Write a balanced chemical equation when barium chloride reacts with sodium sulphate to give insoluble barium sulphate and sodium chloride.

(A)  $ BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) + 2NaCl(aq) $ 

(B)  $ BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) + NaCl(aq) $ 

(C)  $ BaC{l_2}(aq) + NaHS{O_4}(aq) \to BaS{O_4}(s) + 2NaCl(aq) $ 

(D)  $ BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to B{a_2}S{O_4}(s) + 2NaCl(aq) $ 

We can define double displacement reactions as the chemical reactions where one component of each of both the reactants is exchanged with each other to form the product molecules. During this reaction, the cations and anions of two different compounds switch places, forming two different compounds entirely.

For the compounds given in the question, on mixing a solution of barium chloride with sodium sulphate, we observe that a white precipitate of barium sulphate is immediately formed which is insoluble in water. This reaction also forms Sodium chloride as a product which remains dissolved in water and hence cannot be seen. This reaction is ionic in nature. The reactants change into ions when dissolved in water and there is an exchange of ions in solution. This results in the formation of product molecules.

Hence, we can now finally write the reaction as:

 $ BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) + 2NaCl(aq) $ 

In general, the reaction of any compound with barium chloride which provides a white precipitate which remains insoluble in water is considered to be a test for the sulphate anion.