100 mL of a solution of X is titrated with a 0.1 M solution of Y giving the following…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
100 mL of a solution of X is titrated with a 0.1 M solution of Y giving the following titration curve :
Answer the following four questions based on the information provided.
The approximate value of ionization constant (Ka/Kb) of X is
- 5 × 10–9
- 5 × 10–6
- 2 × 10–4
- 3 × 10–5
Answer
(C) 2 × 10 –4
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