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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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181 Medium
181 of 401
Analyzing +4 | -1
The correct order of increasing [H3O+]\left[ {{\mathrm{H}_3}{\mathrm{O}^ + }} \right] in the following aqueous solutions is:
View answer and explanation
Correct option: C
0.01  M  NaNO2<0.01  M  NaCl<0.01  M  H2S<0.01  M  H2SO40.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}
Option C (0.01  M  NaNO2<0.01  M  NaCl<0.01  M  H2S<0.01  M  H2SO40.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}) is correct because it correctly orders the solutions from lowest to highest hydronium ion concentration based on their acidic/basic properties.
182 Medium
182 of 401
Application +4 | -1
For the following gaseous equilibria X,Y\mathrm{X}, \mathrm{Y} and Z\mathrm{Z} at 300 K300 \mathrm{~K}

X:2SO2+O2⇌SO3\mathrm{X}: 2{SO}_2+{O}_2 \rightleftharpoons {SO}_3

Y:PCl5⇌PCl3+Cl2Z:2HI⇌H2+I2\begin{aligned} & \text {Y}: {PCl}_5 \rightleftharpoons {PCl}_3+{Cl}_2 \\ & \mathrm{Z}: 2 {HI} \rightleftharpoons {H}_2+{I}_2 \end{aligned}

ratio of KpK_p and KCK_C in the increasing order is
View answer and explanation
Correct option: C
X<Z<Y\mathrm{X} < \mathrm{Z} < \mathrm{Y}
Option C (X<Z<Y\mathrm{X} < \mathrm{Z} < \mathrm{Y}) is correct because it correctly orders the ratio of KpK_p to KCK_C based on the change in the number of moles of gas (Δng\Delta n_g) for each reaction.
183 Medium
183 of 401
Understanding +4 | -1
Assertion (A): Higher order ionisation constant (Ka2,Ka3)\left(K_{a_2}, K_{a_3}\right) are smaller than the lower order ionisation constant (Ka1)\left(K_{a_1}\right) of a polyprotic acid.

Reason (R): It is more difficult to remove a positively charged proton from a negative ion due to Coulombic forces.
View answer and explanation
Correct option: C
A\mathrm{A} is correct; R\mathrm{R} is incorrect
Option D (A is incorrect; R is correct) is correct because the assertion is incorrect. Higher order ionization constants are smaller due to increased difficulty in removing a proton from a negatively charged species, which is correctly stated in the reason.
184 Medium
184 of 401
Understanding +4 | -1
(A): Water acts as leveling solvent for various acids.

(R): Levelling effect of water is due to its high dielectric constant and strong proton accepting tendency.
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 water's high dielectric constant and proton-accepting ability contribute to its leveling effect on acids.
185 Medium
185 of 401
Remembering +4 | -1
Nucleophiles are:
View answer and explanation
Correct option: B
Lewis bases
Option B (Lewis bases) is correct because nucleophiles are electron-rich species that donate electron pairs, which is the definition of a Lewis base.
186 Medium
186 of 401
Application +4 | -1
The pH\mathrm{pH} of 10−8M  HCl{10^{ - 8}}\mathrm{M\; HCl} solution is
View answer and explanation
Correct option: C
Between 6 and 7
Option C (Between 6 and 7) is correct because the pH of a very dilute HCl solution (10-8 M) is influenced by the autoionization of water, resulting in a pH slightly less than 7.
187 Medium
187 of 401
Remembering +4 | -1
In the problem 13, the dissociation constant of acid is :
View answer and explanation
Correct option: B
1.02×10−41.02 \times {10^{ - 4}}
Option B (1.02×10−41.02 \times {10^{ - 4}}) is correct because the dissociation constant KaK_a for a weak acid can be calculated using the formula Ka=cα2K_a = c\alpha^2, where c is the concentration and α\alpha is the degree of dissociation.
188 Medium
188 of 401
Application +4 | -1
CuSO4⋅5H2O(s)⇌CuSO4⋅3H2O(s)+2H2O(g),KP\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}), K_{\mathrm{P}} for this equilibrium is 1.0×10−4 atm21.0 \times 10^{-4} \mathrm{~atm}^2 at 25∘C25^{\circ} \mathrm{C}. What is the maximum pressure of water vapour (moisture) in the atmosphere, below which the pentahydrate is efflorescent?
View answer and explanation
Correct option: A
7.60 mm7.60 \mathrm{~mm}
Answer A (7.60 mm) is correct because the maximum pressure of water vapor is determined by the square root of the Kp value, converted to mmHg.
189 Medium
189 of 401
Analyzing +4 | -1
Assuming 100% ionisation which will have maximum pH
View answer and explanation
Correct option: C
0.01 M( NH4)3PO40.01 \mathrm{~M}\left(\mathrm{~NH}_4\right)_3 \mathrm{PO}_4
Option C (0.01 M(NH4)3PO40.01 \mathrm{~M}\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4) will have the maximum pH because it produces the highest concentration of NH4+\mathrm{NH}_4^+ ions, leading to the greatest extent of hydrolysis and thus the lowest acidity (highest pH).
190 Medium
190 of 401
Understanding +4 | -1
Statement I: The solubility of gases always increases with inctease in pressure

Statement II : High prespure favours the change where volume decreases
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because the solubility of gases increases with pressure (Henry's Law), and high pressure favors volume decrease (Le Chatelier's principle).
191 Medium
191 of 401
Analyzing +4 | -1
Assertion :- The pH of an aqueous solution of acetic acid remains unchanged on the addition of sodium acetate.

Reason :- The ionization of acetic acid is suppressed by the addition of sodium acetate.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
The correct answer is A (v * r) because, according to Bohr's model, the product of the electron's velocity and the orbit radius is proportional to the orbit number n.
192 Medium
192 of 401
Application +4 | -1
If pKb\mathrm{pK}_{\mathrm{b}} for fluoride ion at 25∘C25^{\circ} \mathrm{C} is 10.83 , the ionisation constant of hydrofluoric acid in water at this temperature is:
View answer and explanation
Correct option: C
6.75×10−46.75 \times 10^{-4}
Option C (6.75×10−46.75 \times 10^{-4}) is correct because it uses the relationship Ka×Kb=KwK_a \times K_b = K_w to find the acid dissociation constant.
193 Medium
193 of 401
Application +4 | -1
100ml100 \mathrm{ml} solution (I) of buffer containing 0.1(M)HA0.1(\mathrm{M}) \mathrm{HA} and 0.2 (M) A−\mathrm{A}^{-}, is mixed with another solution (II) of 100ml100 \mathrm{ml} containing 0.2(M) HA and 0.3(M) A−\mathrm{A^{-}} .
After mixing what is the pH\mathrm{pH} of resulting solution?
Given pKa\mathrm{pK}_{\mathrm{a}} of HA=5\mathrm{HA}=5
View answer and explanation
Correct option: B
5+log⁡5/35+\log 5 / 3
Option B (5+log⁡5/35+\log 5 / 3) is correct because it is calculated by using the Henderson-Hasselbalch equation after determining the total moles of A−A^- and HA after mixing.
194 Medium
194 of 401
Application +4 | -1
The change in pH\mathrm{pH} of water when 0.01 mole of NaOH\mathrm{NaOH} are added in 10 litre water is :
View answer and explanation
Correct option: C
4 units
Option C (4 units) is correct because the pH changes from 7 to 11 when 0.01 mole of NaOH is added to 10 liters of water.
195 Medium
195 of 401
Application +4 | -1
What minimum concentration of SO42−{SO}_4^{2-} is required to precipitate BaSO4{BaSO}_4 from the solution containing 1.0×10−41.0 \times 10^{-4} moles of Ba2+{Ba}^{2+} ? (Given KSPK_{S P} for BaSO4{BaSO}_4 is 4×10−104 \times 10^{-10} )
View answer and explanation
Correct option: C
4×10−6M4 \times 10^{-6} {M}
Option C (4×10−6M4 \times 10^{-6} {M}) is correct because it is the minimum sulfate concentration required to reach the solubility product of barium sulfate.
196 Medium
196 of 401
Application +4 | -1
NH2−COONH4(s)⇌2NH3(g)+CO2(g){NH}_2-{COONH}_4({s}) \rightleftharpoons 2 {NH}_3({g})+{CO}_2({g}). Total pressure at equilibrium is equal to (Kp=4 atm3)\left({K}_{{p}}=4 \mathrm{~atm}^3\right).
View answer and explanation
Correct option: A
3 atm3 \mathrm{~atm}
Option A (3 atm) is correct because it correctly calculates the total pressure at equilibrium using the given KpK_p value.
197 Medium
197 of 401
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.
198 Medium
198 of 401
Understanding +4 | -1
pH of aqueous NaCI solution after electrolysis
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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.
199 Medium
199 of 401
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.
200 Medium
200 of 401
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.
201 Medium
201 of 401
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-.
202 Medium
202 of 401
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.
203 Medium
203 of 401
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.
204 Medium
204 of 401
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.
205 Medium
205 of 401
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.
206 Medium
206 of 401
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.
207 Medium
207 of 401
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.
208 Medium
208 of 401
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.
209 Medium
209 of 401
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-.
210 Medium
210 of 401
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.
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