A red litmus paper when dipped in an aqueous solution of potassium acetate turned blue. Arrange the following steps in proper sequence to explain the above observation. (1) identification of spectator ions (2) identification of free ions which caused the colour change of the litmus paper (3) identification of positive and negative radicals (4) writing the net reaction taking place (5) writing ionic equation for the reaction of positive and negative radicals with water (a) 35142 (b) 31542 (c) 51432 (d) 53142

Identify the positive radical (K+) and negative radical (CH3COO-) in potassium acetate (KCH3COO).

Identify the spectator ion in this case, which is K+ as it does not participate in the reaction affecting the litmus paper.

Understand that the negative radical (CH3COO-) reacts with water (H2O) to form OH-, which is responsible for turning the litmus paper blue. This is because OH- is responsible for making the solution basic.

Now write the ionic equation: CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq).

Finally, write the net reaction by removing the spectator ion: CH3COO-(aq) + H2O(l) <=> CH3COOH(aq) + OH-(aq). Option (b) 31542: In this option, the identification of the spectator ions comes before the identification of the free ions causing the color change. This would be incorrect because, when we identify spectator ions first, we may not have enough information about the free ions causing the color change. Option (c) 51432: This option is not proper because we need to identify the respective positives and negatives radicals of the potassium acetate before identifying spectator ions and writing the chemical equations, alluding to the solution observed from the litmus paper. Option (d) 53142: Again, this option is not the correct sequence because the identification of spectator ions and free ions should come after the identification of positive and negative radicals. Therefore, the correct sequence to explain the given observation is option (a) 35142.