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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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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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631 Hard
631 of 680
Application +4 | -1
The solubility product of SrF2\mathrm{SrF}_2 in water is 8×10−108 \times 10^{-10}. Find its solubility in 0.1 M NaF0.1 \mathrm{~M} \mathrm{~NaF} aqueous solution.
View answer and explanation
Correct option: A
8×10−8M8\times 10^{-8} \mathrm{M}
Option A (8×10−88 \times 10^{-8} M) is correct because it calculates the solubility of SrF₂ in a NaF solution, considering the common ion effect.
632 Medium
632 of 680
Application +4 | -1
For the reaction AB(g)⇌A(g)+B(g),    ABAB\left( g \right) \rightleftharpoons A\left( g \right) + B\left( g \right),\,\,\,\,AB is 33% dissociated at a total pressure of PP. Therefore, PP is related to KP{K_P} by one of the following options.
View answer and explanation
Correct option: D
P=8KpP = 8{K_p}
Option D (P=8KpP = 8{K_p}) is correct because it is derived from the equilibrium constant expression calculated using the given dissociation percentage.
633 Medium
633 of 680
Analyzing +4 | -1
Statement (S): The solubility of HgI2\mathrm{HgI}_2 in water decreases in presence of KI\mathrm{KI}.

Explanation (E): HgI2\mathrm{HgI}_2 is insoluble in water but it becomes soluble in KI(aq)\mathrm{KI}(a q).
View answer and explanation
Correct option: B
S is wrong but E is correct
Option B is correct. HgI2 is sparingly soluble in water, but its solubility increases in the presence of KI due to the formation of a soluble complex ion.
634 Medium
634 of 680
Analyzing +4 | -1
(A): pH\mathrm{pH} of blood is maintained inspite of acidic foods.

(R): Acidity of foods is not so large to change the pH\mathrm{pH} of blood.
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 blood pH is maintained by buffer systems, not because the acidity of food is insignificant.
635 Easy
635 of 680
Application +4 | -1
For an equilibrium reaction involving gases, the forward reaction is 1st order while the reverse reaction is 2nd order. The units of Kb{{K_b}} for the forward equilibrium is:
View answer and explanation
Correct option: A
atm\mathrm{atm}
Option A (atm) is correct because the units of Kp depend on the difference in the stoichiometric coefficients of gaseous products and reactants.
636 Medium
636 of 680
Application +4 | -1
The equilibrium constant for a reaction A+2B⇌2CA+2 B \rightleftharpoons 2 C is 40. The equilibrium constant for reaction C⇌B+12AC \rightleftharpoons B+\frac{1}{2} A is
View answer and explanation
Correct option: B
(140)12\left(\frac{1}{40}\right)^{\frac{1}{2}}
Option B ((140)12\left(\frac{1}{40}\right)^{\frac{1}{2}}) is correct because reversing and halving the reaction will take the inverse square root of the equilibrium constant.
637 Easy
637 of 680
Application +4 | -1
If equilibrium constants of reactions AB and  CD are K1 and  K2AB\,and\,\,CD\,are\,{K_1}\,and\,\,{K_2} respectively, then the equilibrium constant for the reaction D+BA+CD + BA + C will be
View answer and explanation
Correct option: B
1K1K2\frac{1}{{{K_1}{K_2}}}
Option B (1K1K2\frac{1}{{{K_1}{K_2}}}) is correct because the reaction is the reverse of the sum of the reactions with equilibrium constants K1 and K2.
638 Medium
638 of 680
Application +4 | -1
pH\mathrm{pH} of Ca(OH)2\mathrm{Ca}(\mathrm{OH})_2 solution is (Ksp=4×10−15)\left(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-15}\right)
View answer and explanation
Correct option: C
9.39.3
Option C (i=1−α+αn1\mathrm{i}=\frac{1-\alpha+\frac{\alpha}{\mathrm{n}}}{1}) is correct because it represents the van't Hoff factor for association of a solute.
639 Medium
639 of 680
Application +4 | -1
One litre of an aqueous solution contain 0.15 moles of CH3COOH(pka=4.8)C{H_3}COOH\left( {pka = 4.8} \right) and 1.5 mole of CH3COONaC{H_3}COONa . The pH of solution is:
View answer and explanation
Correct option: C
5.8
Option C (5.8) is correct because the pH is calculated using the Henderson-Hasselbalch equation for a buffer solution.
640 Medium
640 of 680
Analyzing +4 | -1
Assertion :- Addition of NH4OH{NH}_4 {OH} to an aqueous solution of BaCl2{BaCl}_2 in the presence of NH4Cl{NH}_4 {Cl} (excess) precipitates Ba(OH)2.{Ba}({OH})_2.

