JEE Main202313 Apr 2023Morning ShiftChemistryIonic EquilibriumActual
25 . 0   mL of 0 . 050   M   Ba NO 3 2 is mixed with 25 . 0   mL of 0 . 020   M   NaF . K sp of BaF 2 is 0 . 5 × 10 – 6 at 298 K . The ratio of Ba 2 + F - 2 and K sp is _ _ _ _ _ _ .
Correct answer
5
Step-by-step solution
Given: Volume of Ba ( NO 3 ) 2 solution = 25 . 0 mL Concentration of Ba ( NO 3 ) 2 solution = 0 . 050 M Volume of NaF solution = 25 . 0   mL Concentration of NaF solution = 0 . 020 M Ba + 2 = 25 × 0 . 05 50 = 0 . 025 M F - = 25 × 0 . 02 50 = 0 . 01 M Ba + 2 F - = 25 × 10 - 7 K sp = 0 . 5 × 10 - 6 = 5 × 10 - 7 Ratio= Ba + 2 F - K s p = 25 × 10 - 7 5 × 10 - 7 = 5