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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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151 Easy
151 of 408
Analyzing +4 | -1
Which one of the following statement is not true?
View answer and explanation
Correct option: C
The  pH  of  1×10−8  M  HCl  is  8\mathrm{The}\;\mathrm{ pH}\;\mathrm{ of}\;1 \times {10^{ - 8}}\;\mathrm{M}\;\mathrm{ HCl}\;\mathrm{ is}\;8
Option C is correct because the pH of a very dilute strong acid like 1×10−8M  HCl1 \times 10^{-8} M \;HCl cannot be 8. It must be less than 7, considering the contribution of water's autoionization.
152 Medium
152 of 408
Understanding +4 | -1
The buffering action of an acidic buffer is maximum when its pHpH is equal to
View answer and explanation
Correct option: D
pKa\mathrm{p}{K_a}
Option D (pKa\mathrm{p}{K_a}) is correct because the buffering action is maximum when the pHpH is equal to the pKapK_a of the weak acid.
153 Medium
153 of 408
Understanding +4 | -1
For which of the following reactions,Kp=Kc{K_p} = {K_c} ?
View answer and explanation
Correct option: B
N2+O2⇋2NO{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}
Option B (N2+O2⇋2NO{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}) is correct because in this reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products, making Δn=0\Delta n = 0.
154 Easy
154 of 408
Understanding +4 | -1
Which is a basic salt?
View answer and explanation
Correct option: D
2PbCO3Pb(OH)22\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}
Option D (2PbCO3Pb(OH)22\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}) is a basic salt because it contains hydroxide ions in its formula.
155 Easy
155 of 408
Application +4 | -1
The pH of millimolar HCl\mathrm{HCl} is
View answer and explanation
Correct option: B
3
Option B (3) is correct because millimolar HCl means 10-3 M, and pH is the negative logarithm of the H+ concentration.
156 Easy
156 of 408
Understanding +4 | -1
If the value of equilibrium constant for a particular reaction is 1.6×10121.6 \times {10^{12}}, then at equilibrium the
system will contain
View answer and explanation
Correct option: B
mostly products
Option B (mostly products) is correct because a very large equilibrium constant indicates that the reaction strongly favors the formation of products at equilibrium.
157 Easy
157 of 408
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 is correct because reversing and adding reactions results in multiplying or inverting the equilibrium constants.
158 Easy
158 of 408
Remembering +4 | -1
For a homogeneous chemical reaction, Kp=KCK_p=K_C when
View answer and explanation
Correct option: A
Δn=0\Delta n=0
Option A (Δn=0\Delta n=0) is correct because Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, so when Δn=0\Delta n = 0, Kp=KcK_p = K_c.
159 Medium
159 of 408
Analyzing +4 | -1
Aqueous solution of which of the following boron compounds will be strongly basic in nature?
View answer and explanation
Correct option: D
Na2 B4O7\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7
Option D (Na2 B4O7\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7) is correct because it is a salt of a weak acid and a strong base, which will hydrolyze in water to produce OH- ions, resulting in a basic solution.
160 Easy
160 of 408
Application +4 | -1
Calculate the pHpH of a solution in which hydrogen ion concentration is 0.005  g−equi/L?0.005\mathrm{ }\;g - equi/L?
View answer and explanation
Correct option: A
2.32.3
Option A (2.3) is correct because pH is calculated as the negative logarithm of the hydrogen ion concentration, and in this case, it's -log(0.005) which is approximately 2.3.
161 Hard
161 of 408
Application +4 | -1
For the equilibrium, 2H2O⇋H3O++OH−2{H_2}O\underset{{}}{\overset{{}}{\leftrightharpoons}}{H_3}{O^ + } + O{H^ - }, the value of ΔG∘\Delta {G^ \circ } at 298 K is approximately
View answer and explanation
Correct option: C
80 kJmol−180\,kJ\mathrm{ }mo{l^{ - 1}}
Option C (80 kJmol−180\,kJ\mathrm{ }mo{l^{ - 1}}) is correct because the Gibbs free energy change is related to the ion product of water (KwK_w) and can be calculated using the formula ΔG∘=−RTln⁡Kw\Delta G^\circ = -RT\ln K_w.
162 Medium
162 of 408
Application +4 | -1
The concentration of fluroacetic acid (Ka{K_a} of acid =2.6×10−3= 2.6 \times {10^{ - 3}}) which is required to get [H+]=1.50×10−3M\left[ {{\mathrm{H}^ + }} \right] = 1.50 \times {10^{ - 3}}M is:
View answer and explanation
Correct option: B
2.37×10−3M2.37 \times {10^{ - 3}}M
Option B (2.37×10−3M2.37 \times 10^{-3} M) is correct because it is calculated using the acid dissociation constant and the given hydrogen ion concentration.
