50 mL of 0.0535 M H 3 PO 4 was titrated with 0.115 M solution of NaOH to the end point…
Chemistry · JEE Main · NTA Exams — Some Basic Concepts in Chemistry
50 mL of 0.0535 M H3PO4 was titrated with 0.115 M solution of NaOH to the end point identified by indicator X. This required 23.1 mL of NaOH. The above titration was repeated using Y as indicator. Now 46.2 mL of same NaOH is used, the number of H3PO4 replaced in presence of X and Y are:
- 1,2
- 2,1
- 1,1
- 2,2
Answer
(B) 2,1
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