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Question

Neutralisation of 0.3504 gm of acid (A) requires 27.24 ml of 0.15 M NaOH, and Mw is found from the mass spectral data to be 172.1 gm mol1.
(a) Calculate the N.E. of (A)
(b) How many ionisable H atoms are there in (A)?

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Solution

(a) mEq of NaOH=27.24×0.15=4.086

mEq.=4.0861000eq.=0.004086eq.
N.E (Ew)=0.35040.004086=85.76gm.eq1

(b) Ionisable H atoms =MwN.E or MwEw
=172.185.76=2eq. mol1

=2 ionisable H atoms mol1

=A is dicarboxylic acid

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