Reason :- Ba(OH)2{Ba}({OH})_2 is insoluble in water.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because the assertion is false as Ba(OH)2 doesn't precipitate in the presence of NH4Cl and NH4OH, and the reason is false because Ba(OH)2 is soluble in water.
641 Medium
641 of 680
Analyzing +4 | -1
Which of the following set of an acid &\& base (one equivalent each) on neutralization will release maximum energy?
View answer and explanation
Correct option: D
HF+NaOH{HF}+{NaOH}
Option D (HF + NaOH) is correct because the neutralization of HF, a weak acid, with a strong base like NaOH releases more energy due to the high hydration energy of the fluoride ion.
642 Medium
642 of 680
Understanding +4 | -1
The hydroxide with highest solubility product is:
View answer and explanation
Correct option: B
Co(OH)2\mathrm{Co}{\left( {\mathrm{OH}} \right)_2}
Option B (Co(OH)2\mathrm{Co}{\left( {\mathrm{OH}} \right)_2}) is correct because Cobalt(II) hydroxide has a higher solubility product compared to the other listed hydroxides.
643 Medium
643 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.
644 Medium
644 of 680
Application +4 | -1
The ionization constant of HCN is 4.5×10−104.5\mathrm{ } \times \mathrm{ }{10^{ - 10}} at 298 K. Hydrolysis constant of sodium cyanide is
View answer and explanation
Correct option: C
2.22×10−−52.22\mathrm{ } \times \mathrm{ }{10^{--5}}
Option C (2.22 x 10⁻⁵) is correct because it is the hydrolysis constant of sodium cyanide calculated using the relationship between Kw, Ka, and Kh.
645 Medium
645 of 680
Application +4 | -1
Solid Ba(NO3)2\mathrm{Ba}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2} is gradually dissolved in a 1.0×10−4  M  Na2CO31.0 \times {10^{ - 4}}\;M\;\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3} solution. At what concentration of Ba2+\mathrm{B}{\mathrm{a}^{2 + }} will a precipitate begin to form?

