Answer to Question #156603 in Physical Chemistry for RAO JUNIOR COLLEGE OF SCIENCE

Question

100 ml of a raw water sample on titration with N/50 H 2 SO 4

required 10 ml of the acid to phenolphthalein end point and

another 10 ml of the acid to methyl orange end point. Determine

the type and extent of alkalinity present in the water sample.

Answer 1

So 10 mL acid of 0.02 N is needed to neutralise the base present in raw water

As H2SO4 is dibasic acid so 0.02 N acid = 0.02\2 = 0.01 M

Now one mole acid is consumed by 2 mol base as it is dibasic.

So mol of acid = (0.01 mol\1000 mL)× 10 mL = 1× 10^-4 mol

So mol of base = 2 ×1×10^-4

Conc of OH- = (2×10^-4 ×1000) \ 100

= 2 × 10^-3 M

POH = -log [OH-] = -log (2×10^-3) = 2.7

PH + POH = 14

PH = 14 - POH = 14 - 2.7 = 11.3

So the water is basic and it's PH value is 11.3