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(a) Define the term neutralisation equivalent of an acid. (b) Find the number of ionisable `H` atoms for a polycarboxyclic acid `(Mw = 210 gm mol^-1)` with a neutralisation equivalent if `70 gm eq^-1`. How many equivalents of `NaOH` would be neutralised by `1 mol` of this acid ? ( c) Find the neutralisation equivalent of mellitic acid, `C_6(COOH)_6`. |
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Answer» (a) `N.E` is the `Ew(gm eq.^-1)` of an acid as determined by titration with standard `NaOH`. (b) The number of ionisable `H` atoms `= 210//70 = 3 eq. mol^-1`. The number of equivalent of `NaOH` = The number of ionisable `H` atom `= 3 eq. mol^-1`. ( c) `N.E` of mellitic acid `= (Mw)/("No. of ionisable H atoms")` =`(342 gm mol^-1)/(6 eq. mol^-1)` =`57 gm eq.^-1`. |
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