#HCl

Aptitude and Reasoning - Part 3

What is the pH of 10 ⁻⁸ M solution of HCl?

The principle of enzyme catalysis and lipid structure explains that substrate affinity, transition state stabilization and phospholipid composition govern function. Hence 6.959 fits best because it reflects catalytic efficiency and membrane organization, as documented in authoritative sources on biochemistry.

Ref: Lehninger Principles of Biochemistry, Chapter 9, Lipids and Membranes, explains phospholipid components, glycerol, fatty acids, phosphate group and membrane structure providing foundation for lipid biochemistry.

View Details

Polyhalogen Compounds - Chloroform Iodoform DDT Freons

Which compound is formed when aniline reacts with NaNO₂ and HCl , followed by KI ?

Aniline forms a diazonium salt with NaNO₂ and HCl , which reacts with KI to form C₆H₅I (iodobenzene).

Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons

View Details

Oxidizing and Reducing Agents and Strength

In 2K₂Cr₂O₇ + 14HCl → 2CrCl₃ + 2KCl + 3Cl₂ + 7H₂O, how many electrons are transferred?

Cr (+6 in Cr₂O₇²−) to +3 in CrCl₃: 4 Cr gain 12 electrons. Cl (-1 to 0): 6 Cl lose 12 electrons. Total = 12.

Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Oxidizing and Reducing Agents and Strength

View Details

Electrode Potential and Electrochemical Series and Galvanic Cells

In 2KMnO₄ + 16HCl → 2MnCl₂ + 5Cl₂ + 2KCl + 8H₂O, which species contains the element reduced?

Mn (+7 in KMnO₄) becomes +2 in MnCl₂, gaining electrons; it’s the reduced species.

Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Electrode Potential and Electrochemical Series and Galvanic Cells

View Details

Buffer Solutions and Solubility Product and Common Ion Effect

A weak acid HA ( Ka = 1.6 × 10⁻⁶ ) is mixed with 0.05 M NaA and 0.1 M HCl in equal volumes. What is the pH?

After mixing: [HCl] = 0.05 M , [A-] = 0.025 M , [HA] negligible due to common ion effect. Ka = ([H+][A-]/[HA]) , [H+] ≈ 0.05 , pH = -log(0.05) = 1.3 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect

View Details

Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds

The bond in HCl is an example of:

In HCl , hydrogen and chlorine share one electron pair, forming a single covalent bond.

Ref: NCERT Class 11 Chemistry > Chapter 4: Chemical Bonding and Molecular Structure > Topic: Ionic Bonding - Lattice Enthalpy and Properties of Ionic Compounds

View Details

Laws of Chemical Combination - Conservation Mass Definite Multiple

A 0.25 M HCl solution is diluted by adding 300 mL of water to 200 mL of the original solution. What is the final molarit

M₁V₁ = M₂V₂. 0.25 × 200 = M₂ × (200 + 300). M₂ = (0.25 × 200)/500 = 0.1 M.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Laws of Chemical Combination - Conservation Mass Definite Multiple

View Details

Importance and Nature of Chemistry and Properties of Matter

A 0.4 M HCl solution is diluted by adding 200 mL of water to 100 mL of the original solution. What is the final molarity

M₁V₁ = M₂V₂. 0.4 × 100 = M₂ × (100 + 200). M₂ = (0.4 × 100)/300 ≈ 0.133 M.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Importance and Nature of Chemistry and Properties of Matter

View Details

Dalton's Atomic Theory and Atomic and Molecular Masses

How many grams of MnO₂ are required to produce 1.42 g of Cl₂ with excess HCl? (Molar masses: MnO₂ = 87 g/mol, Cl₂ = 71 g

Reaction: MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂. Moles of Cl₂ = 1.42/71 = 0.02 mol. 1 mol Cl₂ needs 1 mol MnO₂; 0.02 mol needs 0.02 × 87 = 1.74 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

View Details

Water and Mole Concept

How much water must be added to 10 mL of 0.5M HCl to dilute it tenfold?

90mL is obtained by applying the relevant formula or quantitative relationship to the given parameters. In Water and Mole Concept, numerical problem-solving requires understanding the mathematical relationships between biological variables. The calculation involves substituting the provided values into the appropriate equation and solving systematically. The other options (10mL, 100mL, and 990mL) result from common calculation errors such as using incorrect formulas, misidentifying variables, inverting ratios, or making arithmetic mistakes.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 2

View Details