Respuesta :
Here we have to calculate the amount of [tex]SO_{4}^{2-}[/tex] ion present in the sample.
In the sample solution 0.122g of [tex]SO_{4}^{2-}[/tex] ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + [tex]SO_{4}^{2-}[/tex]
(Na)₂SO₄=2Na⁺ + [tex]SO_{4}^{2-}[/tex]
Thus, BaCl₂+ [tex]SO_{4}^{2-}[/tex] = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of [tex]SO_{4}^{2-}[/tex] ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion ([tex]SO_{4}^{2-}[/tex]) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of [tex]SO_{4}^{2-}[/tex] ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of [tex]SO_{4}^{2-}[/tex] ion is present. So in 1 g of BaSO₄ [tex]\frac{96.06}{233.3}=0.411[/tex] g of [tex]SO_{4}^{2-}[/tex] ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of [tex]SO_{4}^{2-}[/tex] ion is present.
