Question

The concentration of SO42–(aq) in a sample of river water can be determined using a precipitation titration in which a salt of Ba2+(aq) is the standard solution and BaSO4(s) is the precipitate formed. What is the concentration of SO42– (aq) in a 42.6 mL sample if 5.00 mL of a 0.00100 M Ba2+(aq) solution is needed to precipitate all the SO42–(aq) in the sample?

Answer 1

Ba2+(aq) + SO4^2- ---------> BaSO4(s)

1mole. 1mole. 1mole

137.33g 96g 233.33g

0.0010M of Ba2+ means 0.0010 mole in 1 litre

Therefore, in 5 ml No of mole of Ba2+ =( 0.0010/1000)×5 = 5×10^-6mole

1mole of Ba2+ react with 1mole of Ba2+

Therefore , 5 × 10^-6 M Ba2+ react with 5 × 10^-6 mole

Molar mass of SO4^2- is 96g

Therefore, mass of SO4^2- = 5×10^-6 Mole × 96g/mole = 0.00048 g = 0.48mg

0.48 mg present in 42.6ml

Therefore, concentration of SO4^2- = ( 0.48mg/42.6ml)×1000ml = 11.26 mg/litre