hydroxide has been neutralised will be
View answer and explanation
Correct option: A
9.25
Option A (9.25) is correct because at half neutralization, pH = pKa, and pKa + pKb = 14.
291 Medium
291 of 401
Understanding +4 | -1
For the gaseous phase reaction, 2NO⇋N2+O22NO\leftrightharpoons{N_2} + \mathrm{ }{O_2}, ΔH∘=−43.5  k  cal  mol−1,\Delta {H^ \circ } = \mathrm{ } - \mathrm{ }43.5\mathrm{ }\;k\;cal\mathrm{ }\;mo{l^{ - 1}},  which statement is correct for, N2(g)+O2(g)⇋2NO(g)?{N_2}\left( g \right)\mathrm{ } + \mathrm{ }{O_2}\left( g \right)\mathrm{ }\leftrightharpoons\mathrm{ }2NO\left( g \right)?
View answer and explanation
Correct option: C
KK decreases as temperature decreases
Option C (KK decreases as temperature decreases) is correct because the reverse reaction (formation of NO) is endothermic, so decreasing the temperature will shift the equilibrium to the left, decreasing K.
292 Medium
292 of 401
Application +4 | -1
For the reaction A( g)+3B( g)⇌2C( g)A({~g})+3 B({~g}) \rightleftharpoons 2 C({~g}) at 27∘C,227^{\circ} {C}, 2 moles of A,4A, 4 moles of BB and 6 moles of CC are present in 2 litre vessel. If KcK_c for the reaction is 1.2 , the reaction will proceed in :
View answer and explanation
Correct option: A
forward direction
Option A (forward direction) is correct because the reaction quotient Qc is less than Kc, indicating that the reaction will shift towards the products to reach equilibrium.
293 Medium
293 of 401
Analyzing +4 | -1
Consider the reactions
(i) CO(g)+H2O(g)⇌K1CO2(g)+H2(g)\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \stackrel{K_1}{\rightleftharpoons} \mathrm{CO}_2(g)+\mathrm{H}_2(g)
(ii) CH4(g)+H2O(g)⇌K2CO(g)+3H2(g)\mathrm{CH}_4(g)+\mathrm{H}_2 \mathrm{O}(g) \stackrel{K_2}{\rightleftharpoons} \mathrm{CO}(g)+3 \mathrm{H}_2(g)

(iii) CH4(g)+2H2O(g)⇌K3CO2(g)+4H2(g)\mathrm{CH}_4(g)+2 \mathrm{H}_2 \mathrm{O}(g) \stackrel{K_3}{\rightleftharpoons} \mathrm{CO}_2(g)+4 \mathrm{H}_2(g)

Which of the following is correct relation?
View answer and explanation
Correct option: C
K3=K1⋅K2K_3=K_1 \cdot K_2
Option C (K3=K1⋅K2K_3=K_1 \cdot K_2) is correct. Reaction (iii) is the sum of reactions (i) and (ii), so its equilibrium constant is the product of the equilibrium constants of (i) and (ii).
294 Medium
294 of 401
Application +4 | -1
Of the following which change will shift the reaction towards the product?

I2( g)⇌2I(g),ΔHr0(298 K)=+150 kJ{I}_2({~g}) \rightleftharpoons 2 {I}({g}), \Delta H_r^0(298 {~K})=+150 {~kJ}
View answer and explanation
Correct option: C
Increase in temperature
Option C (Increase in temperature) is correct because the reaction is endothermic, and increasing the temperature favors the forward reaction, shifting the equilibrium towards the product.
295 Medium
295 of 401
Understanding +4 | -1
Which of the following ions can exist in solution?
View answer and explanation
Correct option: D
all of these
Option D (all of these) is correct because all the listed ions, H3O+\mathrm{H}_3 \mathrm{O}^{+}, H3O2−\mathrm{H}_3 \mathrm{O}_2^{-}, and H5O2+\mathrm{H}_5 \mathrm{O}_2^{+}, can exist in solution.
296 Medium
296 of 401
Application +4 | -1
In a 5.76 L5.76 \mathrm{~L} vessel, 0.5 moles of H2\mathrm{H}_2 gas and 0.5 moles of I2\mathrm{I}_2 vapours are allowed to react to form HI(g)\mathrm{HI}(\mathrm{g}) at 447∘C,447^{\circ} \mathrm{C}, then the total pressure of gases at equilibrium would be (R=0.08 L−atm/K−mol)(R=0.08 \mathrm{~L}-\mathrm{atm} / \mathrm{K}-\mathrm{mol})
View answer and explanation
Correct option: B
10 atm10 \mathrm{~atm}
Option B (10 atm) is correct because the total pressure at equilibrium is calculated using the ideal gas law, considering that the number of moles remains constant due to Δng=0\Delta n_g = 0.
297 Medium
297 of 401
Application +4 | -1
If solubility of gas 'x' in water is 0.5 g/L at 1 bar pressure then solubility at 4 bar pressure will be
View answer and explanation
Correct option: B
2 g/L
Option B (Al{Al}) is correct because Aluminum is more reactive than Zinc and will displace Zinc from the ZnCl2 solution.
298 Medium
298 of 401
Analyzing +4 | -1
Statement I 1 mole A(g)A(g) and 1 mole B(g)B(g) give 0.5 mole of C(g)C(g) and 0.5 mole D(g)D(g) at equilibrium.

A(g)+B(g)⇌C(g)+D(g)A(g)+B(g) \rightleftharpoons C(g)+D(g)

On taking 2 moles each of A(g)A(g) and B(g)B(g), percentage dissociation A(g)A(g) and B(g)B(g) is also doubled.

Statement II Equilibrium constant, Kc=1K_c=1
View answer and explanation
Correct option: D
Statement II is incorrect but Statement I is correct
Option C (ΔGreaction ∘<0,Ecell <0\Delta \mathrm{G}_{\text {reaction }}^{\circ} < 0, \mathrm{E}_{\text {cell }} < 0) is correct because the standard Gibbs free energy change is negative, indicating spontaneity under standard conditions, but the cell potential is negative under the given non-standard conditions.
299 Medium
299 of 401
Application +4 | -1
Determine the degree of dissociation of 0.05 M NH4OH0.05 \mathrm{~M} \, {NH}_4 {OH} at 25∘C25^{\circ} \mathrm{C} in a solution of pH=10\mathrm{pH}=10.
View answer and explanation
Correct option: A
0.2%0.2 \%
Option A (0.2 %) is correct because it is the degree of dissociation calculated using the given pH and concentration of NH4OH.
300 Medium
300 of 401
Analyzing +4 | -1
In 2HI(g)⇌H2(g)+I2(g),22 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}), 2 mole HI,4 mole H2\mathrm{HI}, 4 \mathrm{~mole}\, \mathrm{H}_2 and 4 mole of I2\mathrm{I}_2 are found at equilibrium. If some hydrogen is now introduced into the container such that at equilibrium concentration of HI\mathrm{HI} and I2\mathrm{I}_2 become equal. Find the Number of mole of H2\mathrm{H}_2 added.
View answer and explanation
Correct option: D
None of these
Option D is correct because the number of moles of H2 added to make the concentrations of HI and I2 equal is not among the provided options. The equilibrium constant is calculated correctly, and the subsequent calculations lead to a value not listed in the options.
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