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The pH of a 0.1 M `CH_(3)COOH` which is `2%` ionised in aqueous solution isA. `2.0`B. `1.0`C. `log((0.1)/(0.001))`D. 2.7 |
Answer» Correct Answer - D `{:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Eqm. conc.",0.1(1-0.02),,0.1 xx 0.02 ,,0.1 xx 0.02):}` `pH=-log[H^(+)]=-log (2 xx 10^(-3))=2.7` |
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