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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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331 Medium
331 of 680
Analyzing +4 | -1
In which of the following solvents, will AgBr\mathrm{AgBr} have highest solubility?
View answer and explanation
Correct option: B
10−3 M NH4OH10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}
Option B (10−3 M NH4OH10^{-3} \mathrm{~M} \mathrm{~NH}_4 \mathrm{OH}) is correct because AgBr forms a complex with ammonia, increasing its solubility, while the other options introduce a common ion effect or don't promote complex formation.
332 Medium
332 of 680
Understanding +4 | -1
pH of aqueous NaCI solution after electrolysis
View answer and explanation
Correct option: A
increases
Option A (increases) is correct because during the electrolysis of aqueous NaCl, H2H_2 gas is produced at the cathode and Cl2Cl_2 gas is produced at the anode, leaving behind OH−OH^- ions in the solution, which increases the pH.
333 Medium
333 of 680
Application +4 | -1
10 mL10 \mathrm{~mL} of 10−4NH2SO410^{-4} \mathrm{NH}_2 \mathrm{SO}_4 is diluted with water to have 1 dm3\mathrm{dm}^3 solution : The pH\mathrm{pH} of this solution is :
View answer and explanation
Correct option: B
slightly more than 6
The correct answer is slightly more than 6 because after dilution, the pH is determined by considering both the sulfuric acid and the autoionization of water.
334 Medium
334 of 680
Analyzing +4 | -1
Statement I : The hydrolysis of an ester in acidic medium does not change with pressure.

Statement II : Pressure does not show effect on equilibrium reaction taking place in solution
View answer and explanation
Correct option: A
Statement I, II are correct
The correct answer is Statement I, II are correct because the hydrolysis of an ester in acidic medium is a solution-phase reaction, and pressure has negligible effect on reactions in solution.
335 Hard
335 of 680
Analyzing +4 | -1
Statement (S): Acidic nature of boron trihalides is in the order :
BF3<BCl3<BBr3<BI3\mathrm{BF}_3<\mathrm{BCl}_3<\mathrm{BBr}_3<\mathrm{BI}_3

Explanation (E): Basic nature of nitrogen trihalides is in the order :
NF3>NCl3>NBr3>NI3\mathrm{NF}_3>\mathrm{NCl}_3>\mathrm{NBr}_3>\mathrm{NI}_3
View answer and explanation
Correct option: A
S is correct but E is wrong
Option A (S is correct but E is wrong) is correct because the acidic nature of boron trihalides increases down the group due to increasing pπ-pπ back bonding, but the basic nature of nitrogen trihalides decreases down the group.
336 Medium
336 of 680
Analyzing +4 | -1
STATEMENT I : pHp H of aqueous solution of NH4Cl\mathrm{NH}_4 \mathrm{Cl} is greater than 7.

STATEMENT II : Solution of a salt of weak base and strong acid is basic.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B (N2>N2+>N2−\mathrm{N}_2 > \mathrm{N}_2^{+} > \mathrm{N}_2^{-}) is correct because N2 has the highest bond order (3), while N2+ and N2- have lower bond orders (2.5), with N2+ being slightly more stable than N2-.
337 Easy
337 of 680
Analyzing +4 | -1
Indicate the correct answer out of the following for the reaction,

NH4Cl+H2O  ⇋NH4OH+HCl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + {\mathrm{H}_2}\mathrm{O}\;\leftrightharpoons\mathrm{N}{\mathrm{H}_4}\mathrm{OH} + \mathrm{HCl}
View answer and explanation
Correct option: C
The reaction is retarded by the addition of hydrogen ion
The reaction is retarded by the addition of hydrogen ion because adding H+H^+ shifts the equilibrium to the left, according to Le Chatelier's principle.
338 Medium
338 of 680
Application +4 | -1
Given Ka{K}_{\mathrm{a}} values are 5.76×10−105.76 \times 10^{-10} and 4.8×10−104.8 \times 10^{-10} for NH4+{NH}_4^{+}and HCN{HCN} respectively. What is the equilibrium constant for the following reaction ?

