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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Hard

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Showing 91-118 of 118 questions
91 Hard
91 of 118
Analyzing +4 | -1
Assertion :- At 25∘C25^{\circ} \mathrm{C} the pH of 10−8 M HCl10^{-8} \mathrm{~M} \, {HCl} is 8.

Reason :- pH of acidic solution is always below 7 at 25∘C.25^{\circ} \mathrm{C}.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
The correct answer is D because both the assertion and the reason are false. The pH of 10-8 M HCl is slightly less than 7, and acidic solutions always have a pH below 7 at 25°C.
92 Hard
92 of 118
Analyzing +4 | -1
Acetic acid and propionic acid have Ka{K_a} values 1.75×10−51.75 \times {10^{ - 5}} and 1.3×10−51.3 \times {10^{ - 5}} respectively at a certain temperature. An equimolar solution of a mixture of the two acids is partially neutralized byNaOH.\mathrm{NaOH}\mathrm{.}. How is the ratio of the contents of acetate and propionate ions related to the Ka{K_a} values and the molarity?
View answer and explanation
Correct option: A
(α1−α)=1.751.3×(β1−β),0\left( {\frac{\alpha }{{1 - \alpha }}} \right) = \frac{{1.75}}{{1.3}} \times \left( {\frac{\beta }{{1 - \beta }}} \right),0 where α\alpha and β\beta are ionised fractions of the acids
Option A is correct because it correctly relates the ratio of ionized fractions to the ratio of the Ka values of the two acids.
93 Hard
93 of 118
Application +4 | -1
KSP{K_{SP}} of MX4M{X_4} and solubility of MX4M{X_4} (S mol/L), is related by
View answer and explanation
Correct option: A
S=[KSP/256]1/5S = {\left[ {{K_{SP}}/256} \right]^{1/5}}
Option A (S=[KSP/256]1/5S = {\left[ {{K_{SP}}/256} \right]^{1/5}}) is correct because it correctly relates the solubility product (KSPK_{SP}) and solubility (S) for the given salt MX4MX_4.
94 Hard
94 of 118
Application +4 | -1
If KspK_{s p} of ZnS\mathrm{ZnS} is 1.0×10−211.0 \times 10^{-21}, the minimum volume of water is required for the dissolution of 0.097 g ZnS0.097 \mathrm{~g} \mathrm{~ZnS} is :
View answer and explanation
Correct option: B
3.16×1073.16 \times 10^7 litre
Option B (3.16×1073.16 \times 10^7 litre) is correct because it calculates the minimum volume of water required for the dissolution of 0.097 g of ZnS using the Ksp value.
95 Hard
95 of 118
Application +4 | -1
For the reaction taking place at certain temperature NH2COONH4(s)⇌2NH3(g)+CO2(g)\mathrm{NH}_2 \mathrm{COONH}_4(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})+\mathrm{CO}_2(\mathrm{g}) If equilibrium pressure os 3X3 \mathrm{X} bar then ΔtG∘\Delta_\mathrm{t} \mathrm{G}^{\circ} would be
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated ΔtG∘\Delta_\mathrm{t} \mathrm{G}^{\circ} is −RTln⁡4−3RTln⁡X-\mathrm{RT} \ln 4 - 3\mathrm{RT} \ln \mathrm{X}, which doesn't match any of the given options.
96 Hard
96 of 118
Application +4 | -1
A(g)+3B(g)⇌4C(g)A\left( g \right) + 3B\left( g \right) \rightleftharpoons 4C\left( g \right) starting concentration of AA is equal to BB, equilibrium concentration of AA and CC are same then KC={K_C} =
View answer and explanation
Correct option: D
8
Option D (8) is correct because, at equilibrium, the equilibrium constant Kc is calculated using the given concentrations of A, B, and C.
97 Hard
97 of 118
Understanding +4 | -1
Assertion :- Boric acid behaves as a weak monobasic acid.

Reason :- Boric acid contains hydrogen bonds in its structure.