163 Medium
163 of 408
Understanding +4 | -1
In the following reaction HC2O4−(aq)+PO4−3(aq)HPO42−(aq)+C2O42−(aq)H{C_2}O_4^ - \left( {aq} \right) + PO_4^{ - 3}\left( {aq} \right)HPO_4^{2 - }\left( {aq} \right) + {C_2}O_4^{2 - }\left( {aq} \right) which are Bronsted bases
View answer and explanation
Correct option: D
PO4−3  and  C2O4−2PO_4^{ - 3}{\mkern 1mu} \,\mathrm{and}{\mkern 1mu} \,{C_2}O_4^{ - 2}
Option D (PO4−3 and C2O4−2PO_4^{ - 3}\,\mathrm{and}\,{C_2}O_4^{ - 2}) is correct because Bronsted bases are proton acceptors, and PO4−3PO_4^{ - 3} and C2O4−2{C_2}O_4^{ - 2} accept protons in the reaction.
164 Medium
164 of 408
Remembering +4 | -1
Select the correct expression
View answer and explanation
Correct option: D
ΔG∘=−2.303 RT log⁡K\Delta G^{\circ}=-2.303 \,{RT} \,\log {K}
Option D (ΔG∘=−2.303 RT log⁡K\Delta G^{\circ}=-2.303 \,{RT} \,\log {K}) is correct because it represents the standard Gibbs free energy change in terms of the equilibrium constant K.
165 Medium
165 of 408
Analyzing +4 | -1
Consider the reaction,
N2(g)+3H2(g)⇋2NH3(g){N_2}\left( g \right) + 3{H_2}\left( g \right)\underset{{}}{\overset{{}}{\leftrightharpoons}}2N{H_3}\left( g \right)
The equilibrium constant of the above reaction is Kp.{K_p}. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that pNH3<<ptotal{p_{N{H_3}}} < < {p_{\mathrm{total}}} at equilibrium)
View answer and explanation
Correct option: C
Kp1/2P216\frac{{K_p^{1/2}{P^2}}}{{16}}
The correct answer is C (Kp1/2P216\frac{{K_p^{1/2}{P^2}}}{{16}}) because it correctly calculates the partial pressure of ammonia at equilibrium when pure ammonia dissociates.
166 Easy
166 of 408
Remembering +4 | -1
In the titration of Na2CO3  and  HCl\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}\;\mathrm{and}\;\mathrm{ HCl}, the indicator used is
View answer and explanation
Correct option: A
Methyl orange
Option A (Methyl orange) is correct because methyl orange is a suitable indicator for titrations involving a strong acid (HCl) and a weak base (Na2CO3), as its color change occurs in the acidic pH range.
167 Hard
167 of 408
Understanding +4 | -1
Consider the following reaction, C(s\mathrm{C}(\mathrm{s}, diamond )=C(s)=\mathrm{C}(\mathrm{s}, graphite )+)+ heat and choose the correct option.
View answer and explanation
Correct option: B
An increase in temperature and pressure will shift the equilibrium to the left.
Option B is correct because the reaction is exothermic, and graphite is denser than diamond, so increasing temperature and pressure will shift the equilibrium to the left.
168 Medium
168 of 408
Application +4 | -1
Which may be added to one litre of water to act as a buffer?
View answer and explanation
Correct option: D
One mole of HC2H3O2\mathrm{H}{\mathrm{C}_2}{\mathrm{H}_3}{\mathrm{O}_2} and 0.50.5 mole of NaOH\mathrm{NaOH}
Option D is correct because a buffer solution is formed by a weak acid and its conjugate base, which is produced when NaOH partially neutralizes the weak acid.
169 Easy
169 of 408
Remembering +4 | -1
Kc{K_c} for m1A+m2B=n1C+n2D{m_1}A + {m_2}B = {n_1}C + {n_2}D is given by:
View answer and explanation
Correct option: C
Kc=[C]n1[D]n2[A]m1[B]m2{K_c} = \frac{{{{\left[ C \right]}^{{n_1}}}{{\left[ D \right]}^{{n_2}}}}}{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}
Option C (Kc=[C]n1[D]n2[A]m1[B]m2{K_c} = \frac{{{{\left[ C \right]}^{{n_1}}}{{\left[ D \right]}^{{n_2}}}}}{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}) is correct because it represents the equilibrium constant expression with products raised to their stoichiometric coefficients divided by reactants raised to their stoichiometric coefficients.
170 Hard
170 of 408
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 true regarding pH calculations, ionic product of water, and degree of dissociation.