NH4+(aq.)+CN−(aq.)⇌NH3(aq.)+HCN(aq.){NH}_4^{+}(aq.) +{CN}^{-}(aq.) \rightleftharpoons {NH}_3 (aq.) +{HCN} (aq.)
View answer and explanation
Correct option: B
1.2
Option B (1.2) is correct because the equilibrium constant for the reaction is the ratio of the Ka values of the ammonium ion and HCN.
339 Medium
339 of 680
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.
340 Hard
340 of 680
Application +4 | -1
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+{Ag}^{+}and Pb2+{Pb}^{2+} at a concentration is 0.10 M0.10 \mathrm{~M}. Aqueous HCl{HCl} is added to this solution until the Cl−{Cl}^{-}concentration is 0.10 M0.10 \mathrm{~M}. What will the concentration of Ag+{Ag}^{+}and Pb2+{Pb}^{2+} be at equilibrium?
(Ksp for AgCl=1.8×10−10,Ksp for PbCl2=1.7×10−5)\left({K}_{{sp}} \text { for } {AgCl}=1.8 \times 10^{-10}, {K}_{{sp}} \text { for } {PbCl}_2=1.7 \times 10^{-5}\right)
View answer and explanation
Correct option: A
[Ag+]=1.8×10−9 M\left[{Ag}^{+}\right]=1.8 \times 10^{-9} \mathrm{~M}, [Pb2+]=1.7×10−3 M\left[{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{~M}
Option A (+3, +4, +5) is correct because these are the calculated oxidation states of P in the respective compounds based on standard oxidation rules.
341 Medium
341 of 680
Understanding +4 | -1
In which of the following case, the reaction will proceed in backward direction:

H2(g)+I2(g)⇌2HI(g),Kc=0.25 at 298 K\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \mathrm{K_c}=0.25 \text { at } 298 \mathrm{~K}
View answer and explanation
Correct option: B
Qc=0.45Q_c=0.45
Option B (Qc=0.45Q_c=0.45) is correct because if Qc>KcQ_c > K_c, the reaction will proceed in the backward direction to reach equilibrium.
342 Medium
342 of 680
Analyzing +4 | -1
Which one is not correct statement about buffer mixture?
View answer and explanation
Correct option: D
0.5 millimole of NaCN+1\mathrm{NaCN}+1 millimole of HCl\mathrm{HCl} in 1 litre
Option D (0.5 millimole of NaCN+1\mathrm{NaCN}+1 millimole of HCl\mathrm{HCl} in 1 litre) is not a buffer because the HCl completely neutralizes the NaCN, leaving excess strong acid.
343 Hard
343 of 680
Understanding +4 | -1
(A): BaCO3\mathrm{BaCO}_3 is more soluble in HNO3\mathrm{HNO}_3 than in plain water.

(R): Carbonate is a weak base and reacts with H+\mathrm{H}^{+}from the strong acid, causing the barium salt to dissociate.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because BaCO3BaCO_3 solubility increases in HNO3HNO_3 due to the reaction of carbonate with H+H^+, causing dissociation of the barium salt.
344 Medium
344 of 680
Application +4 | -1
An acid base indicator which is a weak acid has a pKa\mathrm{p} K_{\mathrm{a}} value =5.5=5.5. At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half way between those of its acid and conjugate base forms? The pKa\mathrm{p} K_{\mathrm{a}} value of acetic acid =4.75.=4.75.
[Antilog⁡(0.75)=5.62,Antilog(0.79)=6.3,Antilog⁡(0.69)=4.93].[\operatorname{Antilog}(0.75)=5.62, \text{Antilog}(0.79)= 6.3, \operatorname{Antilog}(0.69)=4.93].
View answer and explanation
Correct option: C
5.62:15.62: 1
Option C (5.62:15.62: 1) is correct because it is derived from the Henderson-Hasselbalch equation, where the ratio of conjugate base to acid is calculated when the indicator shows a color halfway between its acid and conjugate base forms.
345 Hard
345 of 680
Application +4 | -1
2.5  mL  of  25  M2.5\;mL\;of\;\frac{2}{5}\;M weak monoacidic base Kb=1×10−12    at  25∘C{K_b} = 1 \times {10^{ - 12}}\;\;\mathrm{at}\;25^\circ C is titrated with 215  M  HCl\frac{2}{{15}}\;M\;\mathrm{HCl} in water at 25∘C25^\circ C The concentration of H+{\mathrm{H}^ + } at equivalence point is