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 (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because boric acid is a weak monobasic acid and contains hydrogen bonds, but the hydrogen bonds don't directly explain its weak acidity.
98 Hard
98 of 118
Application +4 | -1
What mass of 95% pure CaCO3CaC{O_3} will be required to neutralise 50 mL of 0.5 m HCl solution according to the following reaction?
CaCO3+2HCl(aq)→CaCl2(aq)+CO2(g)+2H2O(l)CaC{O_3} + 2HC{l_{(aq)}} \to CaC{l_{2(aq)}} + C{O_{2(g)}} + 2{H_2}{O_{(l)}}
[Calculate upto second place of decimal point]
View answer and explanation
Correct option: C
1.32 g
Option C (1.32 g) is correct because it accounts for the purity of the CaCO3CaCO_3 required to neutralize the HCl solution.
99 Hard
99 of 118
Application +4 | -1
The silver ion in a solution is precipitated by the addition of
chloride ion.
Ag+(aq)+Cl−(aq)⇌AgCl(s),K=1×1010\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s), K=1 \times 10^{10}.
The final volume of the solution is 500 mL500 \mathrm{~mL}. At a point when not more than 0.10 mg0.10 \mathrm{~mg} of Ag+\mathrm{Ag}^{+}ion remains unprecipitated [Cl−][\mathrm{Cl^-}] ion is
View answer and explanation
Correct option: A
5.4×10−5M5.4 \times 10^{-5} \mathrm{M}
Option C (5.4×10−4M5.4 \times 10^{-4} \mathrm{M}) is correct because it calculates the chloride ion concentration using the solubility product (KspK_{sp}) and the given silver ion concentration.
100 Hard
100 of 118
Application +4 | -1
In a 10.0 L10.0 \mathrm{~L} container, an equilibrium was established between SO3,SO2\mathrm{SO}_3, \mathrm{SO}_2 and O2\mathrm{O}_2 gases, by starting with SO3\mathrm{SO}_3 only. The density of equilibrium mixture was found to be 16 g/litre16 \mathrm{~g/litre} at a temperature of 9000.0821 K.\frac{900}{0.0821} \mathrm{~K}. If the degree of dissociation of SO3\mathrm{SO}_3 is 40%,40 \%, then the value of KPK_{{P}} (in atm) for the reaction 2SO3(g)⇌2SO2(g)+O2(g)2 \mathrm{SO}_3(\mathrm{g})\rightleftharpoons 2 \mathrm{SO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) is
View answer and explanation
Correct option: A
16 atm16 \mathrm{~atm}
The correct answer is 16 atm. The equilibrium constant KPK_P is calculated using partial pressures derived from the degree of dissociation and total pressure.
101 Hard
101 of 118
Analyzing +4 | -1
A buffer solution is 0.25 M−CH3COOH+0.15 M−CH3COONa,0.25 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COOH}+0.15 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COONa}, saturated in H2S(0.1 M)\mathrm{H}_2 \mathrm{S}(0.1 \mathrm{~M}) and has [Mn2+]=0.04 M,Ka(CH3COOH)=2.0×10−5,\left[\mathrm{Mn}^{2+}\right]=0.04 \mathrm{~M}, K_{\mathrm{a}}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=2.0 \times 10^{-5}, Ka(H2S)=1.0×10−21K_{\mathrm{a}}\left(\mathrm{H}_2 \mathrm{S}\right)=1.0 \times 10^{-21} and Ksp(MnS)=2.5×10−13.K_{\mathrm{sp}}(\mathrm{MnS})=2.5 \times 10^{-13}.
Which buffer component should be increased in concentration and to which minimum value to just start precipitation of MnS?\mathrm{MnS}?
View answer and explanation
Correct option: C
CH3COONa,1.25 M\mathrm{CH}_3 \mathrm{COONa}, 1.25 \mathrm{~M}
Option C (CH3COONa, 1.25 M) is correct because increasing the concentration of CH3COONa to 1.25 M will decrease the [H+] enough to start the precipitation of MnS.
