Question

. A 20 ml water sample required 15 ml of 0.05 N HCl for titration with phenolphthalein indicator.

If same water sample (20 ml) is titrated with 0.05 N HCl in the presence of methyl orange indicator

then 35 ml of acid is used. Determine the type of alkalinity and calculate the amount of alkalinity in

terms of CaCO3 equivalent.​

Answer 1

Answer:

Explanation:

water sample contains carbonates and bicarbonates

when reacted with phenopthalein indicator

equivalent of  caco3 = 0.75

moles*1000 = 0.75

w=0.075 gm

when treated with methyl orange

equivalent of  ca(hco3)formed + equivalent of ca(hco3) initial = 1.75

equivalent of  ca(hco3)2  = 1

w/162*2*1000=1

w=0.081 gm

mass of caco3 = 0.8 gm

0.8 gm in 20 gm water

2500 gm in 10^6 gm water

therefore 2500 ppm