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| 1. |
25 ml of 0.1 M acetic acid is titrated with 0.1NaoH solution. The pH of the solution atequivalence point will be (log 5 0.7CH3COOH = 4.76) |
| Answer» 25 ml of 0.1 M acetic acid is titrated with 0.1NaoH solution. The pH of the solution atequivalence point will be (log 5 0.7CH3COOH = 4.76) | |