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Degree of hydrolysis of `0.25 M CH_(3)COOHNa` is `10%` what will be the degree of hydrolysis if concentration of `CH_(3)COONa` is made `0.01M`A. `10%`B. `1%`C. `50%`D. `75%` |
Answer» Correct Answer - 3 `(h_(1))/(h_(2))=sqrt((c_(2))/(c_(1)))rArr(0.1)/(h_(2))=sqrt((0.01)/(0.25))rArrh=0.5rArr50%(3)` |
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