Testtube B: Has a solution of KMnO4 + NaOH+ NaHSO3----> K2MnO4 (green coloured-solution)+H2O+Na2SO4
If too much HSO3^1- was added to this solution above, what would be the resulting colour and why
To determine the resulting color when too much HSO3^- is added to the given solution, we need to understand the chemical reactions that take place.
The reaction given is:
KMnO4 + NaOH + NaHSO3 → K2MnO4 (green-colored solution) + H2O + Na2SO4
In this reaction, KMnO4 (potassium permanganate) is combined with NaOH (sodium hydroxide) and NaHSO3 (sodium bisulfite) to form K2MnO4 (potassium manganate), H2O (water), and Na2SO4 (sodium sulfate). The resulting solution is green in color due to the formation of K2MnO4.
If an excess amount of HSO3^- is added to the solution, it will affect the balance of the reaction and potentially change the resulting color. HSO3^- is a reducing agent, which means it has the ability to donate electrons and reduce other compounds.
In the given reaction, KMnO4 acts as an oxidizing agent, while HSO3^- acts as a reducing agent. The net reaction involves the reduction of Mn in KMnO4 to Mn in K2MnO4 and the oxidation of HSO3^- to SO4^-2.
However, if an excess of HSO3^- is added, it can continue to reduce Mn in K2MnO4, converting it back to the lower oxidation state of Mn. As a result, the green color of K2MnO4 will be diminished or disappear due to the reduction of Mn.
The exact color that might result from this excess of HSO3^- depends on the concentration and other factors. Generally, the color may shift to a colorless or pale yellow solution, indicating the reduction of Mn and the loss of the green color.
In summary, an excess amount of HSO3^- in the given solution would diminish or eliminate the green color of K2MnO4, resulting in a colorless or pale yellow solution.