Question

If a 0.50 M solution of K2SO4 is slowly poured into a beaker containing 0.25 Mbarium nitrate and 0.30 M lead (II) nitrate at 25oC, what will be the first precipitate that forms

Answer 1

Answer: The first precipitate formed is Barium Sulphate

Step-by-step explanation:

This is because the reaction that takes place will be as follows:

K2SO4 + Ba (NO3)2 ------> BaSO4 + K2(NO3)2

And K2SO4 + Pb(NO3)2 -----> PbSO4 + K2(NO3)2

Considering the concentration of Barium Nitrate which is lower than that of Lead Nitrate it tends to attain equilibrium faster and the Barium salt or precipitate is formed....