A sample of 50.0 mL of 0.10 M NH 3 (K b = 1.8 × 10 –5 ) is titrated with 0.10 M HCl…
Chemistry · JEE Main · NTA Exams — Equilibrium
A sample of 50.0 mL of 0.10 M NH3 (Kb = 1.8 × 10–5) is titrated with 0.10 M HCl. Calculate the pH at the equivalence point.
- 6.98
- 2.87
- 7.78
- 5.28
Answer
(D) 5.28
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