NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
0.25 mol of formic acid H C O 2 H is dissolved in enough water to make one litre of solution. The pH of that solution is 2.19. The K a of formic acid is
Options
- A6.5 × 1 0 - 3
- B4.3 × 1 0 - 4
- C1.7 × 1 0 - 4
- D5.3 × 1 0 - 2
Correct answer
C. 1.7 × 1 0 - 4
Step-by-step solution
pH = 2.19 = - log [ H + ] H + = 1 0 - 2.19 = Antilog - 2.19 = 6.46 × 1 0 - 3 M K a H C O O H = H + H C O O - H C O O H Here H + = H C O O - = 6.46 × 1 0 - 3 M H C O O H = 0.25 mol/1L = 0.25 M and K a H C O O H = ? So K a ( H C O O H ) = ( 6.46 × 1 0 - 3 ) 0.25 2 = 1.7 × 1 0 - 4 .