Question
To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3PO3), the volume of 0.1 M aqueous KOH solution required is
-
10 mL
-
60 mL
-
40 mL
-
20 mL
Solution
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