Advance Study Assignment: Properties of Systems in Chemical Equilibrium Methyl orange, HMO, is a common acid-base indicator. In solution, it ionizes according to the equation: HMO(aq) → H+(aq) + MO-(aq) If methyl orange is added to distilled water, the solution turns yellow. If a drop or two of 6 M HCl is added to the yellow solution, it turns red. If a few drops of 6 M NaOH are added to the red solution, the color reverts to yellow. Why does adding 6 M HCl to the yellow solution of methyl orange tend to cause the color to change to red? (Note that in solution, HCl exists as H+ and Cl- ions) Why does adding 6 M NaOH to the red solution tend to make it turn back to yellow? (Note that in solution, NaOH exists as Na+ and OH- ions) How does increasing [OH-] shift Reaction 3 in the discussion section? How would the resulting change in [H+] affect the dissociation reaction of HMO
The Correct Answer and Explanation is:
When methyl orange (HMO) is added to distilled water, it partially dissociates according to the equilibrium:
HMO(aq) ⇌ H⁺(aq) + MO⁻(aq)
This is an acid-base equilibrium where HMO is the weak acid and MO⁻ is its conjugate base. The color of the solution depends on the ratio of HMO to MO⁻:
- Red color indicates more HMO (acidic conditions).
- Yellow color indicates more MO⁻ (basic conditions).
1. Why does adding 6 M HCl turn the yellow solution red?
Adding HCl introduces a high concentration of H⁺ ions to the solution. According to Le Chatelier’s Principle, increasing [H⁺] shifts the equilibrium to the left, favoring the formation of HMO:
HMO ⇌ H⁺ + MO⁻ ← (shift to the left)
As more HMO is formed and MO⁻ decreases, the solution turns red, the color associated with the undissociated HMO.
2. Why does adding 6 M NaOH turn the red solution back to yellow?
Adding NaOH introduces a high concentration of OH⁻ ions, which react with H⁺ to form water:
OH⁻ + H⁺ → H₂O
This decreases [H⁺] in solution, disturbing the equilibrium. Again, per Le Chatelier’s Principle, the system shifts to the right to produce more H⁺:
HMO ⇌ H⁺ + MO⁻ → (shift to the right)
This increases [MO⁻] (the conjugate base), causing the solution to turn yellow, the color of MO⁻.
3. How does increasing [OH⁻] shift Reaction 3?
Reaction 3 refers to the dissociation of HMO:
HMO ⇌ H⁺ + MO⁻
When [OH⁻] increases, [H⁺] is reduced as OH⁻ neutralizes H⁺. The equilibrium shifts to the right to compensate, increasing the dissociation of HMO.
4. How does this affect HMO dissociation?
With reduced [H⁺], the system favors more dissociation of HMO to restore equilibrium. Thus, [HMO] decreases, [MO⁻] increases, and the color changes accordingly—turning yellow.
Summary:
- HCl adds H⁺, shifts equilibrium left, forms more HMO, color turns red.
- NaOH adds OH⁻, removes H⁺, shifts equilibrium right, forms more MO⁻, color turns yellow.
- Increasing [OH⁻] promotes more HMO dissociation by lowering [H⁺], shifting equilibrium to the right.
