Question
Use the equation: FeSCN^2+ (aq) >< Fe^3+ (aq) + SCN^- (aq)
In the reaction FeSCN^2+the ions are red and Fe^3+ are pale yellow and SCN^- ions are colorless. A solution at equilibrium is yellow-orange. The concentration of SCN^- ions are increased by a moderate amount. Why will the solution likely turn orange
In the reaction FeSCN^2+the ions are red and Fe^3+ are pale yellow and SCN^- ions are colorless. A solution at equilibrium is yellow-orange. The concentration of SCN^- ions are increased by a moderate amount. Why will the solution likely turn orange
Answers
GPT-4o mini
In the given equilibrium reaction:
\[
\text{FeSCN}^{2+} (aq) \rightleftharpoons \text{Fe}^{3+} (aq) + \text{SCN}^- (aq)
\]
FeSCN²⁺ has a red color, Fe³⁺ is pale yellow, and SCN⁻ is colorless. At equilibrium, the solution has a yellow-orange color due to the presence of both Fe³⁺ and SCN⁻ ions.
When the concentration of SCN⁻ ions is increased, according to Le Chatelier's principle, the system will respond to counteract this change by shifting the equilibrium position to the left, towards the formation of more FeSCN²⁺.
However, because the reaction forms FeSCN²⁺, which is red, there will still be a significant amount of yellow Fe³⁺ ions present in solution. The increased SCN⁻ ions will lead to a situation where Fe⁺ ions, which are coloring in their ratios, will still result in a dominant presence of yellow coloration from Fe³⁺.
Ultimately, the apparent color will shift towards orange as the equilibrium shifts towards more FeSCN²⁺ (red), but the pale yellow from Fe³⁺ still persists. Thus, the combination of these colors gives the solution a yellow-orange appearance.
In summary, the solution likely turns orange due to:
1. Increased SCN⁻ shifts the equilibrium to the left, producing more red FeSCN²⁺.
2. The presence of Fe³⁺ ions (yellow) ensures that the resulting mixture does not appear fully red, thus creating an orange tint.
Therefore, the slight increase in SCN⁻ leads to more competition between red (from FeSCN²⁺) and yellow (from Fe³⁺), resulting in an overall orange appearance.
\[
\text{FeSCN}^{2+} (aq) \rightleftharpoons \text{Fe}^{3+} (aq) + \text{SCN}^- (aq)
\]
FeSCN²⁺ has a red color, Fe³⁺ is pale yellow, and SCN⁻ is colorless. At equilibrium, the solution has a yellow-orange color due to the presence of both Fe³⁺ and SCN⁻ ions.
When the concentration of SCN⁻ ions is increased, according to Le Chatelier's principle, the system will respond to counteract this change by shifting the equilibrium position to the left, towards the formation of more FeSCN²⁺.
However, because the reaction forms FeSCN²⁺, which is red, there will still be a significant amount of yellow Fe³⁺ ions present in solution. The increased SCN⁻ ions will lead to a situation where Fe⁺ ions, which are coloring in their ratios, will still result in a dominant presence of yellow coloration from Fe³⁺.
Ultimately, the apparent color will shift towards orange as the equilibrium shifts towards more FeSCN²⁺ (red), but the pale yellow from Fe³⁺ still persists. Thus, the combination of these colors gives the solution a yellow-orange appearance.
In summary, the solution likely turns orange due to:
1. Increased SCN⁻ shifts the equilibrium to the left, producing more red FeSCN²⁺.
2. The presence of Fe³⁺ ions (yellow) ensures that the resulting mixture does not appear fully red, thus creating an orange tint.
Therefore, the slight increase in SCN⁻ leads to more competition between red (from FeSCN²⁺) and yellow (from Fe³⁺), resulting in an overall orange appearance.