Question

Neutralisation of $0.3504gm$ of acid $\left(A\right)$ requires $27.24ml$ of $0.15MNaOH$, and $Mw$ is found from the mass spectral data to be $172.1gmmo{l}^{-1}$.
$\left(a\right)$ Calculate the $N.E.$ of $\left(A\right)$
$\left(b\right)$ How many ionisable $H$ atoms are there in $\left(A\right)$?

Answer 1

(a) mEq of $NaOH=27.24\times 0.15=4.086$

$mEq.=\frac{4.086}{1000}eq.=0.004086eq.$

N.E (Ew)$=\frac{0.3504}{0.004086}=85.76gm.{eq}^{-1}$

(b) Ionisable H atoms $=\frac{Mw}{N.E}$ or $\frac{Mw}{Ew}$

$=\frac{172.1}{85.76}=2eq.$ ${mol}^{-1}$

$=2$ ionisable $H$ atoms ${mol}^{-1}$

$=A$ is dicarboxylic acid