Question

Solutions of A, B, C, D, E have pH 1, 5, 7, 8, 14.
(A)- Which basic solution shows presence of only ions?
(B)- Which solution can evolve Sulphur dioxide readily when treated with Potassium bisulphite?
(C)- Which solution has both ions and molecules?
(D)- Which solution can evolve ammonia when treated with ammonium salts?
(E)- Name the type of reaction and state the products obtained when Solution A reacts with Solution E.

Answer 1

In the given solutions, from the pH we can determine the nature of the solution to be acidic or basic, depending on the concentration of the hydrogen ions.

Complete step by step answer:
In case (A), the basic solution which forms the strong base, in which all the molecules have completely ionised, will show the presence of only ions in the solution. The range of the pH for strong bases is 10 to 14. Therefore, we have solution E with pH = 14.
Similarly, in case (C), solution B and D with pH = 5 and 8 respectively, are weak acid and base, showing partial ionisation. Thus, having both the molecules and the ions present in the solution.
In case (B), an acidic solution will evolve Sulphur dioxide readily when treated with Potassium bisulphite (acts as base). So, we have solution A and B.
In case (D), the reaction of the ammonium salt with a base will produce ammonia gas. So, we have solution D.
In case (E), in the reaction of a solution A, that is an acid with pH =1, with the solution E, that is a base with pH = 14. A neutralisation reaction takes place to form salt and water.

Note: The solution with pH = 7, is neutral which is neither acidic nor basic. It has the concentration of the hydrogen ion and the hydroxide ion equal in the solution.
In the range of the pH from 1 to 7, the concentration of the hydrogen ion is more. So, the solution is acidic, whereas for the range from 7 to 14, the concentration of hydroxide is greater than the hydrogen ions. So, the solution is basic.