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NEET CHEMISTRYEquilibriumEasy

Question

The salt solution that is basic in nature is:

A

Ammonium chloride

B

Ammonium sulphate

C

Ammonium nitrate

D

Sodium acetate

Step-by-Step Solution

The acidic or basic nature of a salt solution depends on the relative strengths of its parent acid and base .

Sodium acetate (CH3COONa\text{CH}_3\text{COONa}) is a salt formed from a weak acid (acetic acid, CH3COOH\text{CH}_3\text{COOH}) and a strong base (sodium hydroxide, NaOH\text{NaOH}). In an aqueous solution, it completely ionises, and the acetate ion undergoes anionic hydrolysis to produce hydroxyl ions: CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) This process results in an increase in the concentration of OH\text{OH}^- ions, making the solution alkaline or basic in nature (pH > 7) .

Conversely, ammonium chloride, ammonium sulphate, and ammonium nitrate are salts formed from a weak base (NH4OH\text{NH}_4\text{OH}) and strong acids (HCl\text{HCl}, H2SO4\text{H}_2\text{SO}_4, and HNO3\text{HNO}_3, respectively). These undergo cationic hydrolysis to release H+\text{H}^+ ions, making their solutions acidic (pH < 7) .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Equilibrium. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYEquilibriumsolutionnature

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Consider the following reaction: $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$. At equilibrium, the concentrations of $[\text{A}_2] = 3.0 \times 10^{–3} \text{ M}$; $[\text{B}_2] = 4.2 \times 10^{–3} \text{ M}$ and $[\text{AB}] = 2.8 \times 10^{–3} \text{ M}$. The value of $K_c$ for the above-given reaction in a sealed container at $527^\circ\text{C}$ is:

A.3.9
B.0.6
C.4.5
D.2
MediumSolve

Amongst the given options, which of the following molecules/ions acts as a Lewis acid?

A.$\text{OH}^-$
B.$\text{NH}_3$
C.$\text{H}_2\text{O}$
D.$\text{BF}_3$
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Boric acid is an acid because its molecule

A.contains replaceable H⁺ ion
B.gives up a proton
C.accepts OH⁻ from water releasing proton
D.combines with proton from water molecule
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The following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations. pH of which one of them will be equal to 1?

A.$60 \text{ mL } \frac{M}{10} \text{ HCl } + 40 \text{ mL } \frac{M}{10} \text{ NaOH}$
B.$55 \text{ mL } \frac{M}{10} \text{ HCl } + 45 \text{ mL } \frac{M}{10} \text{ NaOH}$
C.$75 \text{ mL } \frac{M}{5} \text{ HCl } + 25 \text{ mL } \frac{M}{5} \text{ NaOH}$
D.$100 \text{ mL } \frac{M}{10} \text{ HCl } + 100 \text{ mL } \frac{M}{10} \text{ NaOH}$
MediumSolve

The tendency of $BF_3$, $BCl_3$ and $BBr_3$ to behave as Lewis acid decreases in the sequence:

A.$BCl_3 > BF_3 > BBr_3$
B.$BBr_3 > BCl_3 > BF_3$
C.$BBr_3 > BF_3 > BCl_3$
D.$BF_3 > BCl_3 > BBr_3$
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A compound $\text{BA}_2$ has $K_{sp} = 4 \times 10^{-12}$. Solubility of this compound will be:

A.$10^{-3} \text{ mol/L}$
B.$10^{-4} \text{ mol/L}$
C.$10^{-5} \text{ mol/L}$
D.$10^{-6} \text{ mol/L}$
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What is the molarity of the saturated solution if the solubility product for a salt of type AB is $4 \times 10^{-8}$?

A.$2 \times 10^{-4} \text{ mol/L}$
B.$16 \times 10^{-16} \text{ mol/L}$
C.$2 \times 10^{-16} \text{ mol/L}$
D.$4 \times 10^{-4} \text{ mol/L}$
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In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains $\text{Ag}^+$ and $\text{Pb}^{2+}$ at a concentration of $0.10 \text{ M}$. Aqueous $\text{HCl}$ is added to this solution until the $\text{Cl}^-$ concentration is $0.10 \text{ M}$. What will the concentration of $\text{Ag}^+$ and $\text{Pb}^{2+}$ at equilibrium? ($K_{sp}$ for $\text{AgCl} = 1.8 \times 10^{-10}$, $K_{sp}$ for $\text{PbCl}_2 = 1.7 \times 10^{-5}$)

A.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-4} \text{ M}$
B.$[\text{Ag}^+] = 1.8 \times 10^{-7} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-6} \text{ M}$
C.$[\text{Ag}^+] = 1.8 \times 10^{-11} \text{ M}; [\text{Pb}^{2+}] = 8.5 \times 10^{-5} \text{ M}$
D.$[\text{Ag}^+] = 1.8 \times 10^{-9} \text{ M}; [\text{Pb}^{2+}] = 1.7 \times 10^{-3} \text{ M}$
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