Answer:
a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositive nature of the alkali metals.
While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositive nature of the group1 elements which also indicate their strong reducing agent.
b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.
c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.
Explanation:
a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositivity nature of the alkali metals.
While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositivity nature of the group1 elements which also indicate their strong reducing agent.
b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.
c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.