Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

Calculate dissociation constant for an acid HA having [ H^ + ]...

Calculate dissociation constant for an acid HA\text{HA} having [H+][ H^{+} ] equals to 0.1M\text{0}\text{.1}\text{M} (Given: degree of dissociation of HA\text{HA} is \text{1%})

Options

  1. A\text{1}{\text{0}^{\text{-8}}
  2. B\text{1}{\text{0}^{\text{-6}}
  3. C\text{1}{\text{0}^{\text{-3}}
  4. D\text{1}{\text{0}^{\text{-7}}

Correct answer

C. \text{1}{\text{0}^{\text{-3}}

Step-by-step solution

ka=α2C1−α{k}_{a} = \dfrac{{\alpha}^{2}C}{1 - \alpha} , α\alpha is very small ;
∴1−α≈1∴ 1 - \alpha \approx 1
C \alpha=0.1\text{C \alpha}\text{=}\text{0}\text{.1}
ka=(C \alpha) \alpha{k}_{a} \text{=}\text{(C \alpha ) \alpha}
=0.01×0.1=10−3= 0.01 \times 0.1 = {10}^{- 3}
Practice Ionic Equilibrium on Quantrex Academy →

More from Ionic Equilibrium

Given is a concentrated solution of a weak electrolyte A_xB_y of concentration 'c' and… JEE Main 2026M_3 A_2 is a sparingly soluble salt of molar mass y g mol ^ -1 and solubility x g L ^ -1… JEE Main 2026Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL… JEE Main 202620 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization.… JEE Main 2026The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1… JEE Main 2026The first and second ionization constants of a weak dibasic acid H_2 A are 8.1 × 10^ -8… JEE Main 2026At 25°C , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH… JEE Main 2026The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298… JEE Main 2026 Full Ionic Equilibrium list All NTA Abhyas JEE Main PYQs