Question

$a.$ Define the term neutralisation equivalent of an acid.
$b.$ Find the number of ionisable $H$ atoms for a polycarboxyclic acid $(Mw=210gmmo{l}^{-1})$ with a neutralisation equivalent of $70gme{q}^{-1}$. How many equivalents of $NaOH$ would be neutralized by $1mol$ of this acid?
$c.$ Find the neutralisation equivalent of mellitic acid, $C_6(COOH{)}_6$.

Answer 1

(a) The neutralisation equivalent (NE) is the equivalent weight (g/eq) of an acid as determined by titration with standardized NaOH.
(b) The no. of ionizable H's of an acid is the no. of equivalents per mole, and is equal to Mw/NE.
Here, the no. of equivalents (No. of ionizable H's)$=\frac{210ℊ/㏖}{70ℊ/eq}=3eq/㏖$
The no. of equivalent of NaOH equals the no. of ionizable H's, in this case 3eq/mol.
(c) The neutralization equivalent of mellitic acid is
$NE=\frac{Mw}{No.ofionizable{H}^{\prime }s}=\frac{342ℊ/㏖}{6eq/㏖}=57ℊ/eq$