Question

Acrylic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) is used in the manufacture of paints and plastics. The \(\mathrm{p} K_{\mathrm{a}}\) of acrylic acid is \(4.25 .\) (a) Calculate the \(\mathrm{pH}\) and the concentrations of all species \(\left(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{2}^{-}, \mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.150 \mathrm{M}\) acrylic acid. (b) Calculate the percent dissociation in \(0.0500 \mathrm{M}\) acrylic acid.

Short answer

pH is 2.54; % dissociation is 3.34%.

Step-by-step solution

Write the Dissociation Equation

Acrylic acid dissociates in water according to the equation: \[ \text{HC}_3\text{H}_3\text{O}_2 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{C}_3\text{H}_3\text{O}_2^- (aq) \] The ionization constant expression for this equilibrium is given by: \[ K_a = \frac{[\text{H}^+][\text{C}_3\text{H}_3\text{O}_2^-]}{[\text{HC}_3\text{H}_3\text{O}_2]} = 10^{-4.25} \]

Set Up the ICE Table for 0.150 M Solution

Initial concentrations (I): \([\text{HC}_3\text{H}_3\text{O}_2] = 0.150\, M\), \([\text{H}^+] = [\text{C}_3\text{H}_3\text{O}_2^-] = 0\) ewline Change (C): \([-x, +x, +x]\) ewline Equilibrium concentrations (E): \([\text{HC}_3\text{H}_3\text{O}_2] = 0.150 - x\), \([\text{H}^+] = [\text{C}_3\text{H}_3\text{O}_2^-] = x\)

Calculate "x" Using the Ka Expression

Substitute the equilibrium concentrations into the ionization constant equation: \[ K_a = \frac{x^2}{0.150 - x} \] With \(K_a = 10^{-4.25} = 5.62 \times 10^{-5}\). Assume \(x\) is small, so \(0.150 - x \approx 0.150\). ewline Solve for \(x\) to get \(x = \sqrt{5.62 \times 10^{-5} \times 0.150}\).

Solve for x and Determine [H+], [C3H3O2-], [HC3H3O2]

Calculate \(x\): \[ x = \sqrt{5.62 \times 10^{-5} \times 0.150} = 2.91 \times 10^{-3} \] Then, \([\text{H}^+] = [\text{C}_3\text{H}_3\text{O}_2^-] = 2.91 \times 10^{-3} \, M\) ewline \([\text{HC}_3\text{H}_3\text{O}_2] = 0.150 - 2.91 \times 10^{-3}\approx 0.147\, M\)

Calculate pH

Use the hydrogen ion concentration to calculate the pH: \[ \text{pH} = -\log(2.91 \times 10^{-3}) \approx 2.54 \]

Calculate [OH-] Using Kw

The ion product of water \(K_w = 1.0 \times 10^{-14}\): \[ [\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{2.91 \times 10^{-3}} \approx 3.44 \times 10^{-12} \, M \]

Set Up the ICE Table for 0.0500 M Solution

Repeat the ICE table setup for 0.0500 M acrylic acid with the same steps from Step 2 and 3. Initial concentrations: \([\text{HC}_3\text{H}_3\text{O}_2] = 0.0500 \, M\).

Calculate x in 0.0500 M Solution

\[ x = \sqrt{5.62 \times 10^{-5} \times 0.0500} = 1.67 \times 10^{-3} \] Thus, \([\text{H}^+] = [\text{C}_3\text{H}_3\text{O}_2^-] = 1.67 \times 10^{-3} \, M\).

Determine Percent Dissociation

Percent dissociation is given by: \[ \text{Percent Dissociation} = \left(\frac{x}{\text{Initial } [\text{HC}_3\text{H}_3\text{O}_2]} \right) \times 100 = \left(\frac{1.67 \times 10^{-3}}{0.0500} \right) \times 100 \approx 3.34\% \]

Key concepts

pH Calculation

Determining the pH of a solution is a crucial step in understanding its acidity or basicity. pH is calculated using the formula:

• \[ \text{pH} = -\log[\text{H}^+] \]

In this particular problem, we are dealing with acrylic acid, which partially ionizes in water.The ionization constant, \( K_a \), offers insight into how much the acid dissociates. By calculating the concentration of hydrogen ions \([\text{H}^+]\), we use this formula to find the pH.
In our example with a 0.150 M solution of acrylic acid, the concentration of \([\text{H}^+]\) was found to be \( 2.91 \times 10^{-3} \text{ M} \). By applying the pH formula, we get:

• \[ \text{pH} = -\log(2.91 \times 10^{-3}) \approx 2.54 \]

This indicates that acrylic acid is relatively acidic.

ICE Table in Chemistry

An ICE table (Initial, Change, Equilibrium) is an invaluable tool used to analyze chemical equilibria. For weak acids like acrylic acid, it helps to visualize the dissociation process and equilibrium concentrations.
Initial concentrations: Before any reaction, we know the concentration of acrylic acid and assume zero dissociation for \([\text{H}^+]\) and \([\text{C}_3\text{H}_3\text{O}_2^-]\).

• Initial: \([\text{HC}_3\text{H}_3\text{O}_2] = \text{0.150 M}, [\text{H}^+] = 0, [\text{C}_3\text{H}_3\text{O}_2^-] = 0\)

Change: As the acid dissociates, the changes are represented as \(-x\), \(+x\), and \(+x\) respectively.

• Change: \([-x, +x, +x]\)

Equilibrium: At equilibrium, these quantities combine to reflect the actual concentrations.

• Equilibrium: \([\text{HC}_3\text{H}_3\text{O}_2] = 0.150 - x\), \([\text{H}^+] = x\), \([\text{C}_3\text{H}_3\text{O}_2^-] = x\)

This clear structure allows us to solve the equilibrium expression \( K_a = \frac{x^2}{0.150 - x} \) effectively.

Percent Dissociation

To gauge how much an acid dissociates in a solution, we calculate the percent dissociation. This measure is crucial for understanding the strength of the acid in question.
The formula for percent dissociation is:

• \[ \text{Percent Dissociation} = \left( \frac{x}{\text{Initial } [\text{HC}_3\text{H}_3\text{O}_2]} \right) \times 100 \]

From our experiment with acrylic acid at a concentration of 0.0500 M, \( x \) was calculated to be \(1.67 \times 10^{-3} \text{ M} \). Substituting these values gives:

• \[ \text{Percent Dissociation} = \left( \frac{1.67 \times 10^{-3}}{0.0500} \right) \times 100 \approx 3.34\% \]

This value signifies that a small fraction of the acrylic acid molecules dissociate, characterizing it as a weak acid.

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between the undissociated acid and its ions in solution. For acrylic acid, the equilibrium can be expressed by the dissociation equation:

• \[\text{HC}_3\text{H}_3\text{O}_2 \rightleftharpoons \text{H}^+ + \text{C}_3\text{H}_3\text{O}_2^-\]

The equilibrium state is achieved when the rate of the forward reaction (dissociation) equals the rate of the reverse reaction (re-association). The position of this equilibrium can be manipulated by changing concentrations and is quantitatively described using the equilibrium constant \( K_a \).
A small \( K_a \) value, as in the case of acrylic acid \( K_a = 5.62 \times 10^{-5} \), indicates a weak acid, meaning few molecules dissociate into ions. Understanding these components helps predict the behavior of acids and bases in chemical reactions.