Question

A small amount of Methyl Orange is added to a solution containing equimolar amounts of acetic acid and sodium acetate. Is this solution yellow or red?

Short answer

The solution containing equimolar amounts of acetic acid and sodium acetate has a pH of approximately 4.74. Since Methyl Orange is red in acidic solutions (pH < 3.1) and yellow in basic solutions (pH > 4.4), and the calculated pH is higher than 4.4, the Methyl Orange indicator in this solution will be yellow.

Step-by-step solution

Write the equilibrium expression for acetic acid

We first need to write down the equilibrium expression for the weak acid (acetic acid), to have an idea of the ions present in the solution. Acetic acid (CH3COOH) dissociates into acetate ions (CH3COO-) and hydrogen ions (H+) as follows: \[CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}\]

Find the acid dissociation constant, Ka, for acetic acid

We now need to find the acid dissociation constant, Ka, for acetic acid. Ka is a measure of the strength of an acid in solution. The higher the Ka value, the stronger the acid. For acetic acid, the Ka is approximately 1.8 x 10^{-5}.

Use the Henderson-Hasselbalch equation to determine the pH of the solution

The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of the undissociated acid ([HA]) and its conjugate base ([A-]), and the acid dissociation constant, Ka. The equation is given as: \[pH = pKa + log\left(\frac{[A^{-}]}{[HA]}\right)\] We are given that the solution contains equimolar amounts of acetic acid and sodium acetate, meaning the concentrations of CH3COOH and CH3COO- are equal. Thus the ratio of their concentrations is 1, which makes the logarithm term equal to zero. Plugging in the values, we have: \[pH = -log(1.8 \times 10^{-5})\]

Calculate the pH of the solution

Now we can calculate the pH of the solution: \[pH = -log(1.8 \times 10^{-5}) \approx 4.74\]

Determine the color of the Methyl Orange

With the pH of the solution calculated in Step 4, we can determine the color of the Methyl Orange indicator. Since Methyl Orange is red in acidic solutions (pH < 3.1) and yellow in basic solutions (pH > 4.4), and our calculated pH is approximately 4.74, which is higher than 4.4, the Methyl Orange indicator will be yellow in this solution.

Key concepts

Equilibrium Expression

The concept of an equilibrium expression is rooted in the idea that chemical reactions in solutions often reach a state of balance, where the rates of the forward and reverse reactions are equal. For acetic acid (\(\text{CH}_3\text{COOH}\)), which is a weak acid, this equilibrium involves its dissociation into acetate ions (\(\text{CH}_3\text{COO}^-\)) and hydrogen ions (\(\text{H}^+\)):\[\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+\]This reaction does not go to completion, meaning that both the reactants and products are present in the solution simultaneously. The equilibrium expression for acetic acid can be written as:\[K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}\]Here, \([K_a]\) represents the acid dissociation constant, which quantifies the position of the equilibrium. It's crucial for understanding how much of the acid is dissociated at a given time. The numbers in the equilibrium expression must be concentrations expressed in molarity (moles per liter), reflecting the degree of ionization or dissociation.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an invaluable tool in acid-base chemistry for estimating the pH of buffer solutions. It arises from rearranging the equilibrium expression of an acid-base reaction, specifically the dissociation of a weak acid. For acetic acid and its conjugate base, acetate (\(\text{CH}_3\text{COOH}\) and \(\text{CH}_3\text{COO}^-\)), the equation is:\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]Where:

• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\).
• \([A^-]\) is the concentration of the conjugate base, acetate ions.
• \([HA]\) is the concentration of the undissociated acid, acetic acid.

When both the concentrations \([A^-]\) and \([HA]\) are equal, the ratio becomes 1, and thus \(\log(1) = 0\). Consequently, the \(pH\) directly equals \(pK_a\). This equation illustrates why acetic acid and acetate can maintain a stable pH when small amounts of other acids or bases are added, characterizing their function as a buffer solution.

pH Calculation

Calculating the \(pH\) of a solution is a fundamental step in understanding a solution's acidity or basicity. In the context of acetic acid and acetate, the pH is determined using the Henderson-Hasselbalch equation. Given that both the concentrations of \(\text{CH}_3\text{COOH}\) and \(\text{CH}_3\text{COO}^-\) are equal, the equation simplifies to:\[pH = -\log(K_a) = -\log(1.8 \times 10^{-5})\]The operation \(\log\) is used because the \(pH\) scale is logarithmic, meaning each pH unit represents a tenfold difference in \([H^+]\). Calculating the logarithm provides:\[pH \approx 4.74\]This pH value indicates the solution is slightly acidic, which is an essential piece of information for anticipating the behavior of the solution under different conditions, such as when reacting with indicators like Methyl Orange.

Acid Dissociation Constant (Ka)

The acid dissociation constant, known as \(K_a\), is a measure of the strength of an acid in a solution. For acetic acid, \(K_a\) has a value of approximately \(1.8 \times 10^{-5}\). This magnitude indicates that acetic acid is a weak acid. To contextualize, weak acids only partially dissociate in solution. \(K_a\) values allow chemists to predict the extent of dissociation:

• A higher \(K_a\) signifies a stronger acid, which dissociates more completely providing more \([H^+]\) ions.
• A lower \(K_a\) represents a weaker acid, indicating less dissociation.

Therefore, for acetic acid, the weak dissociation means most of the acetic acid molecules remain undissociated in the solution. Understanding \(K_a\) is key to predicting how an acid will behave in different chemical environments, influencing reactions, solubility, and its role in buffer systems like in this exercise.