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When `0.2M` solution of acetic acid is neutralised with `0.2M NaOH` in `500 mL` of water, the `pH` of the resulting solution will be: `[pK_(a)` of acetic acid `= 4.74]`A. `12.67`B. `7.87`C. `8.87`D. `7` |
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Answer» Correct Answer - C `CH_(3)COOH +NaOH rarr CH_(3)COONA +H_(2)O` `rArr` At equivalence point, a salt of weak acid, strong base is formed. `rArr pH = 7 +(1)/(2)(pK_(a) +logC)` Here `C=` concentration of salt `= (0.2 xx 500)/(500+500) =0.1M` [Check that `500mL` of `CH_(3)COOH` is also required] `rArr pH 7+(1)/(2) (4.75 + log0.1) = 8.87` |
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