NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
If 200 mL of a 0 .031 molar solution of H 2 SO 4 are added to 84 mL of a 0 .150 M KOH solution, what is the pH of the resulting solution?
Options
- A12.4
- B1.7
- C2.2
- D10.85
Correct answer
D. 10.85
Step-by-step solution
m m o l o f H + i n i t i a l = 200 × 0.031 × 2 = 12.4 m m o l o f O H - i n i t i a l = 84 × 0.15 = 12.6 m m o l o f O H - l e f t a f t e r n e u t r a l i s a t i o n = 0.2 OH − final   =   0 .2 284   =   7   ×   10 − 4   M pOH = 3 .15 and pH = 10 .85