Kw=1×10−14  at  25∘C{K_w} = 1 \times {10^{ - 14}}\;\mathrm{at}\;25^\circ C
View answer and explanation
Correct option: D
2.7×10−2  M2.7 \times {10^{ - 2}}\;M
Option D (2.7×10−2  M2.7 \times {10^{ - 2}}\;M) is correct because it accurately calculates the hydrogen ion concentration at the equivalence point considering the hydrolysis of the resulting salt.
346 Hard
346 of 680
Application +4 | -1
The solubility product of As2S3  is  2.8×10−72\mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3}\;\mathrm{is}\;2.8 \times {10^{ - 72}}. What is the solubility of As2S3\mathrm{A}{\mathrm{s}_2}{\mathrm{S}_3}?
View answer and explanation
Correct option: A
1.92×10−15  mol/L1.92 \times {10^{ - 15}}\;mol/L
Option C (40 ml) is correct because the volume of KMnO4 required is calculated based on the stoichiometry of the redox reaction.
347 Medium
347 of 680
Understanding +4 | -1
The vapour density of completely dissociated NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} would be:
View answer and explanation
Correct option: B
Half of that of ammonium chloride
Option B is correct. Upon complete dissociation, NH4Cl breaks into NH3 and HCl, doubling the number of moles. Therefore, the experimental molecular weight is half the normal molecular weight, and the vapor density is also halved.
348 Medium
348 of 680
Understanding +4 | -1
Statement I : As per Lechutelier's principle, addition of heat to an equilibrium solid ⇌\rightleftharpoons liquid results in decrease in the amount of solid.

Statement II : Reaction is endothermic, so on heating forward reaction is favoured.
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because Le Chatelier's principle states that adding heat to an endothermic reaction will favor the forward reaction, thus decreasing the amount of solid.
349 Easy
349 of 680
Application +4 | -1
For a chemical reaction ;
A(g)+B(ℓ)⇌D(g)+E(g){A}(g)+{B}(\ell) \rightleftharpoons {D}(g)+{E}(g)