102 Hard
102 of 118
Application +4 | -1
At 700 K700 \mathrm{~K} and 350 bar, a 1:31: 3 mixture of N2(g)\mathrm{N}_2(g) and H2(g)\mathrm{H}_2(g) reacts to form an equilibrium mixture containing X(NH3)=0.50X\left(\mathrm{NH}_3\right)=0.50. Assuming ideal behaviour KpK_p for the equilibrium reaction, N2(g)+3H2(g)⇌2NH3(g)\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) is
View answer and explanation
Correct option: D
3.1×10−43.1 \times 10^{-4}
The correct answer is D (3.1×10−43.1 \times 10^{-4}) because it is calculated using the equilibrium constant expression in terms of partial pressures, considering the mole fractions and total pressure.
103 Hard
103 of 118
Analyzing +4 | -1
2A=B,Keq =X2 A=B, K_{\text {eq }}=X
2B⇌C;Keq =Y2 B \rightleftharpoons C ; K_{\text {eq }}=Y
D⇌C,Keq =ZD\rightleftharpoons C, {K}_{\text {eq }}=Z
Equilibrium constant for the reaction D⇌4AD \rightleftharpoons 4 A is
View answer and explanation
Correct option: B
ZX2Y\frac{Z}{X^2 Y}
Option B is correct. By manipulating the given equilibrium reactions and their constants, the equilibrium constant for D ⇌\rightleftharpoons 4A is found to be Z/(X2Y).
104 Hard
104 of 118
Analyzing +4 | -1
Benzoic acid (C6H5COOH,12 g mol−1)\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}, 12 \mathrm{~g} \mathrm{~mol}^{-1}\right) forms dimer in benzene solvent
C6H5COOH⇌12(C6H5COOH)2\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \rightleftharpoons \frac{1}{2}\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right)_2
Vapour density of a mixture of AA and BB is found to be 91.5 . If to initially (A)=1.0M,Kc(A)=1.0 \mathrm{M}, K_c at equilibrium is
View answer and explanation
Correct option: A
1.74
Option A (1.74) is correct because the equilibrium constant Kc is calculated based on the degree of association and the initial concentration of benzoic acid.
105 Hard
105 of 118
Analyzing +4 | -1
An amount of 2.0 M2.0 \mathrm{~M} solution of Na2CO3\mathrm{Na}_2 \mathrm{CO}_3 is boiled in a closed container with excess of CaF2.\mathrm{CaF}_2. Very little amount of CaCO3\mathrm{CaCO}_3 and NaF\mathrm{NaF} are formed. If the solubility product of CaCO3\mathrm{CaCO}_3 is 'xx' and the molar solubility of CaF2\mathrm{CaF}_2 is 'yy', the molar concentration of F−\mathrm{F}^{-}in the resulting solution after equilibrium is attained is
View answer and explanation
Correct option: C
8y3x\sqrt{\frac{8 y^3}{x}}
Option C (8y3x\sqrt{\frac{8 y^3}{x}}) is correct because it correctly applies the solubility product principle and the equilibrium relationship between the concentrations of fluoride and carbonate ions.
106 Hard
106 of 118
Application +4 | -1
If solubility product of Zr3(PO4)4Z{r_3}{(P{O_4})_4} is denoted by Ksp{K_{sp}} and its molar solubility is denoted by SS, then which of the following relation between SS and Ksp{K_{sp}} is correct?
View answer and explanation
Correct option: B
S = (Ksp6912)1/7S\, = \,{\left( {\frac{{{K_{sp}}}}{{6912}}} \right)^{1/7}}
Option B (S = (Ksp6912)1/7S\, = \,{\left( {\frac{{{{K_{sp}}}}}{{{{6912}}}}} \right)^{1/7}}) is correct because it represents the correct relationship between molar solubility (S) and the solubility product (Ksp) for Zr3(PO4)4.
