InterviewSolution
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InterviewSolution
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A definite volume of an aqueous `N//20` acetic acid `(pK_(a) = 4.74)` is titrated with a strongs base. It is found that `75` equal-sized drops of `NaOH` added from a burette effect the complete neutralisation. Find the `pH` when an acid solution is neutralised to the extent of `20%, 40%`, and `80%`, respectively. |
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Answer» During the titration of weak acid with swtrong base, if weak acid is left bwhind, acidic buffer is formed. i. `20%` neutralistion of `W_(A)` means `20%` salf of `W_(A)//S_(B)` is formed and `80% W_(A)` is left. `:. pH = pK_(a) + "log"(["Salt"])/(["Acid")]` `= 4.74 +log ((20)/(80))` `=4.74 + log ((1)/(4))` `= 4.74 - log2^(2)` `= 4.74 - 2 xx 0.3 = 4.14` ii. `40%` neutralisation of `W_(A)` means `40%` salt of `W_(A)//S_(B)` is formed and `60% W_(A)` is left. `:. pH = 4.64 +log ((40)/(60))` `= 4.74 +log 2 - log3` `= 4.74 +0.3 - 0.48 = 4.56` iii. `90%` neutralisation of `W_(A)` means `90%` salt of `W_(A)//S_(B)` is formed and `10% W_(A)` is left. `:. pH = 4.74 + log ((90)/(10))` `= 4.74 + log 3^(2)` `= 4.74 +2 xx 0.48 = 5.70` |
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