Hypothetically at what temperature, Kp=Kc{K}_{{p}}={K}_{{c}} (when, R=0.08ℓ−atm/mole−K{R}=0.08 \ell-{atm} / {mole}-{K} )
View answer and explanation
Correct option: C
T=12.5 K{T}=12.5 {~K}
Option C (T=12.5 K) is correct because when Kp = Kc, RT = 1, and solving for T gives 1/R = 1/0.08 = 12.5 K.
350 Medium
350 of 680
Analyzing +4 | -1
Two buffers, X\mathrm{X} and Y\mathrm{Y} of pH,4.0\mathrm{pH}, 4.0 and 6.0,6.0, respectively, are prepared from acid HA\mathrm{HA} and the salt NaA.\mathrm{NaA}. Both the buffers are 0.50 M0.50 \mathrm{~M} in HA.\text{HA}. What would be the pH\mathrm{pH} of the solution obtained by mixing equal volumes of the two buffers? The value of KaK_{{a}} for HA=1.0×10−5(log⁡5.05=0.7).\mathrm{HA}=1.0 \times 10^{-5}(\log 5.05=0.7).
View answer and explanation
Correct option: D
5.7
Option D (5.7) is correct because mixing the two buffers results in a pH that can be calculated using the Henderson-Hasselbalch equation after considering the new concentrations of HA and A-.
351 Medium
351 of 680
Understanding +4 | -1
A: Increasing the concentration of H2\mathrm{H}_2 will increase magnitude of equilibrium constant of the reaction H2+I2⇌2HI\mathrm{H}_2+\mathrm{I}_2 \rightleftharpoons 2 \mathrm{HI}
R: Value of KC\mathrm{K}_{\mathrm{C}} depends upon the concentration of reactants and products taken.
View answer and explanation
Correct option: D
If both A and R are incorrect
Option D (If both A and R are incorrect) is correct because the equilibrium constant depends on temperature, not the concentration of reactants or products.
352 Medium
352 of 680
Application +4 | -1
The pH\mathrm{pH} of 0.005 M−NaOH0.005 \mathrm{~M}-\mathrm{NaOH} solution is (log⁡2=0.3)(\log 2=0.3)
View answer and explanation
Correct option: D
11.7
Option D (11.7) is correct because it is the pH calculated from the pOH of the NaOH solution using the relationship pH + pOH = 14.
353 Hard
353 of 680
Application +4 | -1
A mixture contains 0.1 M HCOOH(Ka=1.8×10−4)0.1 \mathrm{~M} \mathrm{~HCOOH}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) and 0.1 M HOCN(Ka=3.3×10−4)0.1 \mathrm{~M} \mathrm{~HOCN}\left(K_{\mathrm{a}}=3.3 \times 10^{-4}\right). Thus, molar concentration of H3O+\mathrm{H}_3 \mathrm{O}^{+}is
View answer and explanation
Correct option: A
7.14×10−3M7.14 \times 10^{-3} \mathrm{M}
Option A (7.14×10−3M7.14 \times 10^{-3} \mathrm{M}) is correct because it represents the calculated hydrogen ion concentration in the mixture of two weak acids.
354 Medium
354 of 680
Application +4 | -1
In a reversible reaction A+B⇌2CA+B \rightleftharpoons 2 C, the equilibrium concentrations of (A), (B) and (C) are 2M,2M2 \mathrm{M}, 2 \mathrm{M} and 4M4 \mathrm{M} respectively then KC\mathrm{K}_{\mathrm{C}} for the reaction is
View answer and explanation
Correct option: C
4
The correct answer is C (4) because the equilibrium constant Kc is calculated by dividing the square of the concentration of C by the product of the concentrations of A and B.
355 Easy
355 of 680
Analyzing +4 | -1
Which one of the following is least likely to act as a Lewis base?
View answer and explanation
Correct option: A
I+{\mathrm{I}^ + }
Option A (I+{\mathrm{I}^ + }) is correct because Lewis bases are electron pair donors, and I+ is electron deficient, thus it is least likely to act as a Lewis base.
356 Easy
356 of 680
Analyzing +4 | -1
The pH\mathrm{pH} of 0.1  M\mathrm{0}\mathrm{.1}\;\mathrm{M} solution of the following salts increases in the order
View answer and explanation
Correct option: B
HCl<NH4Cl<NaCl<NaCN\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN
Option B is correct because HCl is a strong acid, NH4Cl is an acidic salt, NaCl is a neutral salt, and NaCN is a basic salt, resulting in the pH increasing in the order HCl < NH4Cl < NaCl < NaCN.
357 Hard
357 of 680
Understanding +4 | -1
Blue colour disappears in which solution by passing SO2{SO}_2
View answer and explanation
Correct option: B
I2+{I}_2+ starch
Option B is correct. SO2 decolorizes iodine in the presence of starch, as it reduces iodine to iodide ions.
358 Easy
358 of 680
Understanding +4 | -1
Which acts both as Lowry Bronsted acid and base?
View answer and explanation
Correct option: D
HSO4−\mathrm{HSO}_4^ -
The correct answer is HSO4−\mathrm{HSO}_4^ - because it can both donate a proton (acting as an acid) and accept a proton (acting as a base).
359 Easy
359 of 680
Understanding +4 | -1
An example of a salt dissolved in water to give acidic solution is
View answer and explanation
Correct option: A
Ammonium chloride
Option A (Ammonium chloride) is correct because ammonium chloride (NH4Cl\mathrm{NH}_4\mathrm{Cl}) is a salt of a weak base (NH4OH\mathrm{NH}_4\mathrm{OH}) and a strong acid (HCl), resulting in an acidic solution upon dissolution in water.
360 Medium
360 of 680
Analyzing +4 | -1
Statement (S): HCl acts as weak base in liquid HF but weak acid in acetic acid
Explanation (E): The tendency to act as acid or base also depends upon the nature of other substance to accept or donate a proton.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation for S
Option C (Both S and E are correct and E is correct explanation for S) is correct because the acidity or basicity of a substance depends on the relative ability to donate or accept protons, which is influenced by the solvent.
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