posted by John on .
What volume (in mL) of 0.45 M Na2SO4 solution is needed to precipitate all the barium, as BaSO4(s), from 14.5 mL of 0.15 M Ba(NO3)2 solution?
Ba(NO3)2(aq) + Na2SO4(aq)= BaSO4(s) + 2 NaNO3(aq)
mols Ba(NO3)2 = M x L = ?
mols Na2SO4 = mols Ba(NO3)2 (look at the coefficients in the balanced equation)
M Na2SO4 = mols Na2SO4/L Na2SO4. You now mols and M, solve for L and convert to mL.