Problem 1
Question
Methyl orange, \(\mathrm{HMO}\), is a common acid-base indicator. In solution it ionizes according to the equation: $$\begin{array}{c} \mathrm{HMO}(\mathrm{aq})=\mathrm{H}^{+}(\mathrm{aq})+\mathrm{MO}^{-}(\mathrm{aq}) \\\ \mathrm{red}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{yellow} \end{array}$$ a. Why does adding \(6 \mathrm{M}\) \(\mathrm{HCI}\) to the yellow solution of methyl orange tend to cause the color to change to red? (Note that in solution \(\mathrm{HCl}\) exists as \(\mathrm{H}^{+}\) and \(\mathrm{Cl}^{-}\) ions.) b. Why does adding \(6 \mathrm{M}\) \(\mathrm{NaOH}\) to the red solution tend to make it turn back to yellow? Note that in solution \(\mathrm{NaOH}\) exists as \(\mathrm{Na}^{+}\) and \(\mathrm{OH}^{-}\) ions. (Hint: How does increasing \(\left[\mathrm{OH}^{-}\right]\) shift Reaction 3 in the discussion section? How would the resulting change in \(\left[\mathrm{H}^{+}\right]\) affect the dissociation reaction of \(\mathrm{HMO}\)? Explain these as part of your answer.)
Step-by-Step Solution
Verified Answer
Adding HCl turns the solution red by shifting equilibrium left; NaOH turns it yellow by shifting equilibrium right.
1Step 1: Understanding the ionization of methyl orange
Methyl orange ionizes in solution as \( \mathrm{HMO} \rightarrow \mathrm{H}^{+} + \mathrm{MO}^{-} \). The color depends on the dominant form: \( \mathrm{HMO} \) is red, and \( \mathrm{MO}^{-} \) is yellow.
2Step 2: Effect of adding HCl to a yellow solution
When \( \mathrm{HCl} \) is added to the solution, it increases \( [\mathrm{H}^{+}] \). According to Le Chatelier's principle, the equilibrium shifts left to reduce the change by recombining \( \mathrm{H}^{+} \) with \( \mathrm{MO}^{-} \), increasing \( \mathrm{HMO} \) concentration and turning the solution red.
3Step 3: Effect of adding NaOH to a red solution
Adding \( \mathrm{NaOH} \) increases \( [\mathrm{OH}^{-}] \) which reacts with \( \mathrm{H}^{+} \) to form water, reducing \( [\mathrm{H}^{+}] \). This causes the equilibrium to shift right, converting \( \mathrm{HMO} \) to \( \mathrm{MO}^{-} \) and turning the solution yellow.
4Step 4: Summary of the chemical effects
Adding \( \mathrm{HCl} \) shifts the equilibrium left, favoring \( \mathrm{HMO} \) and a red color. Adding \( \mathrm{NaOH} \) shifts the equilibrium right, favoring \( \mathrm{MO}^{-} \) and a yellow color.
Key Concepts
Le Chatelier's PrincipleChemical EquilibriumMethyl Orange Ionization
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry, describing how a system at equilibrium responds to changes. When a stress is applied to a system at equilibrium, the system shifts in a direction that counteracts that stress. For example, if there's an increase in concentration of reactants, the equilibrium shifts to produce more products, and vice versa.
In the case of methyl orange, when hydrochloric acid (HCl) is added, the concentration of hydrogen ions \( \mathrm{H}^{+} \) increases. This serves as a stress to the equilibrium of the methyl orange ionization reaction. To counter this, the reaction shifts to the left, forming more \( \mathrm{HMO} \), which is red in color.
In contrast, when sodium hydroxide (NaOH) is added, \( \mathrm{OH}^{-} \) ions react with \( \mathrm{H}^{+} \) to form water, decreasing the \( \mathrm{H}^{+} \) concentration. According to Le Chatelier's Principle, the system will then shift to the right, favoring the formation of \( \mathrm{MO}^{-} \) ions, which are yellow.
In the case of methyl orange, when hydrochloric acid (HCl) is added, the concentration of hydrogen ions \( \mathrm{H}^{+} \) increases. This serves as a stress to the equilibrium of the methyl orange ionization reaction. To counter this, the reaction shifts to the left, forming more \( \mathrm{HMO} \), which is red in color.
In contrast, when sodium hydroxide (NaOH) is added, \( \mathrm{OH}^{-} \) ions react with \( \mathrm{H}^{+} \) to form water, decreasing the \( \mathrm{H}^{+} \) concentration. According to Le Chatelier's Principle, the system will then shift to the right, favoring the formation of \( \mathrm{MO}^{-} \) ions, which are yellow.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant, though reactions continue to occur.
Methyl orange in solution reaches equilibrium through the reversible ionization reaction: \( \mathrm{HMO} \rightleftharpoons \mathrm{H}^{+} + \mathrm{MO}^{-} \).
Methyl orange in solution reaches equilibrium through the reversible ionization reaction: \( \mathrm{HMO} \rightleftharpoons \mathrm{H}^{+} + \mathrm{MO}^{-} \).
- As the forward reaction proceeds, \(\mathrm{HMO}\) dissociates into \(\mathrm{H}^{+}\) and \(\mathrm{MO}^{-}\).
- As the reverse reaction occurs, \(\mathrm{H}^{+}\) and \(\mathrm{MO}^{-}\) recombine to form \(\mathrm{HMO}\).
Methyl Orange Ionization
Methyl orange is a well-known acid-base indicator because of its ability to change color depending on the pH of a solution.
The ionization of methyl orange can be represented by the equation: \( \mathrm{HMO} \rightarrow \mathrm{H}^{+} + \mathrm{MO}^{-} \). Each form of methyl orange exhibits a unique color: \( \mathrm{HMO} \) is red, while \( \mathrm{MO}^{-} \) is yellow. The interaction between these species allows methyl orange to show a visible color change as the pH of the solution transforms.
The ionization of methyl orange can be represented by the equation: \( \mathrm{HMO} \rightarrow \mathrm{H}^{+} + \mathrm{MO}^{-} \). Each form of methyl orange exhibits a unique color: \( \mathrm{HMO} \) is red, while \( \mathrm{MO}^{-} \) is yellow. The interaction between these species allows methyl orange to show a visible color change as the pH of the solution transforms.
- In low pH environments (acidic), high concentrations of \( \mathrm{H}^{+} \) cause more \( \mathrm{HMO} \) to form, resulting in a red color.
- In higher pH environments (basic), the concentration of \( \mathrm{H}^{+} \) decreases, allowing \( \mathrm{MO}^{-} \) to dominate, and the color shifts to yellow.