Question

To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H₃PO₃), the value of 0.1 M aqueous KOH solution required is (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

Answer 1

C.

40 mL

H3PO3 is a dibasic acid (containing two ionizable protons attached to O directly)

H3PO3 ⇌ 2H+ + HPO42-

0.1 M H3PO3 = 0.2 N H3PO3

0.1 M KOH = 0.1 N KOH

N1V1 = N2V2

(KOH) = (H3PO3)

0.1 V1 = 0.2 x 20

V1 = 40 mL

H3PO3 is a dibasic acid (containing two ionizable protons attached to O directly)

H3PO3 ⇌ 2H+ + HPO42-

0.1 M H3PO3 = 0.2 N H3PO3

0.1 M KOH = 0.1 N KOH

N1V1 = N2V2

(KOH) = (H3PO3)

0.1 V1 = 0.2 x 20

V1 = 40 mL