Question

100ml solution of an acid (M.W. = 82) containing 39g of acid per litre was completely neutrlaized by 95mL of
aqueous NaOH solution containing 40g of NaOH per litre. The basicity of acid is

Answer 1

Answer:no. Of moles of NAOH= 40g/40gmol- = 1mol

No. Of moles of acid= 39/82

For neutralisation,

NI VI =N2 V2

(Molarity)1 × acidity × VI = (Molarity)2 ×basicity × V2

=No.of moles of NAOH/1l X100= no.of moles of acid/1l × acidity× V2

=1/1×100= 32/82×basicity×95

Basicity= (82×100)/(95×39)

=2.21

=2approximately

Explanation: