Silver nitrate solution when added to a colorless aqueous solution E forms a white precipitate which dissolves in excess of E. If the white precipitate is heated with water, it turns black and the supernatant solution gives a white precipitate with acidified barium nitrate solution. Therefore, E is

A

$$\text{Na}_2\text{S}$$

B

$$\text{Na}_2\text{S}_2\text{O}_3$$

C

$$\text{Na}_2\text{SO}_3$$

D

$$\text{Na}_2\text{SO}_4$$