(Ksp  for  BaCO3=5.1×10−9)({K_{sp}}\;\mathrm{for}\;\mathrm{BaC}{\mathrm{O}_3} = 5.1 \times {10^{ - 9}})
View answer and explanation
Correct option: B
5.1×10−5  M5.1 \times {10^{ - 5}}\;M
Option B (5.1×10−5  M5.1 \times {10^{ - 5}}\;M) is correct because it's the concentration of Ba2+Ba^{2+} calculated using the KspK_{sp} and the concentration of CO32−CO_3^{2-}.
646 Medium
646 of 680
Application +4 | -1
KPK_{\mathrm{P}} for formation of ethane from hydrogen and ethylene is 5.5×1018 atm−15.5 \times 10^{18} \mathrm{~atm}^{-1} and KPK_{\mathrm{P}} for formation of ethylene from hydrogen and acetylene is 5×1026 atm−15 \times10^{26} \mathrm{~atm}^{-1} at 323 K.323 \mathrm{~K}. What is KpK_{\mathrm{p}} for the reaction between hydrogen and acetylene to form ethane at 323 K?323 \mathrm{~K}?
View answer and explanation
Correct option: A
2.75×1045 atm−22.75 \times 10^{45} \mathrm{~atm}^{-2}
Option A (2.75×1045 atm−22.75 \times 10^{45} \mathrm{~atm}^{-2}) is correct because it is the equilibrium constant calculated by multiplying the equilibrium constants of the two given reactions.
647 Medium
647 of 680
Analyzing +4 | -1
The pH\mathrm{pH} of 0.01(M)KOH0.01(\mathrm{M}) \mathrm{KOH} is 12 ; if the temperature of the given KOH\mathrm{KOH} solution is increased which of the following would occur?
View answer and explanation
Correct option: B
pH\mathrm{pH} would be decreased but pOH\mathrm{pOH} remain constant
Option B (pH would be decreased but pOH remain constant) is correct because the concentration of OH- remains the same, but Kw increases with temperature, leading to a decrease in pH.
648 Easy
648 of 680
Application +4 | -1
If the pHpH of a solution is 2, its normality will be:
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the normality depends on the acid's nature (strong or weak) and its dissociation.
649 Medium
649 of 680
Application +4 | -1
At 448∘C448^{\circ} \mathrm{C}, the equilibrium constant (KC)\left(K_C\right) for the reaction H2(g)+I2(g)⇌2HI(g)\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) is 50.5.50.5 . If we start with 2×10−22 \times 10^{-2} mol of HI,1×10−2 mol\mathrm{HI}, 1 \times 10^{-2} \mathrm{~mol} of H2\mathrm{H}_2 and 3×10−2 mol I23 \times 10^{-2} \mathrm{~mol} \, \mathrm{I}_2 in a 2 L2 \mathrm{~L} vessel, then at 448∘C448^{\circ} \mathrm{C}
View answer and explanation
Correct option: B
Reaction will proceed in the forward direction
Option B (Reaction will proceed in the forward direction) is correct because the reaction quotient (Q) is less than the equilibrium constant (K), indicating the reaction needs to shift towards products.
650 Easy
650 of 680
Understanding +4 | -1
A cylinder provided with a piston has some PCl5\mathrm{PCl}_5 which is in equilibrium with PCl3\mathrm{PCl}_3 and Cl2\mathrm{Cl}_2. The system is compressed with the help of piston. Indicate the correct statement.
View answer and explanation
Correct option: C
PCl3\mathrm{PCl}_3 and Cl2\mathrm{Cl}_2 will combine to form PCl5\mathrm{PCl}_5
The correct answer is C. Compressing the system favors the reverse reaction, combining PCl3PCl_3 and Cl2Cl_2 to form PCl5PCl_5.
651 Hard
651 of 680
Application +4 | -1
What is the molarity of F−\mathrm{F}^{-}in a saturated solution of InF⁡3\operatorname{InF}_3 ? (Ksp=7.9×10−10)\left(K_{s p}=7.9 \times 10^{-10}\right)
View answer and explanation
Correct option: D
7.0×10−37.0 \times 10^{-3}
Option D (7.0×10−37.0 \times 10^{-3}) is not the correct answer. Option B (8.3×10−38.3 \times 10^{-3}) is the correct answer because it is calculated using the solubility product (KspK_{sp}) expression for InF⁡3\operatorname{InF}_3.
652 Easy
652 of 680
Understanding +4 | -1
Which one of the following species acts as both Bronsted acid and base?
View answer and explanation
Correct option: C
HPO42−\mathrm{HPO}_4^{2 - }
Option C (HPO42−\mathrm{HPO}_4^{2 - }) is correct because it can both accept a proton (acting as a base) and donate a proton (acting as an acid).
653 Medium
653 of 680
Application +4 | -1
At 273 K273 \mathrm{~K} and 1 atm,1 L1 \mathrm{~atm}, 1 \mathrm{~L} of N2O4(g)\mathrm{N}_2 \mathrm{O}_4(g) decomposes to NO2(g)\mathrm{NO}_2(g) as given,

N2O4(g)⇌2NO2(g)\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)

At equilibrium, original volume is 25%25 \% less than the existing volume. Percentage decomposition of N2O4(g)\mathrm{N}_2 \mathrm{O}_4(g) is thus,
View answer and explanation
Correct option: D
0.3333
The correct answer is 33.33% because the percentage decomposition can be calculated from the change in volume at equilibrium, which is related to the degree of dissociation.
654 Easy
654 of 680
Remembering +4 | -1
Soda water has a pH value:
View answer and explanation
Correct option: A
Less than 7
Option A (Less than 7) is correct because soda water contains dissolved CO2, which forms carbonic acid, making it acidic.
655 Medium
655 of 680
Analyzing +4 | -1
What is the solubility product of Al(OH)3{Al}({OH})_3 in water? Given:

Al(OH)4−(aq)⇌Al3+(aq)+4OH−(aq);K=1.3×10−34{Al}({OH})_4^{-}(\mathrm{aq}) \rightleftharpoons {Al}^{3+}(\mathrm{aq})+4 {OH}^{-}(\mathrm{aq}); K=1.3 \times 10^{-34}

Al(OH)3(s)+OH−(aq)⇌Al(OH)4−(aq);K=38.5{Al}({OH})_3(\mathrm{s})+{OH}^{-}(\mathrm{aq}) \rightleftharpoons {Al}({OH})_4^{-}(\mathrm{aq}) ; K=38.5
View answer and explanation
Correct option: B
5×10−335 \times 10^{-33}
The correct answer is B (5×10−335 \times 10^{-33}) because the solubility product is calculated by multiplying the equilibrium constants of the given reactions.
656 Medium
656 of 680
Application +4 | -1
Choose the correct statement(s)

A) If the solubility of Sb2S3{Sb}_2 S_3 is 1.0×10−5 mol/L1.0 \times 10^{-5} {~mol} / {L} at 298 K298 {~K}, its solubility product will be 108×10−25108 \times 10^{-25}

B) pH{pH} of an aqueous solution having [H+]=10−8M\left[{H}^{+}\right]=10^{-8} {M} is 8 .

C) Ammonia is a leveling solvent for stronger acids like HCl,HBr,HI{HCl}, {HBr}, {HI} while glacial acetic acid is differentiating solvent.
View answer and explanation
Correct option: D
A, B and C
Option D (A, B and C) is correct because all three statements are correct regarding solubility product, pH calculation, and solvent leveling/differentiating effects.
657 Easy
657 of 680
Understanding +4 | -1
According to Le-chatelier's principle, adding heat to a solid ⇌\rightleftharpoons liquid equilibrium will cause the
View answer and explanation
Correct option: D
Amount of solid to decrease
Option D (Amount of solid to decrease) is correct because, according to Le Chatelier's principle, adding heat to an endothermic equilibrium (solid to liquid) will shift the equilibrium towards the liquid phase, thus decreasing the amount of solid.
658 Medium
658 of 680
Application +4 | -1
When 0.01 moles of sodium hydroxide are added to a buffer solution, its pH\mathrm{pH} changes from 4.745 to 4.815 . The buffer capacity of the buffer solution is
View answer and explanation
Correct option: A
0.140.14
The correct answer is 0.14 because the buffer capacity is calculated by dividing the number of moles of base added by the change in pH.
659 Easy
659 of 680
Analyzing +4 | -1
Which of the following will not function as a buffer solution?

(i) NaCl\mathrm{NaCl} and NaOH\mathrm{NaOH}

(ii) NaOH\mathrm{NaOH} and NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

(iii) CH3COONH4\mathrm{C}{\mathrm{H}_3}\mathrm{COON}{\mathrm{H}_4} and HCl\mathrm{HCl}

(iv) Borax and boric acid
View answer and explanation
Correct option: A
(i), (ii), (iii)
Option D ((i), (ii), (iii), (iv)) is incorrect, it should be Option A ((i), (ii), (iii)). NaCl and NaOH, NaOH and NH4OH, and CH3COONH4 and HCl will not function as buffer solutions because they do not contain a weak acid/base and its conjugate.
660 Medium
660 of 680
Application +4 | -1
Ksp{K_{sp}} for sodium chloride is 36  mol2/litre236\;\mathrm{mo}{\mathrm{l}^2}/\mathrm{litr}{\mathrm{e}^2}. The solubility of sodium chloride is:
View answer and explanation
Correct option: C
6M6M
Option C (6M6M) is correct because the solubility is equal to the square root of the solubility product for a binary salt like NaCl.
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