107 Hard
107 of 118
Analyzing +4 | -1
A volume of 1.0 L1.0 \mathrm{~L} of solution which was in equilibrium with solid mixture of AgCl{AgCl} and Ag2CrO4{Ag}_2 {CrO}_4 was found to contain 1×10−41 \times 10^{-4} moles of Ag+{Ag}^{+} ions, 1.0×10−61.0 \times 10^{-6} moles of Cl−{Cl}^{-} ions and 8.0×10−48.0 \times 10^{-4} moles of CrO42−{CrO}_4{ }^{2-} ions. The Ag+{Ag}^{+} ions are added slowly to the above mixture (keeping the volume constant) till 8.0×10−78.0 \times 10^{-7} moles of AgCl{AgCl} got precipitated. How many moles of Ag2CrO4{Ag}_2 {CrO}_4 were precipitated simultaneously?
View answer and explanation
Correct option: A
7.68×10−47.68 \times 10^{-4}
Option A (7.68×10−47.68 \times 10^{-4}) is correct because it calculates the amount of Ag2CrO4Ag_2CrO_4 precipitated based on the change in chloride concentration and the solubility product constants.
108 Hard
108 of 118
Analyzing +4 | -1
Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R.
Assertion A: During the boiling of water having temporary hardness, Mg(HCO3)2Mg{(HC{O_3})_2} is converted to MgCO3MgC{O_3}.
Reason R The solubility product of Mg(OH)2Mg{(OH)_2} is greater than that of MgCO3MgC{O_3}.
In the light of the above statements, choose the most appropriate answer from the options given below
View answer and explanation
Correct option: D
A is false but R is true.
Option D is correct because the assertion is false (Mg(HCO3)2 converts to Mg(OH)2 during boiling), and the reason is true (Ksp of Mg(OH)2 is greater than that of MgCO3).
109 Hard
109 of 118
Analyzing +4 | -1
At a certain temperature, KPK_{\mathrm{P}} for the dissociation of solid CaCO3\mathrm{CaCO}_3 is 4.5×10−24.5 \times 10^{-2} atm and for the reaction C(s)+CO2(g)⇌2CO(g),\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}), the KPK_{\mathrm{P}} value is 2.0 atm.2.0 \mathrm{~atm}. The pressure of carbon monoxide at this temperature, when solid carbon CaO\mathrm{CaO} and CaCO3\mathrm{CaCO}_3 are mixed together and allowed to attain equilibrium is
View answer and explanation
Correct option: B
0.30 atm0.30 \mathrm{~atm}
Option B (0.30 atm) is correct because it is the partial pressure of CO calculated from the equilibrium constant.
110 Hard
110 of 118
Application +4 | -1
SO3(g)\mathrm{SO}_3(g) is partially dissociated into SO2(g)\mathrm{SO}_2(g) and O2(g)\mathrm{O}_2(g) at 900 K900 \mathrm{~K} and 1 atm1 \mathrm{~atm} as:
SO3(g)⇌SO2(g)+12O2(g)\mathrm{SO}_3(g) \rightleftharpoons \mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g)

The density of the equilibrium mixture was found to be 0.866 g dm−30.866 \mathrm{~g} \mathrm{~dm}^{-3}. Thus, degree dissociation of SO3( g)\mathrm{SO}_3(\mathrm{~g}). is
View answer and explanation
Correct option: C
0.5
Option C (50%) is correct because the degree of dissociation is calculated using the given density and the equilibrium constant expression.
111 Hard
111 of 118
Analyzing +4 | -1
Equilibrium constants K1{K_1} and K2{K_2}  for the following equilibria are related as :
NO(g)+12O2(g)⇋NO2(g);K1\mathrm{NO}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right);{K_1}
2NO2(g)⇋2NO(g)+O2(g);K22\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right);{K_2}
View answer and explanation
Correct option: A
K2=1K12{K_2} = \frac{1}{{K_1^2}}
The correct answer is A (K2=1K12{K_2} = \frac{1}{{K_1^2}}) because the second reaction is the reverse of the first reaction multiplied by 2, so the equilibrium constant is the inverse squared.
