Chemical Equilibrium And Acids – Bases (VSAQs)

Chemistry-1 | 7. Chemical Equilibrium and Acids – Bases – VSAQs:
Welcome to VSAQs in Chapter 7: Chemical Equilibrium and Acids – Bases. This page includes the important FAQs for Very Short Answer Questions. Answers are provided in simple English and follow the exam format. This approach helps in focusing on key details and aiming for top marks in your final exams.

VSAQ-1 : What is homogeneous equilibrium? Write two homogeneous reactions.

Homogeneous Equilibrium

Homogeneous equilibrium occurs when all the reactants and products in a chemical reaction are in the same physical state (e.g., all in the gas phase, all in the liquid phase, or all in the solid phase).

Examples of Homogeneous Reactions:

  1. Nitrogen Synthesis: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$
  2. Water Decomposition: $$2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g)$$

In these reactions, all the involved substances are in the gaseous phase.

VSAQ-2 : What is heterogeneous equilibrium? Write two heterogeneous reactions.

Heterogeneous Equilibrium

Heterogeneous equilibrium refers to a chemical equilibrium in which the reactants and products exist in different physical states (e.g., some in the gas phase, some in the liquid phase, and some in the solid phase).

Examples of Heterogeneous Reactions:

  1. Ice Melting: $$H_2O(s) \rightleftharpoons H_2O(l)$$
  2. Formation of Rust: $$4Fe(s) + 3O_2(g) \rightleftharpoons 2Fe_2O_3(s)$$

In these reactions, there is a mixture of solid and liquid phases in the first example and a mixture of solid and gas phases in the second example, illustrating heterogeneous reactions.

VSAQ-3 : Define Ionic product of water. What is the value at room temperature?

Ionic Product of Water (Kw)

The ionic product of water, denoted as Kw, represents the product of the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in pure water at a specific temperature.

Value at Room Temperature: At room temperature, typically around 25°C or 298 K, the value of Kw is approximately 1.0 x 10^(-14) mol^2/L^2.

VSAQ-4 : Define Basicity of acid and Acidity of base.

  1. Basicity of Acid: The basicity of an acid is the count of replaceable H+ ions it can release.
  2. Acidity of Base: The acidity of a base is the number of replaceable OH– ions it can release.

VSAQ-5 : What is Lewis acid? Give one example?

  1. Lewis Acid: A Lewis acid is a substance capable of accepting a pair of electrons to form a covalent bond.
  2. Example: BF3 is an example of a Lewis acid.

VSAQ-6 : What is a Bronsted base? Give one example?

  1. Bronsted Base: A Bronsted base is a substance capable of accepting a proton (H+ ion).
  2. Example: NH3 (ammonia) is an example of a Bronsted base.

VSAQ-7 : All Bronsted bases are Lewis bases. Explain.

All Bronsted bases are indeed Lewis bases because in order to accept a proton (H+ ion), they must donate an electron pair, which qualifies them as Lewis bases.

VSAQ-8 : What is conjugate acid-base pair? Give example.

Conjugate Acid-Base Pair

A conjugate acid-base pair is a pair of substances consisting of an acid and a base that differ by the transfer of one proton (H+ ion).

Example: In the reaction: $$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$$

The pairs NH4 + & NH3 and H2O & OH– are conjugate acid-base pairs.

VSAQ-9 : Define p^Hof a solution. Calculate the pH of 0.001M HCl

pH of a Solution

pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration.

pH Calculation Example:

To find the pH of a 0.001M HCl solution:

  1. Identify the hydrogen ion concentration: [H+] = 0.001M.
  2. Use the formula: $$pH = -\log_{10} [H+]$$
  3. Plug in the concentration: $$pH = -\log_{10}(0.001)$$
  4. Calculate: $$-(-3 \log_{10} 10) = 3$$

Result: The pH of the 0.001M HCl solution is 3.

VSAQ-10 : Calculate the p^Hof 10-8 M HCl

Calculating the pH of a Solution

Calculation Example: To calculate the pH of a 10−810−8 M HCl solution:

  1. Find the hydrogen ion concentration [H+]: [H+] = 10−8 M.
  2. Calculate [H+] with slight dilution: [H+] = 10−7 × (0.1 + 1) = 1.1 × 10−7.
  3. Calculate pH using the formula: pH = −log10​[H+].
  4. Plug in the concentration: pH = −log10​(1.1×10−7).
  5. Calculate: pH ≈ 6.959.

Result: The pH of the solution is approximately 6.959.

VSAQ-11 : The Concentration of Hydrogen ion in a sample of soft drink is 3.8 x 10-3 M. What is its pH?

Calculating pH of a Soft Drink

Calculation Example: To find the pH of a soft drink with a hydrogen ion concentration of 3.8 x 10^(-3) M:

  1. Use the formula: $$pH = -\log_{10}[H+]$$
  2. Plug in the concentration: $$pH = -\log_{10}(3.8 x 10^(-3))$$
  3. Calculate: $$pH \approx -( \log_{10} 3.8 + \log_{10} 10^(-3))$$
  4. Simplify: $$pH \approx -( \log_{10} 3.8 – 3 \cdot \log_{10} 10)$$
  5. Further simplify: $$pH \approx 3 \cdot \log_{10} 10 – \log_{10} 3.8$$
  6. Calculate: $$pH \approx 3 – 0.58 = 2.42$$

Result: The pH of the soft drink is approximately 2.42.

VSAQ-12 : Calculate the pH of 0.05M H2SO4 solution

  1. H2SO4 dissociates into two moles of H+ ions for every mole of the compound.
  2. Calculate the hydrogen ion concentration [H+] by multiplying the molarity of H2SO4 by 2 (due to the two moles of H+ ions per mole of H2SO4): [H+] = 0.05 M * 2 = 0.10 M.
  3. Use the formula for pH: pH = -log10 [H+].
  4. Plug in the concentration: pH = -log10(0.10).
  5. Calculate: pH ≈ -(-1) = 1.

The pH of the 0.05M H2SO4 solution is approximately 1.

VSAQ-13 : Calculate the pH of 0.001M NaOH

pH Calculation for 0.001M NaOH Solution:

  1. Recognize that NaOH is a strong and monoacidic base.
  2. Calculate the hydroxide ion concentration [OH-]: [OH-] = Molarity x Acidity = 0.001 x 1 = 10−3.
  3. Calculate the pOH using the formula: $$pOH = -\log_{10}[OH-] = -\log_{10}(10^{-3})$$
  4. Simplify pOH: $$pOH = -( \log_{10}10^{-3}) = 3$$
  5. Use the relationship between pH and pOH: $$pH + pOH = 14$$
  6. Calculate pH: $$pH = 14 – pOH = 14 – 3 = 11$$

Result: The pH of the 0.001M NaOH solution is 11.