Problem 18 - Chemistry: The Central Science

Question

(a) Calculate the percentionization of \(0.085 \mathrm{M}\) lactic acid \(\left(K_{a}=1.4 \times 10^{-4}\right) .\) (b) Calculate the percent ionization of \(0.095 M\) lactic acid in a solution containing \(0.0075 M\) sodium lactate.

Step-by-Step Solution

Answer

The percent ionization of 0.085 M lactic acid is approximately 4.06% and the percent ionization of 0.095 M lactic acid in a solution containing 0.0075 M sodium lactate is approximately 1.71%.

Step 1: Set up the equilibrium expression for lactic acid

Lactic acid is a weak acid that can ionize into a lactate ion and a hydrogen ion according to the following reaction: \[HA \rightleftharpoons H^{+} + A^{-}\] The given \(K_a\) value is the acid dissociation constant for this reaction, which can be represented by the equilibrium expression: \[K_a = \frac{[H^{+}][A^{-}]}{[HA]}\] In part (a), since only lactic acid is present, we will need to use the initial concentration of lactic acid and the given \(K_a\) value to calculate the concentrations of ions at equilibrium. In part (b), the initial concentration of sodium lactate will affect the ionization of lactic acid.

Step 2: Calculate the percent ionization for 0.085 M lactic acid (Part a)

Let \(x\) represent the change in concentrations of lactic acid, hydrogen ions, and lactate ions at equilibrium: \[HA \rightleftharpoons H^{+} + A^{-}\] \[- x \ \ \ \ + x \ \ \ + x\] The equilibrium concentrations are: \[0.085 - x, \ x, \ x\] Now we can plug these concentrations into the equilibrium expression: \[1.4 \times 10^{-4} = \frac{x\cdot x}{0.085 - x}\] Solve for x using quadratic formula or iteration method. Considering the small value of \(K_a\), it's reasonable to assume that \(x\) is much smaller than 0.085, so: \[x^2 \approx 1.4 \times 10^{-4} \cdot 0.085\] \x\[x \approx \sqrt{1.19 \times 10^{-5}}\] \[x \approx 0.00345\] Now, we can calculate the percent ionization: \[\% \ ionization = \frac{x}{0.085} \times 100\%\] \[\% \ ionization \approx \frac{0.00345}{0.085} \times 100\% \approx 4.06\%\] So, the percent ionization of 0.085 M lactic acid is approximately 4.06%.

Step 3: Calculate the percent ionization for 0.095 M lactic acid with 0.0075 M sodium lactate (Part b)

In this case, the presence of sodium lactate means there is already some amount of lactate ions in the solution. We'll again let \(x\) represent the change in concentrations of lactic acid and hydrogen ions at equilibrium: \[HA \rightleftharpoons H^{+} + A^{-}\] \[- x \ \ \ \ \ \ + x \ \ \ \ \ \ \ +0\] The equilibrium concentrations are: \[0.095 - x, \ x, \ 0.0075\] Now we can plug these concentrations into the equilibrium expression: \[1.4 \times 10^{-4} = \frac{x\cdot (x+0.0075)}{0.095 - x}\] This time, solve for x using a quadratic formula, or we can still assume \(x << 0.095\), so: \[1.4 \times 10^{-4} \approx \frac{x\cdot (x+0.0075)}{0.095}\] \[x^2 + 0.0075x \approx 1.33 \times 10^{-5}\] Solve for x using your preferred method: \[x \approx 0.00162\] Now, we can calculate the percent ionization: \[\% \ ionization = \frac{x}{0.095} \times 100\%\] \[\% \ ionization \approx \frac{0.00162}{0.095} \times 100\% \approx 1.71\%\] So, the percent ionization of 0.095 M lactic acid in a solution containing 0.0075 M sodium lactate is approximately 1.71%.

Key Concepts

Weak Acids

Understanding weak acids, like lactic acid, is crucial to many biological and chemical processes. A weak acid doesn't completely dissociate in water. Instead, it partially releases hydrogen ions resulting in an equilibrium between the acid and its ions. This behavior is different from strong acids, which completely dissociate in solution.
When considering a weak acid like lactic acid, we express its dissociation using the chemical reaction:

• HA ↔ H⁺ + A^-

Here, HA represents the undissociated acid, H⁺ is the hydrogen ion, and A^- the conjugate base. The extent of ionization of a weak acid in solution determines how many hydrogen ions are available, which also affects the pH of the solution. Since weak acids do not fully dissociate, their percent ionization is usually just a few percent. This makes them distinct from strong acids which have a much higher ionization percentage. Understanding this fundamental property is essential for predicting the behavior of weak acids in different environmental conditions.

Equilibrium Expressions

Equilibrium expressions are a powerful tool for understanding the balance between products and reactants in reversible reactions. For acids, these expressions help to quantify how much an acid dissociates in solution. The equilibrium expression for a weak acid like lactic acid is derived from its dissociation reaction.For example, the equilibrium constant (K_a) is derived using the expression:

• K_a = \[ \frac{[H⁺][A^-]}{[HA]} \]

This enters the realm of chemical equilibrium where concentrations of reactants and products remain steady over time. It tells us how far the dissociation reaction proceeds. A small K_a value like that for lactic acid indicates that the acid largely remains in its original form, with limited ionization. Solving problems involving equilibrium expressions typically requires setting up an expression based on initial concentrations, changes due to dissociation, and final equilibrium concentrations. It forms the basis for calculating important properties such as the Percent Ionization.

Acid Dissociation Constant

The acid dissociation constant, noted as K_a, plays a pivotal role in characterizing the strength of weak acids. It is a measure of the acid's ability to donate protons (H⁺ ions) to the solution. For a given weak acid, the value of K_a can predict how much the acid will dissociate at a particular concentration. A small K_a value implies that the acid doesn't release protons easily, meaning it is weaker. Conversely, a larger K_a would suggest a stronger acid with a higher degree of ionization. This constant is used as a tool to compare different acids and their capacities to affect solution pH. When calculating percent ionization or working with buffer solutions, a good grasp of K_a allows chemists to make informed assumptions about the extent of a reaction's progress and adjust their calculations accordingly, ensuring precision and accuracy.

Buffer Solutions

Buffer solutions are special types of mixtures that resist changes in pH when small amounts of acids or bases are added. They are formed by combining a weak acid with its corresponding salt, or a weak base with its corresponding salt. In the case of lactic acid, adding sodium lactate creates a buffer solution. Buffers function by utilizing the equilibrium between the weak acid and its conjugate base. When hydrogen ions are added to a solution, the base part of the buffer reacts to neutralize them, minimizing pH changes. Similarly, if base is added, the weak acid in the buffer neutralizes the added hydroxide ions. Their ability to maintain a steady pH is essential in many environments: biological systems rely on buffers to regulate pH, and laboratory settings use buffers to stabilize reactions. Understanding buffers and their components allows one to predict and control the extent of reactions that can occur within a given system.