Problem 18 - Chemistry The Central Science In Si Units

Question

(a) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .(\mathbf{b})\) Calculate the percent ionization of \(0.250 \mathrm{M}\) lactic acid in a solution containing \(0.050 \mathrm{M}\) sodium lactate.

Step-by-Step Solution

Answer

(a) Lactic acid ionizes 2.36%. (b) With sodium lactate, it ionizes 0.28%.

Step 1: Write the Ionization Reaction

Lactic acid, \ \( CH_3CHOHCOOH \, \, \) ionizes in water to form hydrogen ions and lactate ions. The ionization reaction is: \[ CH_3CHOHCOOH_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3CHOHCOO^-_{(aq)} \]

Step 2: Set Up the Expression for Ka

The acid dissociation constant is given by: \[ K_a = \frac{{[H^+][CH_3CHOHCOO^-]}}{{[CH_3CHOHCOOH]}} \]Where \( K_a = 1.4 \times 10^{-4} \) for lactic acid.

Step 3: Setting Initial Concentrations and Changes

Initial concentration of lactic acid is 0.250 M, and it partially ionizes. Let \( x \) be the concentration of \( H^+ \) ions at equilibrium. Initially: \[ [CH_3CHOHCOOH] = 0.250 \, M \] \[ [H^+] = 0 \, M \] \[ [CH_3CHOHCOO^-] = 0 \, M \] Changes at equilibrium: \[ [CH_3CHOHCOOH] = 0.250 - x \] \[ [H^+] = x \] \[ [CH_3CHOHCOO^-] = x \]

Step 4: Substitute into Ka Expression for Case (a)

Substitute the equilibrium concentrations into the expression for \( K_a \): \[ K_a = \frac{{x^2}}{{0.250 - x}} = 1.4 \times 10^{-4} \] Assume \( x \ll 0.250 \), so \( 0.250 - x \approx 0.250 \). \[ \frac{{x^2}}{{0.250}} = 1.4 \times 10^{-4} \] \[ x^2 = 1.4 \times 10^{-4} \times 0.250 \] Calculate \( x \): \[ x = \sqrt{3.5 \times 10^{-5}} \approx 5.9 \times 10^{-3} \]

Step 5: Calculate Percent Ionization for Case (a)

Percent ionization is \[ \frac{x}{\text{initial concentration}} \times 100 = \frac{5.9 \times 10^{-3}}{0.250} \times 100 = 2.36\% \]

Step 6: Set Up ICE Table for Case (b)

For case (b), the solution consists of 0.250 M lactic acid and 0.050 M sodium lactate. Initial:\[ [CH_3CHOHCOOH] = 0.250 \, M \]\[ [H^+] = 0 \, M \]\[ [CH_3CHOHCOO^-] = 0.050 \, M \]Equilibrium changes:\[ [CH_3CHOHCOOH] = 0.250 - x \]\[ [H^+] = x \]\[ [CH_3CHOHCOO^-] = 0.050 + x \]

Step 7: Substitute into Ka Expression for Case (b)

Substitute into the \( K_a \) expression: \[ K_a = \frac{x(0.050 + x)}{0.250 - x} = 1.4 \times 10^{-4} \] Assume \( x \ll 0.050 \), hence \( 0.050 + x \approx 0.050 \) and \( 0.250 - x \approx 0.250 \). Substitute in: \[ \frac{x \cdot 0.050}{0.250} = 1.4 \times 10^{-4} \]Solving for \( x \): \[ x = \frac{1.4 \times 10^{-4} \times 0.250}{0.050} = 7.0 \times 10^{-4} \]

Step 8: Calculate Percent Ionization for Case (b)

Percent ionization is: \[ \frac{x}{0.250} \times 100 = \frac{7.0 \times 10^{-4}}{0.250} \times 100 \approx 0.28\% \]

Key Concepts

Acid Dissociation Constant

The acid dissociation constant, represented as \( K_a \), is a vital parameter in understanding the strength of an acid. It measures the extent to which an acid can donate protons in a solution.
In mathematical terms, it is given by the expression:

• \( K_a = \frac{{[H^+][A^-]}}{{[HA]}} \)

Here, \([H^+]\) is the concentration of hydrogen ions, \([A^-]\) is the concentration of anions formed, and \([HA]\) is the concentration of the non-dissociated acid.
A larger \( K_a \) value indicates a stronger acid because it shows a higher degree of ionization.
In the case of lactic acid with \( K_a = 1.4 \times 10^{-4} \), this small value shows that lactic acid is a weak acid, not fully ionizing in solution.

Ionization Equilibrium

Ionization equilibrium refers to the state where the rate of ionization of a weak acid equals the rate of recombination of ions in the solution.
At this stage, the concentrations of ions and the non-dissociated acid remain constant. This balance can be disrupted by changes in concentration, temperature, or the presence of other substances.
For lactic acid, the ionization equilibrium is expressed as:\[ CH_3CHOHCOOH_{(aq)} \rightleftharpoons H^+_{(aq)} + CH_3CHOHCOO^-_{(aq)} \]This equilibrium equation shows how lactic acid partially dissociates into hydrogen ions \( H^+ \) and lactate ions \( CH_3CHOHCOO^- \).
As conditions change, such as adding more acid or a common ion, the equilibrium can shift according to Le Châtelier's principle, affecting percent ionization.

Weak Acids

Weak acids, like lactic acid, do not completely dissociate in water. Instead, they partially ionize, leading to a dynamic balance between ionized and non-ionized forms.
The degree of ionization is small and characterized by a low \( K_a \) value, reflective of its minor dissociation tendency.
Key characteristics of weak acids include:

• They have a \( K_a \) much less than 1.
• Only a small fraction of their molecules release \( H^+ \) into the solution.
• The rate of recombination of \( H^+ \) and \( A^- \) is significant, leading to equilibrium.

In calculating percent ionization, the weak nature of acids is crucial, dictating assumptions like \( x \ll 0.250 \) in problem-solving, where \( x \) represents equilibrium concentrations.

Common Ion Effect

The common ion effect occurs when a salt containing an ion in common with the weak acid is added to the solution. This effect reduces the ionization of the acid.
For example, when adding sodium lactate to lactic acid, the concentration of the lactate ion \( CH_3CHOHCOO^- \) increases.
This stresses the equilibrium by increasing the concentration of ions formed by dissociation, so the solution compensates by decreasing the ionization of the acid.

• This is seen in the shift of equilibrium towards the non-ionized acid, reducing the percent ionization.
• Calculated using \( K_a = \frac{x(0.050 + x)}{0.250 - x} = 1.4 \times 10^{-4} \) where \( x \) represents \([H^+]\) in the presence of sodium lactate.

The effect is crucial in buffering solutions and controlling pH, showing the practical impacts of interaction between weak acids and their salts.