NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
For the indicator, HIn; the ratio In - HIn is 7 .0 at pH of 4 .3 . K In for the indicator is [ Given: log 7 = 0 .845 and Antilog ( − 3.455 ) = 3.5 × 10 − 4 ]
Options
- A3 . 5 × 1 0 - 4
- B3 . 5 × 1 0 - 5
- C3 . 5 × 1 0 - 2
- D3 . 5 × 1 0 - 3
Correct answer
A. 3 . 5 × 1 0 - 4
Step-by-step solution
For weak organic acid indicators pH = pK In + log In − HIn 4 .3 = pK In + log 7 pK In = 4.3 − 0.845 = 3.455 pK In = − log 10 K In K In = Antilog − pK In K In =   Antilog   − 3.455 = 3 .5 × 10 − 4