171 Hard
171 of 408
Application +4 | -1
Ammonium carbonate decomposes as

NH2COONH4(s)⇋2NH3(g)+CO2(g)\mathrm{N}{\mathrm{H}_2}\mathrm{COON}{\mathrm{H}_4}\left( s \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

For the reaction, Kp=2.9×10−5  atm−3{K_p} = 2.9 \times {10^{ - 5}}\;\mathrm{at}{\mathrm{m}^{ - 3}} If we start with 1 mole of the compound, the total pressure at equilibrium would be
View answer and explanation
Correct option: B
0.0582  atm0.0582\;\mathrm{ }atm
Option B (0.0582 atm) is correct because it represents the total pressure at equilibrium calculated using the Kp expression.
172 Easy
172 of 408
Analyzing +4 | -1
Which of the following is the strongest conjugate base?
View answer and explanation
Correct option: B
CH3COO−\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }
The correct answer is B (CH3COO−\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }) because it is the conjugate base of a weak acid (acetic acid), making it a stronger base than the conjugate bases of strong acids.
173 Medium
173 of 408
Application +4 | -1
The pH of a 0.1 M solution of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} (having Kb=1.0×10−5){K_b} = 1.0 \times {10^{ - 5}}) is equal to
View answer and explanation
Correct option: C
11
Option C (11) is correct because NH4OH is a weak base, and the pH can be calculated using the Kb value and the concentration of the base.
174 Easy
174 of 408
Understanding +4 | -1
Boric acid is a
View answer and explanation
Correct option: C
Lewis acid
Option C (Lewis acid) is correct because Boric acid (B(OH)3B(OH)_3) accepts a lone pair of electrons from hydroxide ions, thus acting as a Lewis acid.
175 Easy
175 of 408
Application +4 | -1
In which of the following reaction Kp>Kc{K_p} > {K_c}?
View answer and explanation
Correct option: C
2SO3⇋O2+2SO22\mathrm{S}{\mathrm{O}_3}\leftrightharpoons{\mathrm{O}_2} + 2\mathrm{S}{\mathrm{O}_2}
Option C is correct because Kp > Kc when Δn_g > 0, and in this reaction, Δn_g = (1+2) - 2 = 1.
176 Easy
176 of 408
Understanding +4 | -1
An aqueous solution of ammonium acetate is:
View answer and explanation
Correct option: D
Almost neutral
Option D (Almost neutral) is correct because ammonium acetate is a salt of a weak acid (acetic acid) and a weak base (ammonium hydroxide), and their strengths are comparable, resulting in a nearly neutral solution.
177 Medium
177 of 408
Understanding +4 | -1
In the reaction, H2(g)+Cl2(g)⇋2HCl(g){\mathrm{H}_2}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HCl}\left( \mathrm{g} \right)
View answer and explanation
Correct option: B
Kp=Kc{K_p} = {K_c}
Option B (Kp=Kc{K_p} = {K_c}) is correct because the change in the number of moles of gas (Δn\Delta n) is zero in this reaction, making KpK_p equal to KcK_c.
178 Medium
178 of 408
Application +4 | -1
The equivalent conductance of M32\frac{M}{{32}} solution of a weak monobasic acid is 8.0  mhos  cm28.0\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2} and at infinite dilution is 400  mhos  cm2.400\;\mathrm{mhos}\;\mathrm{c}{\mathrm{m}^2}. The dissociation constant of this acid is:
View answer and explanation
Correct option: B
1.25×10−51.25 \times {10^{ - 5}}
Option B (1.25×10−51.25 \times 10^{-5}) is correct because it represents the calculated dissociation constant using the given equivalent conductance values.
179 Hard
179 of 408
Application +4 | -1
The molar solubility of Cd (OH)2Cd\,{\left( {OH} \right)_2} is 1.84×10−5 m1.84 \times {10^{ - 5}}\,m in water. The expected solubility of Cd (OH)2Cd\,{\left( {OH} \right)_2} in a buffer solution of pH=12pH = 12 is
View answer and explanation
Correct option: D
2.49×10−10 M 2.49 \times {10^{ - 10}}\,M\,
Option D (2.49×10−10 M 2.49 \times {10^{ - 10}}\,M\,) is correct because it calculates the solubility of Cd(OH)2 in a buffer solution at pH 12 by considering the hydroxide ion concentration and the solubility product.
180 Easy
180 of 408
Analyzing +4 | -1
STATEMENT I: When aqueous, solution of CH3COONH4\mathrm{CH}_3 \mathrm{COONH}_4 is diluted, then its degree of hydrolysis increase.

STATEMENT II : Ammonium acetate is the salt of weak acid and weak base, its degree of hydrolysis depend on the concentration.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B (Statement I, II are incorrect) is correct because the degree of hydrolysis for ammonium acetate increases with dilution, and it depends on the ionic product of water, not just concentration.
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