If you mix 200 ml of 1.0x10-4 M Ba(NO3)2 and 500ml of 8.0x10-2M H2SO4, what will be observed?

Question

 For BaSO4, the Ksp=1.1x10-10

Answer 1

The question is, will a precipitate of BaSO4 form under these conditions?

Ba(NO₃)₂ + H₂SO₄ ==> BaSO₄(s) + 2HNO₃

200 ml + 500 ml = 700 ml = 0.7 L => final volume

moles Ba(NO3)2 = 0.2 L x 1x10^-4 mol/L = 2x10^-5 moles 

moles  H2SO4 = 0.5 L x 8x10^-2 mol/L = 4x10^-2 moles

Final [Ba²⁺] = 2x10^-5 moles/0.7 liters = 2.86x10^-5 M

Final [SO4^2-] = 4x10^-2 moles/0.7 liters = 5.71x10^-2 M

Q = (2.86x10^-5)(5.71x10^-2) = 1.6x10^-6

Ksp = 1.1x10^-10

Q >> Ksp therefore a precipitate of BaSO4 will form and what will be observed is the appearance of a white precipitate