(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`.

Explanation

(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`.