112 Hard
112 of 118
Application +4 | -1
Solid Ca(HCO3)2{Ca}\left({HCO}_3\right)_2 decomposes as
Ca(HCO3)2( s)⇌CaCO3( s)+CO2( g)+H2O(g){Ca}\left({HCO}_3\right)_2({~s}) \rightleftharpoons {CaCO}_3({~s})+{CO}_2({~g})+{H}_2 {O}({g})

If the total pressure is 0.2 bar at 420 K420 {~K}, what is the standard free energy change for the given reaction (ΔrG∘)\left(\Delta_r G^{\circ}\right) ?
View answer and explanation
Correct option: D
16.083 kJ/mol16.083 {~kJ} / {mol}
Option D (16.083 kJ/mol16.083 {~kJ} / {mol}) is correct because it is the standard free energy change calculated using the equilibrium constant derived from the total pressure.
113 Hard
113 of 118
Application +4 | -1
The solubility product of Cr(OH)3Cr{\left( {OH} \right)_3} at 298 K is 6.0×10−316.0 \times {10^{ - 31}}. The concentration of hydroxide ions in a saturated solution of Cr(OH)3Cr{\left( {OH} \right)_3} will be
View answer and explanation
Correct option: B
(18×10−31)1/4{\left( {18 \times {{10}^{ - 31}}} \right)^{1/4}}
The correct answer is (18×10−31)1/4{\left( {18 \times {{10}^{ - 31}}} \right)^{1/4}}. The hydroxide ion concentration is calculated from the solubility product using the appropriate stoichiometric relationship.
114 Hard
114 of 118
Analyzing +4 | -1
pH\mathrm{pH} of a saturated solution of silver salt of monobasic acid HA\mathrm{HA} is found to be 9. Select the option(s) which is / are correct. Given : Ka(HA)=10−10\mathrm{K}_{\mathrm{a}}(\mathrm{HA})=10^{-10}
I) Solubility of the silver salt is 1.1×10−5 mol/1.1 \times 10^{-5} \mathrm{~mol} / lit
II) Solubility product of the silver salt is 1.1×10−11 M21.1 \times 10^{-11} \mathrm{~M}^2
III) Concentration of Ag+\mathrm{Ag}^{+} in the solution will be greater than that of A−\mathrm{A}^{-}
View answer and explanation
Correct option: D
I, II &\& III
The correct answer is D (I, II & III) because all three statements regarding the solubility, solubility product, and ion concentrations are correct based on the given information and calculations.
115 Hard
115 of 118
Application +4 | -1
The acid ionization constant of Zn2+{Zn}^{2+} is 2.0×10−10.2.0 \times 10^{-10}. What is the pH\mathrm{pH} of 0.001 M0.001 \mathrm{~M} solution of ZnCl2?{ZnCl}_2? (log⁡2=0.3)(\log 2=0.3)
View answer and explanation
Correct option: C
6.35
Option C (6.35) is correct. The pH of the ZnCl2 solution is calculated by considering the hydrolysis of Zn2+ ions and using the given acid ionization constant to find the [H+] concentration.
116 Hard
116 of 118
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.
117 Hard
117 of 118
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.
118 Hard
118 of 118
Analyzing +4 | -1
Consider the following statements.
I) The pH of a mixture containing 400 mL of   0.1  M  H2SO4\;0.1\;M\;{H_2}S{O_4} and 400 mL of   0.1  M  NaOH\;0.1\;M\;NaOH will be approximately 1.3
II) Ionic product of water is temperature dependent.
III) A monobasic acid with Ka=10−5{K_a} = {10^{ - 5}} has a pH=5.pH = 5. The degree of dissociation of this acid is 50%.
IV) The Le-Chatelier’s principle is not applicable to common-ion effect.
The correct statements are
View answer and explanation
Correct option: D
I, II and III
Option D (I, II and III) is correct because statements I, II, and III are all accurate regarding pH calculations, ionic product of water, and degree of dissociation.
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